Circular Motion Notes Class 11th – IIT JEE |

This article includes complete Circular Motion Notes for Class 11th. These notes will also help you in preparing Exams like IIT JEE, NEET and other Entrance Exams. eSaral Provides you free Detailed Notes with proper images and explanations.

System of Particles and Rotational Motion Class 11 Notes for IIT JEE & NEET

System of Particles and Rotational Motion Class 11 is one of the important chapters when it comes to understanding the basics of circular motion. We all are aware of the definition of rotational motion, the center of mass and center of gravity etc.Rotational motion refers to objects that are rotating in a curved path and comprises torque, moment of inertia, angular velocity, angular displacement, angular acceleration and angular momentum. System of Particles and Rotational Motion Class 11 notes are given below:

eSaral have already come up with an amazing revision series of Physics where you can easily revise your chapter within minutes with all the important formulae and key points. This revision series is free for all.

Click Here for Complete Physics Revision Series by Saransh Gupta Sir (AIR-41)

Computer Science Graduate from IIT Bombay. Cracked IIT-JEE with AIR-41 and AIEEE (JM) with AIR-71 in 2006. He is an author of JEE Mentorship Book “StrateJEE”.
He taught Physics for 5 years at Allen and was loved immensely by all the students. Many of his students bagged success with flying colours in JEE & NEET Exams.

eSaral brings you detailed Class 11th Physics study material.  eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th.  These notes will also help you in your IIT JEE & NEET preparations.

We hope these Physics Notes for Class 11 will help you understand the important topics and remember the key points for the exam point of view.

Get Complete Physics Notes for Physics Class 11 for easy learning and understanding. For free video lectures and complete study material, Download eSaral APP.

At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.

Work Energy and Power Class 11 Physics Notes for IIT JEE | NEET

Work Energy and Power can never be ignored as it covers basic concepts which provides a smooth access to the next some topics in Mechanics. This chapter is a key to prepare for entrance examinations like IIT-JEE & NEET. This is one of the most important topic in mechanics for IIT-JEE, NEET and CBSE perspectives.
The terms work, energy and power have specific meanings in Physics. Work is done when a force produces motion. To do work, energy is required. This energy we acquire from the food which we eat and if work is done by machine, then energy is supplied by fuels or by current.

There are various forms of energy – kinetic energy, potential energy, heat energy, chemical energy, electrical energy, nuclear energy etc.
Work Energy and Power Class 11 Physics notes  are given here:

To watch free video tutorials on Physics topics by Saransh Sir, AIR-41 Install eSaral App

eSaral have already come up with an amazing revision series of Physics where you can easily revise your chapter within minutes with all the important formulae and key points. This revision series is free for all.

Click Here for Complete Physics Revision Series by Saransh Gupta Sir (AIR-41)

Computer Science Graduate from IIT Bombay. Cracked IIT-JEE with AIR-41 and AIEEE (JM) with AIR-71 in 2006. He is an author of JEE Mentorship Book “StrateJEE”.
He taught Physics for 5 years at Allen and was loved immensely by all the students. Many of his students bagged success with flying colours in JEE & NEET Exams.

eSaral brings you detailed Class 11th Physics study material.  eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th.  These notes will also help you in your IIT JEE & NEET preparations.

We hope these Physics Notes for Class 11 will help you understand the important topics and remember the key points for the exam point of view.

Get Complete Physics Notes for Physics Class 11 for easy learning and understanding. For free video lectures and complete study material, Download eSaral APP.

At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.

Class 11 Laws of Motion Notes for IIT JEE & NEET

In this chapter, students will get to learn about moving bodies and the factors that acts on these bodies. Also get to learn about the direction of these moving bodies, their corresponding motion, effect of gravitational force on theses bodies etc. eSaral provides complete comprehensive chapter-wise notes for Class 11 Physics. Here you’ll get Class 11 Laws Of Motion Notes.
These notes are also helpful in JEE and NEET Exams. These notes comprises proper images with diagram. Let’s study in detail about Laws of Motion in this article.

eSaral have already come up with an amazing revision series of Physics where you can easily revise your chapter within minutes with all the important formulae and key points. This revision series is free for all.

Click Here for Complete Physics Revision Series by Saransh Gupta Sir (AIR-41)

Computer Science Graduate from IIT Bombay. Cracked IIT-JEE with AIR-41 and AIEEE (JM) with AIR-71 in 2006. He is an author of JEE Mentorship Book “StrateJEE”.
He taught Physics for 5 years at Allen and was loved immensely by all the students. Many of his students bagged success with flying colours in JEE & NEET Exams.

eSaral brings you detailed Class 11th Physics study material.  eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th.  These notes will also help you in your IIT JEE & NEET preparations.

We hope these Physics Notes for Class 11 will help you understand the important topics and remember the key points for the exam point of view.

Get Complete Physics Notes for Physics Class 11 for easy learning and understanding. For free video lectures and complete study material, Download eSaral APP.

At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.

EMF & Internal Resistance of a cell | Terminal Potential Difference | Types of cells

There are some components that are integral part of a circuit like EMF & Internal resistance of a cell and they are all inter-related. In this article we will explore about electrochemical cells, EMF and internal resistance of a cell. Later we will also see terminal potential difference and types of cell in detail.

Contents:

## Electrochemical Cell

An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or using electrical energy to cause chemical reactions. The electrochemical cells which generate an electric current are called voltaic cells or galvanic cells and those that generate chemical reactions, via electrolysis for example, are called electrolytic cells. A common example of a galvanic cell is a standard 1.5 volt cell meant for consumer use. A battery consists of one or more cells, connected either in parallel, series or series-and-parallel pattern. [source]

An electrochemical cell is a device which by converting chemical energy into electrical energy maintains the flow of charge in a circuit.

It usually consists of two electrodes of different materials and an electrolyte. The electrode at higher potential is called anode and the one at lower potential is cathode.

## Electromotive Force (EMF)

The emf of a cell is defined as work done by cell in moving a unit positive charge in the whole circuit including the cell once.

1. emf E = W/q; SI unit is joule/coulomb or volt.
2. emf is the maximum potential difference between the two electrodes of the cell when no current is drawn from the cell.
3. emf is the characteristic property of cell and depends on the nature of electrodes and electrolyte used in cell.
4. emf is independent of quantity of electrolyte, size of electrodes and distance between the electrodes.

## Internal Resistance of a Cell

The opposition offered by the electrolyte of the cell to the flow of electric current through it is called the internal resistance of the cell. The internal resistance of cell depends on.

1. Distance between electrodes (r ∝ d) larger is the separation between electrodes more is the length of electrolyte through which ions have to move so more is internal resistance.
2. Conductivity or nature of electrolyte (r ∝ 1/σ)
3. Concentration of electrolyte (r ∝ c)
4. Temperature of electrolyte (r ∝ 1/T)
5. Nature and area of electrodes dipped in electrolyte (r ∝ 1/A)

## Terminal Potential Difference

The potential difference between the two electrodes of a cell in a closed circuit i.e. when current is being drawn from the cell is called terminal potential difference.

### (a) When cell is Discharging

When cell is discharging current inside the cell is from cathode to anode.

Current ${\rm{I}} = {{\rm{E}} \over {{\rm{r}} + {\rm{R}}}}$ or $\quad {\rm{E}} = {\rm{IR}} + {\rm{Ir}} = {\rm{V}} +$ Ir $\quad$ or ${\rm{V}} = {\rm{E}} –$ Ir

When current is drawn from the cell potential difference is less than emf of cell. Greater is the current drawn from the cell smaller is the terminal voltage. When a large current is drawn from a cell its terminal voltage is reduced.

### (b) When cell is Charging

When cells are charging current inside the cell is from anode to cathode.

Current ${\rm{I}} = {{{\rm{V}} – {\rm{E}}} \over {\rm{r}}}\quad$ or $\quad {\rm{V}} = {\rm{E}} + {\rm{Ir}}$

During charging terminal potential difference is greater than emf of cell.

### (c) When cell is in Open circuit

In open circuit, ${\rm{R}} = \infty$

${\rm{I}} = {{\rm{E}} \over {{\rm{R}} + {\rm{r}}}} = 0$

So, ${\rm{V}} = {\rm{E}}$

In open circuit terminal potential difference is equal to emf and is the maximum potential difference which a cell can provide.

### (d) When cell is Short-circuited

In short circuit $R = 0$ so $I = {E \over {R + r}} = {E \over r}\quad$ and $\quad V = IR = 0$

In short circuit current from cell is maximum and terminal potential difference is zero.

### (e) Power transferred to load by cell

${\rm{P}} = {{\rm{I}}^2}{\rm{R}} = {{{{\rm{E}}^2}{\rm{R}}} \over {{{({\rm{r}} + {\rm{R}})}^2}}}$

$\matrix{{{\rm{so}}} & {{\rm{P}} = {{\rm{P}}_{\max }}\quad {\rm{ if }}\quad {{{\rm{dP}}} \over {{\rm{dR}}}} = 0} \cr{\rm{P}} & { ={{\rm{P}}_{\max }}\quad {\rm{ if }}\quad {\rm{r}} = {\rm{R}}} \cr}$

Power transferred by cell to load is maximum when $r = R$ and ${P_{\max }} = {{{E^2}} \over {4r}} = {{{E^2}} \over {4R}}$

## Difference between EMF and Potential Difference

1. The current inside a cell is due to motion of both positive and negative ions while outside it depends on nature of circuit elements like conductors, semiconductor, gas or electrolyte.
2. The cell is a source of constant emf and not of constant current because if resistance of circuit changes then current I = E/r + R also changes but emf remains constant.
3. As I = E/r so more current can be drawn from a cell with larger emf and smaller internal resistance.
Example: In lead acid accumulator E = 2.05 V and rmin = 0.1Ω.
4. With use of cell its internal resistance increases appreciably but emf fall slightly. The current delivering ability is reduced.
5. A cell neither creates nor destroys charge but maintains the flow of charge by providing required energy.
6. The emf of cell is taken to be positive in a circuit if current inside a cell is from negative to positive i.e. during discharging otherwise negative.

7. Capacity of a battery is equal to product of current in ampere and time in hour for which a cell can operate. It depends on the amount of electrolyte and size of cell.
Example: capacity 8Ah means we can draw 8A current for one hour or 2A current for 4 hours.

## Type of Cells

### Primary Cells

The cells which cannot be recharged electrically are called primary cells. Here the original state of cell cannot be brought back by passing electrical energy through cell from external source after cell is discharged.

Example: Voltaic cell, Daniel cell, Leclanche cell, Manganese-alkaline cell, Mercury button cell etc.

### Secondary Cells

The cells in which chemical process is reversible are called secondary cells. Here original chemical state of cell can be brought back by passing electrical energy through cell from external source.

Example: Lead acid accumulator, alkali cells etc.

### DEFECTS OF VOLTAIC CELL

Local Action: This is due to impurities of copper, iron carbon etc. in commercial zinc. When zinc rod is dipped in electrolyte the impurities and zinc in contact form small local cells in which small currents are produced resulting in wastage of zinc even when cell is not being used. This defect is called local action. This is overcome by amalgamating zinc rod by mercury. When cell is being used fresh zinc continues to come on surface so chemical reaction continues local action.

Polarization: Polarization is formation of hydrogen gas bubbles on the anode of cell. This causes an increase in internal resistance because layer of Hydrogen is a bad conductor. The hydrogen ions moving towards anode are unable to reach anode and transfer their charge. These positively charged ions set up a field from hydrogen to zinc resulting in back emf weakening the action of cell. This defect may be overcome by using a depolarizer i.e. oxidising agent like MnO$_2$ or CuSO$_2$ which oxidizes hydrogen to water.

The construction of some cells is shown below:

Kirchhoff’s Circuit Laws – KCL & KVL | Symmetric line method

We have studies Ohm’s law but in real life we have to deal with a lot complex circuits. To calculate current and potential difference across them we require something else. So here we will know about Kirchhoff’s circuit laws.

Contents:

## Kirchhoff’s Circuit Laws

Gustav Kirchhoff, a German physicist, in 1825 developed a set of rules or laws to find the voltages and currents circulating in complex circuits where working only with Ohm’s law don’t suffice. There are majorly two laws to calculate current and potential difference along different loops present in a circuit known as Kirchhoff’s circuit laws. For the calculations, to obtain circuit equations these laws are useful.

The first rule, the junction theorem, states that the sum of the currents into a specific junction in the circuit equals the sum of the currents out of the same junction. Electric charge is conserved: it does not suddenly appear or disappear; it does not pile up at one point and thin out at another.

The second rule, the loop equation, states that around each loop in an electric circuit the sum of the emf’s (electromotive forces, or voltages, of energy sources such as batteries and generators) is equal to the sum of the potential drops, or voltages across each of the resistances, in the same loop. All the energy imparted by the energy sources to the charged particles that carry the current is just equivalent to that lost by the charge carriers in useful work and heat dissipation around each loop of the circuit. [source]

## Kirchhoff’s First Law or Kirchhoff’s Current Law (KCL)

In any electrical network, the algebraic sum of currents meeting at a junction is always zero i.e. ∑I = 0

This law is also called as Junction rule.

$$I$$1 – $$I$$2 + $$I$$3 + $$I$$4 – $$I$$5 = 0

or $$I$$1 + $$I$$3 + $$I$$4 = $$I$$2 + $$I$$5

1. The sum of currents flowing towards a junction is equal to sum of currents leaving the junction.
2. By convention the current directed towards the junction is positive while those directed away from the junction is taken as negative.

3. The first law is in accordance with conservation of charge.
4. The charges do not accumulate at a junction. The total charge entering a junction is equal to total charge leaving the junction.

## Kirchhoff’s Second Law or Kirchhoff’s Voltage Law (KVL)

The algebraic sum of all the potential drops and emf’s along any closed path in a network is zero. i.e. ∑V = 0

This law is also called as Loop rule.

1. The second law is in accordance with conservation of energy.
2. According to second law the electric energy given to the charge by a source of emf is lost in passing through resistance.
3. The change in potential in covering a resistance in the direction of current is negative (–IR) while in opposite direction it is positive.
The potential falls along direction of current. The potential fall is taken as negative while potential rise is taken as positive.
4. The emf of source is taken as positive when it is traversed from negative to positive terminal while it is taken as negative when it is traversed from positive to negative irrespective of the direction of current.

5. If there are $n$ loops, there will be $(n – 1)$ equations according to loop rule.
6. The algebraic sum of products of currents and resistances in a closed loop is equal to sum of emf’s applied in the circuit i.e. ∑E =∑IR

## Dissimilar cells in Parallel

Let two cells of emf ${E_1}$ and ${E_2}$ with internal resistances ${r_1}$ and ${r_2}$ be connected in parallel to an external resistance$R$.

Applying loop rule to loop ${\rm{d}}$${\rm{l}}$${\rm{m}}$${\rm{n}}$${\rm{c}}$${\rm{d}} – {\rm{IR}} – {{\rm{I}}_1}{{\rm{r}}_1} + {{\rm{E}}_1} = 0\quad \ldots \ldots (1) Applying loop rule to loop {\rm{a}}$${\rm{l}}$${\rm{m}}$${\rm{n}}$${\rm{b}}$${\rm{a}}$ $– {\rm{IR}} – {{\rm{I}}_2}{{\rm{r}}_2} + {{\rm{E}}_2} = 0\quad \ldots \ldots$ (2)

Applying first rule at junction $\ell {\rm{I}} = {{\rm{I}}_1} + {{\rm{I}}_2}$

Multiply equation 1 by ${r_2}$ and eqn. 2 by ${r_1}$ and put ${I_2} = I – {I_1}$ we get

$– {\rm{IR}}{{\rm{r}}_2} – {{\rm{I}}_1}{{\rm{r}}_1}{{\rm{r}}_2} + {{\rm{E}}_1}{{\rm{r}}_2} = 0\quad$ and $\quad – {\rm{IR}}{{\rm{r}}_1} – \left( {{\rm{I}} – {{\rm{I}}_4}} \right){{\rm{r}}_2}{{\rm{r}}_1} + {{\rm{E}}_2}{{\rm{r}}_1} = 0$

On adding these we get $\quad {E_1}{r_2} + {E_2}{r_1} = IR\left( {{r_1} + {r_2}} \right) + {{\mathop{\rm Ir}\nolimits} _1}{r_2} = I\left( {{r_1} + {r_2}} \right)\left[ {R + {{{r_1}{r_2}} \over {{r_1} + {r_2}}}} \right]$

Or, $\quad {\rm{I}} = {{\left( {{{\rm{E}}_{{{\rm{f}}_2}}} + {{\rm{E}}_2}{{\rm{r}}_1}} \right)/{{\rm{r}}_1} + {{\rm{r}}_2}} \over {{\rm{R}} + {{{{\rm{r}}_1}{{\rm{r}}_2}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}}}} = {{{{\rm{E}}_{{\rm{eq}}}}} \over {{\rm{R}} + {{\rm{r}}_{{\rm{eq}}}}}}$

Two dissimilar cells in parallel are equivalent to a single cell of internal resistance ${{\rm{r}}_{{\rm{eq}}}} = {{{{\rm{r}}_{\rm{r}}} \cdot {{\rm{r}}_2}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}}$and ${\rm{emf}} {{\rm{E}}_{{\rm{eq}}}} = {{{{\rm{E}}_1}{{\rm{r}}_2} + {{\rm{E}}_2}{{\rm{r}}_1}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}} = {{{{\rm{r}}_1}{{\rm{r}}_2}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}}\left[ {{{{{\rm{E}}_1}} \over {{{\rm{r}}_1}}} + {{{{\rm{E}}_2}} \over {{{\rm{r}}_2}}}} \right]$

## Symmetric line method

If a circuit is symmetric about a line then all points lying on line of symmetry are at same potential.

In this method we divide the circuit in two equal parts and double the resistance of one part.

Example: In above question DF is line of symmetry so D and F are at same potential.

Now ${{\rm{R}} \over 2}$ is in parallel to ${\rm{R}}({\rm{CD}}) = {{{\rm{R}} \times {\rm{R}}/2} \over {{\rm{R}} + {\rm{R}}/2}} = {{\rm{R}} \over 3}$

This ${{\rm{R}} \over 3}$ is in series with ${\rm{R}}({\rm{AC}}) = {\rm{R}} + {\rm{R}}/3 = 4{\rm{R}}/3$

This ${{\rm{4R}} \over 3}$ is in parallel with ${\rm{R}}({\rm{AD}}) = 4{\rm{R}}/7\quad$

So ${{\rm{R}}_{{\rm{eq}}}} = {{2 \times 4{\rm{R}}} \over 7} = {{8{\rm{R}}} \over 7}$

## Solved Examples

Q. A circuit has a section ABC. If the potential at points $A, B$ and $C$ are ${V_1},{V_2}\&\,\,{V_3}$ then calculate potential at O.

Sol. Applying junction rule at O

$\quad – {I_1} – {I_2} – {I_3} = 0$ or $\quad {I_1} + {I_2} + {I_3} = 0$

Let ${V_0}$ be potential at O then

${{\rm{V}}_0} – {{\rm{V}}_1} = {{\rm{I}}_1}{{\rm{R}}_1},\quad {{\rm{V}}_0} – {{\rm{V}}_2} = {{\rm{I}}_2}{{\rm{R}}_2},\quad {{\rm{V}}_0} – {{\rm{V}}_3} = {{\rm{I}}_3}{{\rm{R}}_3}$

so $\quad {{\rm{I}}_1} = {{{{\rm{V}}_0} – {{\rm{V}}_1}} \over {{{\rm{R}}_1}}},\quad {{\rm{I}}_2} = {{{{\rm{V}}_0} – {{\rm{V}}_2}} \over {{{\rm{R}}_2}}},{{\rm{I}}_3} = {{{{\rm{V}}_0} – {{\rm{V}}_3}} \over {{{\rm{R}}_3}}}$

adding them we get $\quad {{\rm{V}}_0}\left[ {{1 \over {{{\rm{R}}_1}}} + {1 \over {{{\rm{R}}_2}}} + {1 \over {{{\rm{R}}_3}}}} \right] – \left[ {{{{{\rm{V}}_1}} \over {{{\rm{R}}_1}}} + {{{{\rm{V}}_2}} \over {{{\rm{R}}_2}}} + {{{{\rm{V}}_3}} \over {{{\rm{R}}_3}}}} \right] = 0$

or $\quad {V_0} = {{{V_1}} \over {{R_1}}} + {{{V_2}} \over {{R_2}}} + {{{V_3}} \over {{R_3}}}/{1 \over {{R_1}}} + {1 \over {{R_2}}} + {1 \over {{R_3}}}$

Cells connected in Series, Parallel & Mixed

Like resistors, a circuit can also have a combination of cells connected in series, parallel or even both! And again like resistors, we can calculate current and voltages that can be replaced by an equivalent cell in a circuit which we will learn here.

A circuit composed solely of components connected in series is known as a series circuit; likewise, one connected completely in parallel is known as a parallel circuit.

In a series circuit, the current that flows through each of the components is the same, and the voltage across the circuit is the sum of the individual voltage drops across each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the flowing through each component. [source]

Here we will see cells connected in series, parallel and mixed one by one along with examples.

Contents:

## Cells connected in Series Combination

The cells are said to be connected in series if negative terminal first is connected to positive of second whose negative terminal is connected to positive of third cell. The external resistance is connected between free terminals of first and last cells.

Let ‘n’ identical cells each of emf E and internal resistance r be connected in series. The combination can be replaced by a single cell of emf nE and internal resistance nr. The current flowing through load ${\rm{I}} = {{{\rm{nE}}} \over {{\rm{R}} + {\rm{nr}}}}$

1. If ${\rm{nr}} < < {\rm{R}}$ then ${\rm{I}} = {{{\rm{nE}}} \over {\rm{R}}}$ If equivalent internal resistance nr is less than external resistance R then current in circuit is equal to n times circuit current due to single cell.
2. If nr >> R then I = E/r. If equivalent internal resistance nr is greater than external resistance R then current in circuit is equal to short circuited current obtained from one cell.
3. Maximum current can be drawn from series combination of cells if external resistance is very large as compared to equivalent internal resistance.
4. If in series combination of n cells p cells are reversed than equivalent emf
${E_{{\rm{eq}}}} = ({\rm{n}} – {\rm{p}}){\rm{E}} – {\rm{pE}} = ({\rm{n}} – 2{\rm{p}}){\rm{E}}\quad$ and $\quad {{\rm{r}}_{{\rm{eq}}}} = {\rm{nr}}\quad$
So current ${\rm{I}} = {{({\rm{n}} – 2{\rm{p}}){\rm{E}}} \over {{\rm{nr}} + {\rm{R}}}}$
5. If un-identical cells are connected in series then
${E_{{\rm{eq}}}} = {{\rm{E}}_1} + {{\rm{E}}_2} + \ldots . = \Sigma {{\rm{E}}_{\rm{i}}}$ and ${r_{eq}} = {r_1} + {r_2} + \ldots . = \Sigma {r_i}$
So current ${\rm{I}} = {{\Sigma {{\rm{E}}_{\rm{i}}}} \over {{\rm{R}} + \Sigma {{\rm{r}}_{\rm{i}}}}}$.

## Cells connected in Parallel Combination

The cells are said to be connected in parallel if positive terminals of all the cells are connected together at one point and their negative terminals at another point. The external resistor is connected between these two points.

Let ‘m’ identical cells each of emf E and internal resistance r be connected in parallel. The combination can be replaced by a single cell of emf $E$ and internal resistance ${\rm{r}}/{\rm{m}}.$ The current ${\rm{I}} = {{\rm{E}} \over {{\rm{R}} + {\rm{r}}/{\rm{m}}}}$

1. If ${\rm{r}}/{\rm{m}} < < {\rm{R}}$ then ${\rm{I}} = {{\rm{E}} \over {\rm{R}}}$ . If equivalent internal resistance ${\rm{r}}/{\rm{m}}$ is less than external resistance ${\rm{R}}$ then current in circuit is equal to current produced by a single cell.
2. If ${\rm{r}}/{\rm{m}} > > {\rm{R}}$ then ${\rm{I}} = {{{\rm{mE}}} \over {\rm{r}}}$ . If equivalent internal resistance ${{\rm{r}} \over {\rm{m}}}$ is greater than external resistance then current in circuit is equal to ${\rm{m}}$ times the current produced by a short circuited cell.
3. Thus maximum current can be drawn from parallel combination of cells if external resistance is small as compared to net internal resistance of cells.

## Mixed Grouping

Let there be $n$ identical cells in one row and $m$ rows of cell in parallel. The combination of cells can be replaced by a single cell of emf nE and internal resistance ${{{\rm{ nr }}} \over m}$ .

The current $I = {{nE} \over {R + {{nr} \over m}}} = {E \over {{R \over n} + {r \over m}}}$ where $\quad {\rm{n}} \times {\rm{m}} = {\rm{p}} =$ total number of cells.

1. The current in the circuit is maximum when ${{\rm{R}} \over {\rm{n}}} + {{\rm{r}} \over {\rm{m}}}$ is minimum
So $\quad {{\rm{d}} \over {{\rm{dm}}}}\left( {{{\rm{R}} \over {\rm{n}}} + {{\rm{r}} \over {\rm{m}}}} \right) = 0\quad$
or $\quad {{\rm{d}} \over {{\rm{dm}}}}\left( {{{{\rm{mR}}} \over {\rm{p}}} + {{\rm{r}} \over {\rm{m}}}} \right) = 0\quad$
or $\quad {{\rm{R}} \over {\rm{p}}} – {{\rm{r}} \over {{{\rm{m}}^2}}} = 0$
or $\quad {{\rm{R}} \over {\rm{p}}} = {{\rm{R}} \over {{\rm{mn}}}} = {{\rm{r}} \over {{{\rm{m}}^2}}}\quad$
or $\quad {{\rm{R}} \over {\rm{n}}} = {{\rm{r}} \over {\rm{m}}}$
In mixed grouping of cells current in circuit is maximum if ${{\rm{R}} \over {\rm{n}}} = {{\rm{r}} \over {\rm{m}}}\quad$ and $\quad {{\rm{I}}_{\max }} = {{{\rm{nE}}} \over {2{\rm{R}}}} = {{{\rm{mE}}} \over {2{\rm{r}}}}$
2. In mixed grouping of cells power transferred to the load is maximum where external resistance R is equal to total internal resistance or $R = {{{\rm{ nr }}} \over m}\quad$ or $\quad {R \over n} = {r \over m}$. This shows power transfer is maximum when current is maximum.
$P = {{{E^2}R} \over {{{\left( {{R \over n} + {r \over m}} \right)}^2}}}\quad$ and $\quad {P_{\max }} = {{{n^2}{E^2}} \over {4R}} = {{{m^2}{E^2}R} \over {4{r^2}}}$
3. In mixed combination of cells, current in circuit and power transferred to load become maximum under same condition. This is why it is preferred over series and parallel combination of cells.

## Solved Examples

Q. Three identical cells each of emf 2V and unknown internal resistance are connected in parallel. This combination is connected to a 5$\Omega$ resistor. If the terminal voltage across the cells is 1.5V. What is internal resistance of each cell?

Sol. In parallel combination of three cells ${\rm{I}} = {{\rm{E}} \over {{\rm{R}} + {\rm{r}}/3}}$

Terminal potential difference $V = IR = {{ER} \over {R + r/3}}\quad$ or $\quad {r \over 3} = {{ER – RV} \over V}$

or $\quad r = 3\left( {{{ER – RV} \over V}} \right) = 3\left( {{{2 \times 5 – 5 \times 1.5} \over {1.5}}} \right) = 5\Omega$

Q. Determine currents ${{\rm{I}}_1},{{\rm{I}}_2}$ and ${{\rm{I}}_3}?$

Sol. Using II law in loop abcda $– 2{{\rm{I}}_1} + 24 – 27 – 6{{\rm{I}}_2} = 0\quad$ or $\quad 2{{\rm{I}}_1} + 6{{\rm{I}}_2} = – 3$ $\ldots ..(1)$

Using II law in loop cdfec we get $\quad – 27 – 6{{\rm{I}}_2} + 4{{\rm{I}}_3} = 0\quad$ or $\quad – 6{{\rm{I}}_2} + 4{{\rm{I}}_3} = 27\quad \ldots \ldots (2)$

From junction rule at c we get $\quad {{\rm{I}}_1} = {{\rm{I}}_2} + {{\rm{I}}_3}$ $\ldots ..(3)$

From 2 and 3 we get $\quad – 6{{\rm{I}}_2} + 4\left( {{{\rm{I}}_1} – {{\rm{I}}_2}} \right) = 27\quad$ or $\quad 4{{\rm{I}}_1} – 10{{\rm{I}}_2} = 27\quad \ldots \ldots$ (4)

Solving 1 and 4 we get ${{\rm{I}}_1} = 3{\rm{A}},\quad {{\rm{I}}_2} = – 1.5{\rm{A}}\quad$ so ${{\rm{I}}_3} = {{\rm{I}}_1} – {{\rm{I}}_2} = 4.5{\rm{A}}$

Negative ${{\rm{I}}_2}$ means the direction should have been opposite to shown in figure.

How to read a resistance color code?

Resistors are the most common and important component of any electronic circuit. In daily life we see a lot of circuits having resistances but how do we recognize them? To know the value of a resistance they are color coded with rings which have specific meanings according to the color and the place they are at. Let’s figure out now what Resistance Color Code is and what it all means!!

Contents:

## Resistance Color Code

Resistors are produced for laboratory and domestic uses. Generally they are available in two types:

1. Wire bound Resistors
2. Carbon Resistors

Wire bound resistors are made up by winding the alloy such as Nichrome, Manganin, Constantan etc as their resistivities are relatively insensitive to the temperature. they can be in the range of a fraction of an ohm to few hundred ohms.

The second one, Carbon Resistors are used to make higher range of resistors. These are made up of Carbon. These are in general compact, inexpensive hence used extensively in electronic circuits. These are very small so their values are given by color coding them. Next we are going to see what is this color coding.

## Color coding of Carbon Resistors

Check out the following table where each color is followed by 4 Strips. Here we will see what those 4 colored rings mean wound around a resistor.

### Mnemonics

A useful mnemonic matches the first letter of the color code, in numeric order. Here are two that includes tolerance codes gold, silver, and none:

• Bad beer rots our young guts but vodka goes well – get some now.
• B B ROY of Great Britain had a Very Good Wife who wore Gold and Silver Necklace.

[source]

A Carbon resistor looks like the one shown above. It has 4 rings around it.

The first two rings give first two significant figures of resistance.

The third ring indicates the decimal multiplier i.e. the number of zeroes that will follow the two significant figures.

The fourth ring shows tolerance or percentage accuracy.

Q. What will be the color code for $42{\rm{k}}\Omega \pm 10\%$ carbon resistance.

Sol. According to color code color for digit 4 is yellow, for digit 2 it is red, for 3 colors is orange and 10$\%$ tolerance is represented by silver color.

So color code should be yellow, red, orange and silver.

Q. What is resistance of following resistor?

Sol. Brown color gives multiplier ${10^\prime },$ Gold gives a tolerance of $\pm 5\%$

So resistance of resistor is $47 \times {10^1}\Omega \pm 5\% \quad$ or $\quad 470 \pm 5\% \Omega$

## Body End Dot Color Code

Earlier, resistors were manufactured with this color code.

1. The color of body gives first significant figure.
2. The color of end gives second significant figure.
3. The color of dot gives the decimal multiplier.
4. The color of ring gives the tolerance.

Combination of Resistances – Series & Parallel Combination

We have learn now what a resistance it. Now let’s see how the combination of resistances can be used in a circuit to control the flow of current and voltage division. Majorly the combination of these resistances are of 2 types: Series and Parallel. Here we will be taken both in detail below.

Contents:

## Combination of Resistances

series circuit comprises a path along which the whole current flows through each component. A parallel circuit comprises branches so that the current divides and only part of it flows through any branch. The voltage, or potential difference, across each branch of a parallel circuit is the same, but the currents may vary. In a home electrical circuit, for instance, the same voltage is applied across each light or appliance, but each of these loads draws a different amount of current, according to its power requirements. A number of similar batteries connected in parallel provides greater current than a single battery, but the voltage is the same as for a single battery. [source]

Let’s start with Combination of resistances with series combination at first then parallel combination.

### Series Combination

Resistances are said to be connected in series between two points if they provide only a single path between two points. Resistances are connected in series if same current flows through each resistance when some potential difference is applied across the combination.

Potential difference across each resistance is different and is directly proportional to its resistance ${\rm{V}} \propto {\rm{R}}.$
So $\quad {{\rm{V}}_1} = {\rm{I}}{{\rm{R}}_1};\quad {{\rm{V}}_2} = {\rm{I}}{R_2}\quad$ and $\quad {{\rm{V}}_3} = {\rm{I}}{{\rm{R}}_3}$

1. The series combination obeys law of conservation of energy
So $V = {V_1} + {V_2} + {V_3} = I\left( {{R_1} + {R_2} + {R_3}} \right)$
Equivalent resistance ${{\rm{R}}_{\rm{s}}} = {{\rm{V}} \over {\rm{I}}} = {{\rm{R}}_1} + {{\rm{R}}_2} + {{\rm{R}}_3}$
2. The equivalent resistance is equal to sum of individual resistances.
3. The equivalent resistance is greater than largest of individual resistance.
4. The resistances are connected in series
1. to increase the resistance and
2. to divide large potential difference across many resistances.
5. In ‘n’ identical resistances R are connected in series then the equivalent resistance Rs = nR
6. This combination is used in resistance boxes and sometimes in decorative bulbs.
7. In resistances connected in series if one resistance get open the current in whole circuit will become zero.

### Parallel Combination

Resistances are said to be connected in parallel between two points if it is possible to proceed from one point to another along different paths.

Resistances are said to be in parallel if potential across each resistance is same and equal to applied potential.

Current through each resistance is different and is inversely proportional to resistance of resistor.

$l \propto 1/R.$ So $\quad {I_1} = {V \over {{R_1}}},{I_2} = {V \over {{R_2}}}\quad$ and $\quad {I_3} = {V \over {{R_3}}}$

1. The parallel combination obeys the conservation of charge.
So $\quad {\rm{I}} = {{\rm{I}}_1} + {{\rm{I}}_2} + {{\rm{I}}_3} = {\rm{V}}\left( {{1 \over {{{\rm{R}}_1}}} + {1 \over {{{\rm{R}}_2}}} + {1 \over {{{\rm{R}}_3}}}} \right)$
Reciprocal of equivalent resistance ${1 \over {{{\rm{R}}_{\rm{p}}}}} = {{\rm{I}} \over {\rm{V}}} = {1 \over {{{\rm{R}}_1}}} + {1 \over {{{\rm{R}}_2}}} + {1 \over {{{\rm{R}}_3}}}$
2. The reciprocal of equivalent resistance is equal to sum of reciprocal of individual resistances.
3. The equivalent resistance is smaller than smallest of individual resistance.
4. The resistances are connected in parallel to decrease resistance.
5. If ‘n’ identical resistances $R$ are connected in parallel then equivalent resistance ${R_p} = R/n$
6. This combination is used in household electrical appliances.
7. In resistances connected in parallel if one resistance becomes open then also all others will work as usual.
8. In case of two resistances in parallel ${{{{\rm{I}}_1}} \over {{{\rm{I}}_2}}} = {{{{\rm{R}}_2}} \over {{{\rm{R}}_1}}}\quad$ and $\quad {{\rm{I}}_1} + {{\rm{I}}_2} = {\rm{I}}$
So $\quad {{\rm{I}}_1} = {{{{\rm{R}}_2}{\rm{I}}} \over {{{\rm{R}}_1} + {{\rm{R}}_2}}}\quad$ and $\quad {{\rm{I}}_2} = {{{{\rm{R}}_1}{\rm{I}}} \over {{{\rm{R}}_1} + {{\rm{R}}_2}}}$

## Solved Examples on Combination of Resistances

Q. Find equivalent resistance between the points $A$ and $B$ ?

Sol. Point 1 and 3 are at same potential. Similarly point 2 and 4 are at same potential. Joining resistance between 1 and 2,2 and 3,3 and 4 we find that they all are in parallel.

So equivalent resistance ${1 \over {{R_p}}} = {1 \over {2R}} + {1 \over {2R}} + {1 \over R} = {2 \over R}\quad$ or $\quad {R_p} = R/2$

Q. Five resistors are connected as shown. Find equivalent resistance between the points ${\rm{B}}$ and ${\rm{C}}$ .

Sol. Resistance of arm ${\rm{ADC}} = 3 + 7 = 10\Omega$ .

This is in parallel to 10$\Omega$ of arm AC. Their parallel combination gives equivalent resistance of 5$\Omega$

So effective resistance between ${\rm{B}}$ and ${\rm{C}}$

${{\rm{R}}_{{\rm{eq}}}} = {{14 \times 5} \over {14 + 5}} = {{70} \over {19}} = 3.684\Omega$

Q. When two resistances are joined in series their resistance is 40$\Omega$ and when they are joined in parallel the resistance is 7.5$\Omega$ . Find the individual resistances?

Sol. Let the two resistances be ${R_1}$ and ${R_2}$

When connected in series ${R_1} + {R_2} = 40$

When connected in parallel ${{{R_1}{R_2}} \over {{R_1} + {R_2}}} = 7.5\quad$ so $\quad {R_1}{R_2} = 7.5 \times 40 = 300$

${\left( {{R_1} – {R_2}} \right)^2} = {\left( {{R_1} + {R_2}} \right)^2} – 4{R_1}{R_2} = {40^2} – 4 \times 300 = 400$

So $\quad {R_1} – {R_2} = 20\Omega$

Solving 1 and 2 we have ${R_1} = 30\Omega$ and ${R_2} = 10\Omega$

Q. Twelve equal resistances $R$ are used to generate shape of a cube. Calculate equivalent resistance across the side of cube?

Sol. By symmetry potential at point 4 and 5 is same. Similarly potential at point 3 and 6 is same. The equivalent circuits can be drawn

as :

The equivalent resistance between a and b is ${{\rm{R}}_{{\rm{eq}}}} = {{{\rm{R}} \times {7 \over 5}{\rm{R}}} \over {{\rm{R}} + {7 \over 5}{\rm{R}}}} = {7 \over {12}}{\rm{R}}$

Q. Calculate the current shown by ammeter A in the circuit shown in fig.

Sol. The equivalent circuits can be drawn as

The equivalent resistance of circuit is ${{\rm{R}}_{{\rm{eq}}}} = {{10} \over 3}\Omega$

The current ${\rm{I}} = {{\rm{E}} \over {{{\rm{R}}_{{\rm{eq}}}}}} = {{12} \over {10/3}} = 3.6{\rm{A}}$

Q. The resistance between the points A and B in the following diagram Fig. will be-

$(1)\,4\Omega$

$(2)\,8\Omega$

$(3)\,6\Omega$

$(4)\,2\Omega$

Sol. $\quad {\rm{R}} = {{\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right)} \over 2} + {1 \over 2}{\left[ {{{\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right)}^2} + 4{{\rm{R}}_3}\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right)} \right]^{{1 \over 2}}} \ldots \ldots \ldots$ (1)

${\rm{R}}1 = 1\Omega ,{{\rm{R}}_2} = 0.{{\rm{R}}_3} = 2\Omega$

From eqs. $(1)$ and $(2)$

${\rm{R}} = {1 \over 2} + {1 \over 2}\left[ {1 + 4 \times 2 \times 4{1 \over 2}[1 + 3] = 2\Omega } \right.$

Q. Calculate equivalent resistance between A and B

Sol. After distribution of current using loop rule for

$\quad$ Loop a $– \left( {{\rm{I}} – {{\rm{I}}_1}} \right){\rm{R}} – {{\rm{I}}_2}{\rm{R}} + {{\rm{I}}_1}{\rm{R}} = 0\quad$ or $\quad 2{{\rm{I}}_1} – {{\rm{I}}_2} = {\rm{I}}$

Loop b $– \left( {1 – {{\rm{I}}_1} – {{\rm{I}}_2}} \right){\rm{R}} + {{\rm{I}}_3}{\rm{R}} + {{\rm{I}}_2}{\rm{R}} = 0\quad$ or $\quad {{\rm{I}}_1} + 2{{\rm{I}}_2} + {{\rm{I}}_3} = {\rm{I}}$

Loop ${\rm{c}} – {{\rm{I}}_3}{\rm{R}} – \left( {{\rm{I}} – {{\rm{I}}_1} – {{\rm{I}}_2} + {{\rm{I}}_3}} \right){\rm{R}} + \left( {{{\rm{I}}_1} + {{\rm{I}}_2} – {{\rm{I}}_3}} \right){\rm{R}} = 0\quad$ or $\quad 2{{\rm{I}}_1} + 2{{\rm{I}}_2} – 3{{\rm{I}}_3} = {\rm{I}}$

Solving for ${{\rm{I}}_1},{{\rm{I}}_2}$ and ${{\rm{I}}_3}$ we get,

${{\rm{I}}_1} = {4 \over 7}{\rm{I}},{{\rm{I}}_2} = {{\rm{I}} \over 7},{{\rm{I}}_3} = {{\rm{I}} \over 7}$

Here ${{\rm{I}}_2} = {{\rm{I}}_3}$ so, no distribution of current takes place at D so it is a pseudo junction.

The equivalent circuit becomes.

Resistance of triangle DCE is R parallel to $({\rm{R}} + {\rm{R}}) = {2 \over 3}{\rm{R}}$

Resistance of network other than ${\rm{AB}} = {\rm{R}} + {2 \over 3}{\rm{R}} + {\rm{R}} = {8 \over 3}{\rm{R}}$

Equivalent resistance between ${\rm{AB}}$

${{\rm{R}}_{{\rm{eq}}}} = {{2{\rm{R}} \times {{8{\rm{R}}} \over 3}} \over {2{\rm{R}} + {{8{\rm{R}}} \over 3}}} = {8 \over 7}{\rm{R}}$

Q. Calculate the effective resistance between A and B in following network.

Sol. Ratio of upper resistances 5 : 10 : 15=1 : 2 : 3

Ratio of lower resistances 10 : 20 : 30=1 : 2 : 3

The ratio is same so resistance in middle are not useful.

Equivalent resistance =(5+10+15)| |(10+20+30)

So ${R_{{\rm{ eq }}}} = {{30 \times 60} \over {30 + 60}} = 20\Omega$

Q. Calculate potential difference between points d and b in the circuit.

Sol. Using voltage rule from ${\rm{b}}$ to ${\rm{d}}$ via c we get

${{\rm{V}}_{\rm{b}}} – 1 \times 2 + 4 \times 1 = {{\rm{V}}_{\rm{d}}}\quad$ or $\quad {{\rm{V}}_{\rm{d}}} – {{\rm{V}}_{\rm{b}}} = 2$ volt.

Introduction to Ohm’s Law and Resistance | Resistivity | Thermistors

We now have basic idea about Potential difference and Current. Now let us study the relation between them. How potential difference applied across the conductor and current flowing through it are related to each other if certain physical conditions are kept the same. This relation was given by G. S. Ohm in 1828 even before the physical mechanism responsible for flow of current was discovered. We will start with Introduction to Ohm’s Law along with resistance, resistivity, specific uses of conducting materials, thermistors.

Contents:

## Introduction to Ohm’s Law

If the physical state i.e. temperature, nature of material and dimensions of a conductor remain unchanged then the ratio of potential difference applied across to ends to current flowing through it remains constant.

${\rm{V}} \propto {\rm{I}}$ or $\quad {\rm{V}} = {\rm{IR}}\quad$ where ${\rm{R}} = {{\rm{V}} \over {\rm{I}}}$ is resistance of conductor.

${\rm{I}} = {\rm{neA}}{{\rm{V}}_{\rm{d}}}$

$= {\rm{n}}e{\rm{A}}{{{\rm{eE}}} \over {\rm{m}}}\tau$

$= \left( {{{{\rm{n}}{{\rm{e}}^2}\tau } \over {\rm{m}}}} \right){\rm{AE}}$

$= \left( {{{{\rm{n}}{{\rm{e}}^2}\tau } \over {\rm{m}}}} \right){\rm{A}}{{\rm{V}} \over {\rm{L}}}$

So, ${\rm{R}} = {{\rm{V}} \over {\rm{I}}} = \left( {{{\rm{m}} \over {{\rm{n}}{{\rm{e}}^2}\tau }}} \right){{\rm{L}} \over {\rm{A}}}$

R is resistance of conductor.

This is the basic introduction to Ohm’s Law. Let’s know more about Ohm’s Law and Resistance in detail below.

## Ohm’s Law & Resistance

The property of a substance due to which it opposes the flow of current through it is called resistance.

It is a scalar quantity with unit volt/ampere called ohm $(\Omega )$
Dimensions of ${\rm{R}} = {{\rm{V}} \over {\rm{I}}} = {{\rm{W}} \over {{\rm{qI}}}} = {{{{\rm{M}}^1}{{\rm{L}}^2}{{\rm{T}}^{ – 2}}} \over {{\rm{ATA}}}} = {{\rm{M}}^\prime }{{\rm{L}}^2}{{\rm{T}}^{ – 3}}{{\rm{A}}^{ – 2}}$

The reciprocal of resistance is called conductance ${\rm{G}} = {1 \over {\rm{R}}}$
The SI unit is ohm-1 or mho or siemen (s) and its dimensions are ${M^{ – 1}}{L^{ – 2}}{{\rm{T}}^3}{{\rm{T}}^2}$.

Important Points

1. The substances which obey ohm’s law are called Ohmic or linear conductor. The resistance of such conductors is independent of magnitude and polarity of applied potential difference. Here the graph between I and V is a straight line passing through the origin. The reciprocal of slope of straight line gives resistance ${\rm{R}} = {{\rm{V}} \over {\rm{I}}} = {1 \over {\tan \theta }}$ = constant.
Examples: silver, copper, mercury, carbon, mica etc.

2. The substances which do not obey ohm’s law are called non-Ohmic or non-linear conductors. The I–V curve is not a straight line.
Example: p n diode, transistors, thermionic valves, rectifiers etc.

3. ${\rm{V}} =$ IR defines resistance and hence it is applicable to all ohmic and non-ohmic conductors. If ${\rm{R}}$ is constant or ${\rm{V}} \propto {\rm{I}}$ then it represents ohm’s law.
4. The relation ${\rm{R}} = {{\rm{V}} \over {\rm{I}}}$ is macroscopic form while $\rho = {{\rm{E}} \over {\rm{J}}}$ is microscopic form of Ohm’s Law.

The resistance of a conductor depends on temperature, nature of material, length and area of cross-section.

The temperature dependence of resistance is given by ${\rm{R}} = {{\rm{R}}_0}(1 + \alpha \Delta \theta ),$where $\alpha$ is temperature coefficient of resistance and $\Delta $$\theta is change in temperature. The temperature coefficient of resistance \alpha = {{R – {R_0}} \over {{R_0}\Delta \theta }} is defined as change in resistance per unit resistance at 0^{\circ} \mathrm{C} per degree rise of temperature. The resistance of a given object depends primarily on two factors: What material it is made of, and its shape. For a given material, the resistance is inversely proportional to the cross-sectional area; for example, a thick copper wire has lower resistance than an otherwise-identical thin copper wire. Also, for a given material, the resistance is proportional to the length; for example, a long copper wire has higher resistance than an otherwise-identical short copper wire. [source] For maximum metals \alpha = {1 \over {273}} per ^{\circ} \mathrm{C} so {\rm{R}} = {{\rm{R}}_0}\left( {1 + {{\Delta \theta } \over {273}}} \right) = {{\rm{R}}_0}\left( {{{273 + \Delta \theta } \over {273}}} \right) = {{\rm{R}}_0}{{\rm{T}} \over {273}} So, R \propto T i.e. the resistance of pure metallic conductor is proportional to its absolute temperature. The fractional change in resistance without change in volume or mass are: (a) When change in length is small ( \le 5\% ) fractional change in R is {{\Delta R} \over R} = {{2\Delta L} \over L} (b) When change in radius is small ( \le 5\% ) fractional change in R is {{\Delta R} \over R} = {{ – 4\Delta r} \over r} (c) When change in area is small ( \le 5\% ) fractional change in R is {{\Delta R} \over R} = {{ – 2\Delta A} \over A} ## Resistivity {\rm{R}} \propto {\rm{L}} and {\rm{R}} \propto {1 \over {\rm{A}}} So,\quad {\rm{R}} \propto {{\rm{L}} \over {\rm{A}}}\quad or \quad {\rm{R}} = \rho {{\rm{L}} \over {\rm{A}}} where, \rho is called resistivity or specific resistance. In terms of microscopic quantities E = \rho J so resistivity is numerically equal to ratio of magnitude of electric field to current density. \rho = {{RA} \over L} So, if L = 1{\rm{m}},{\rm{A}} = 1{{\rm{m}}^2} then \rho = {\rm{R}}. Specific resistivity is numerically equal to resistance of substance having unit area of cross-section and unit length. It is a scalar with unit ohm-meter (\Omega – m) and dimensions {M^1 }{L^3}{T^{ – 3}}{{\rm{A}}^{ – 2}} . The reciprocal of resistivity is called conductivity or specific conductance with units mho/m and dimensions {{\rm{M}}^{ – 1}}{{\rm{L}}^{ – 3}}{{\rm{T}}^3}{{\rm{A}}^2},\quad \sigma = {1 \over \rho } = {{{\rm{n}}{{\rm{e}}^2}\tau } \over {\rm{m}}} = ne\mu Important Points 1. The resistivity is independent of shape and size of body and depends on nature of material of body. The resistivity is the property of material while resistance is property of object. 2. The temperature dependence of resistivity is given by relation \rho = {\rho _0}(1 + \infty \Delta \theta ) where \propto is temperature coefficient of resistivity and \Delta \theta is change is temperature. For metals \propto is positive so resistivity increases with temperature while for non-metals \infty is negative so resistivity decreases with temperature. ## Specific uses of conducting materials 1. The heating element of devices like heater, geyser, press etc. are made of nichrome because it has high resistivity and high melting point. It does not react with air and acquires steady state when red hot at 800^{\circ} \mathrm{C} . 2. Fuse wire is made of tin lead alloy because it has low melting point and low resistivity. The fuse is used in series, and melts to produce open circuit when current exceeds the safety limit. 3. Resistances of resistance box are made of Manganin or Constantan because they have moderate resistivity and very small temperature coefficient of resistance. The resistivity is nearly independent of temperature. 4. The filament of bulb is made up of tungsten because it has low resistivity, high melting point of 3300 K and gives light at 2400 K. The bulb is filled with inert gas because at high temperature it reacts with air forming oxide. 5. The connection wires are made of copper because it has low resistance and resistivity. Q. Discuss the effect on the resistance R of wire when mass is kept constant and (a) length is increased n times (b) radius is increased n times (c) cross-sectional area is increased n times. Sol. (a) {\rm{R}} = {{\rho {\rm{L}}} \over {\rm{A}}} = {{\rho {\rm{L}}} \over {\rm{A}}}{{{\rm{xL}}} \over {{\rm{xL}}}} = {{\rho {{\rm{L}}^2}} \over {\rm{V}}} = {{\rho {{\rm{L}}^2}{\rm{d}}} \over {\rm{m}}} where d is density, m is mass and V is volume of wire. When length is increased n times the new resistance {{\rm{R}}^\prime } = {{\rho {{({\rm{nL}})}^2}{\rm{d}}} \over {\rm{m}}} So, \quad {{\rm{R}}^\prime } = {{\rm{n}}^2}{\rm{R}} (b) {\rm{R}} = {{\rho {\rm{L}}} \over {\rm{A}}} = {{\rho {\rm{L}} \times {\rm{A}}} \over {{\rm{A}} \times {\rm{A}}}} = {{\rho {\rm{V}}} \over {{{\rm{A}}^2}}} = {{\rho {\rm{m}}} \over {{\rm{d}}{\pi ^2}{{\rm{r}}^4}}} When radius is increased n times new resistance R = {{\rho m} \over {d{\pi ^2}{{(nr)}^4}}} so {R \over R} = {{{r^4}} \over {{{(nr)}^4}}} or R = {R \over {{n^4}}} (c) {\rm{R}} = {{\rho {\rm{L}}} \over {\rm{A}}} = {{\rho {\rm{L}} \times {\rm{A}}} \over {{\rm{A}} \times {\rm{A}}}} = {{\rho {\rm{V}}} \over {{{\rm{A}}^2}}} = {{\rho {\rm{m}}} \over {{\rm{d}}{{\rm{A}}^2}}} When area is increased n times new resistance {{\rm{R}}^\prime } = {{\rho {\rm{m}}} \over {{\rm{d}}{{({\rm{nA}})}^2}}} so {{{{\rm{R}}^\prime }} \over {\rm{R}}} = {{{{\rm{A}}^2}} \over {{{({\rm{nA}})}^2}}}\quad or \quad {{\rm{R}}^\prime } = {{\rm{R}} \over {{{\rm{n}}^2}}} ## Thermistors A thermistor is a heat sensitive device made of semiconducting material where resistivity changes rapidly with change of temperature. It has a large temperature coefficient. The temperature coefficient may be positive or negative. These can be used over wide range of temperature. ### Important Applications 1. The thermistors are used to detect small temperature changes because they respond rapidly to temperature changes. 2. The thermistors with a large negative temperature coefficient is used in resistance thermometer in very low temperature measurement. 3. These are used in temperature control units of industry. 4. Thermistors are used in protection of windings of generators, motors and transformers. 5. They are used for voltage stabilization, remote sensing and safe guarding filament of picture tube of T.V. Also Read: Electric Currents in Conductors | Mobility We now know an electric charge experiences a force when placed in electric field, if it starts to move it constitutes Electric Current. Here we are going to study the Electric Currents in Conductors when there is no electric field and when the conductor is placed in electric field. We will also learn about Mobility of charge carriers in the end. Contents: ## Introduction to Conductors & Insulators In an atom or a molecule, negatively charged electrons are strongly held by positively charged nucleus therefore the charges are bound to each other and are not free to move. But free charge particles do exist in upper strata of atmosphere called the ionosphere. Matter is made up of molecules like a small water droplet contains approx 1019 molecules. These molecules are closely packed such that the electrons are not attached to only one nuclei. In some materials, these electrons do not move even when electric field is applied, such materials are called insulators. While in some materials (generally metals), some electrons are allowed free movement inside that material. These are known as conductors, they form electric currents in them when electric field is applied. Although an electrical insulator is ordinarily thought of as a nonconducting material, it is in fact better described as a poor conductor or a substance of high resistance to the flow of electric current. Different insulating and conducting materials are compared with each other in this regard by means of a material constant known as resistivity. Electrical insulators are used to hold conductors in position, separating them from one another and from surrounding structures. They form a barrier between energized parts of an electric circuit and confine the flow of current to wires or other conducting paths as desired. [source] Here we will be studying in details about Electric currents in conductors. A few important points about conductors: 1. Conductors are substances through which electric charges can flow easily. 2. They are characterized by presence of a large number of free electrons ({10^{29}} electrons per m3) 3. The number density of free electrons in a conductor is same throughout the conductor. This is because free electrons experience repulsive force between them and conductor allows movement of free electrons. Thus, free electrons are evenly scattered throughout the volume of conductor. 4. These free electrons transport electric charge so are called as Conduction electrons. ## Electric Current in Conductors with or without Electric field ### a) Behavior of Conductor in absence of applied Potential Difference 1. The free electrons present in a conductor gain energy surrounding and move randomly in a conductor. 2. The speed gained by virtue of temperature is called as thermal speed of an electron. {1 \over 2}{\rm{mv}}_{{\rm{ms}}}^2 = {3 \over 2}{\rm{kT}}\quad So, Thermal speed \quad {{\rm{v}}_{{\rm{ms}}}} = \sqrt {{{3{\rm{kT}}} \over {\rm{m}}}} where {\rm{m}} is mass of electron. At room temperature T = 300{\rm{K}}\quad {{\rm{v}}_{{\rm{rms}}}} = {10^5}{\rm{m}}/{\rm{sec}} 3. The average distance traveled by a free electron between two consecutive collisions is called as Mean Free Path \lambda .\left(\lambda-10 \mathrm{A}^{\circ}\right) Mean free path \lambda = {{{\rm{ total distance travelled }}} \over {{\rm{ number of collisions }}}} 4. The time taken by an electron between two successive collisions is called as Relaxation time. \tau .\left( {\tau – {{10}^{ – 14}}{\rm{s}}} \right) Relaxation time \tau = {{{\rm{ total time taken }}} \over {{\rm{ number of collisions }}}} 5. The thermal speed can be written as {v_T} = {\lambda \over \tau } 6. In absence of applied potential difference electrons have random motion. The average displacement and average velocity is zero. There is no flow of current due to thermal motion of free electrons in a conductor. ### b) Behavior of Conductor in presence of applied Potential Difference 1. When two ends of conductors are joined to a battery then one end is at higher potential and another at lower potential. This produces an electric field inside the conductor from point of higher to lower potential i.e. E = {v \over L} 2. The field exerts an electric force on free electrons causing acceleration of each electron. \overrightarrow {\rm{F}} = {\rm{m}}\overrightarrow {\rm{a}} = – {\rm{e}}\overrightarrow {\rm{E}} So acceleration \vec a = {{ – e\vec E} \over m} 3. The average velocity with which the free electrons are drifted towards the positive end of a conductor under the influence of an external electric field is called drift velocity vd. Using \quad \vec v = \vec u + \vec a{\rm{t}}\quad we have {{\vec{v}}_{d}}=\frac{-e\vec{E}}{m}\tau$$\left( {{v_d}~{{10}^{ – 4}}m/sec} \right)$
4. The direction of drift velocity for electrons in a metal is opposite to that of applied field E.
5. Relation between current and drift velocity
Let ‘n’ be number density of free electrons and A be area of cross-section of conductor.
Number of free electrons in conductor of length L = nAL
Total charge on these free electrons $\Delta {\rm{q}} =$ neAl
Current ${\rm{I}} = {{\Delta {\rm{q}}} \over {\Delta {\rm{t}}}} = {\rm{neAL}}{{{{\rm{v}}_{\rm{d}}}} \over {\rm{L}}} = {\rm{neA}}{{\rm{v}}_{\rm{d}}}$ or $\quad {\rm{I}} = {\rm{neA}}{{\rm{v}}_{\rm{d}}}$
The current flowing through a conductor is directly proportional to the drift velocity $\left( {{\rm{I}} \propto {{\rm{v}}_{\rm{d}}}} \right)$
6. The current density ${\rm{J}} = {{\rm{I}} \over {\rm{A}}} = {\rm{ne}}{{\rm{v}}_{\rm{d}}} = {\rm{ne}}\left( {{{{\rm{eE}}} \over {\rm{m}}}} \right)\tau = \left( {{{{\rm{n}}{{\rm{e}}^2}\tau } \over {\rm{m}}}} \right){\rm{E}}$
So, $\quad J \propto E\quad$ or $\quad J = \sigma E$
where, $\sigma = {{{\rm{n}}{{\rm{e}}^2}\tau } \over {\rm{m}}}$ is specific conductivity of conductor which depends on temperature and nature of material.
$\vec J = \sigma \vec E$ is microscopic form of ohm’s law.
7. The drift velocity depends on nature of metal through $\tau$ , applied potential difference, length of conductor.
${v_d} = {{eE} \over m}\tau = {{eV} \over {mL}}\tau$
${{\rm{v}}_{\rm{d}}}$ is independent of radius or area of cross-section of a conductor.
8. The rise of temperature causes increase in ${{\rm{v}}_{{\rm{ms}}}}$ and hence a decrease in $\lambda$ and relaxation time $\tau$ causing a decrease in drift velocity.

## Mobility

Conductivity arises from mobile charge carriers like electrons in metals, positive charged ions and electrons in an ionized gas, positive and negative ions in an electrolyte. So there must be some difference in the movement of these different charge carriers. It is called Mobility.

Mobility of a charge carrier is defined as drift velocity acquired per unit electric field.
Mobility $\mu = {{{v_d}} \over E} = {{e\tau } \over m} = {e \over m}{{m\sigma } \over {n{e^2}}} = {\sigma \over {ne}}$
The unit is ${m^2}{V^{ – 1}}{s^{ – 1}}$ and dimensions are ${M^{ – 1}}{T^2}{A^1}$

The mobility depends on applied potential difference, length of conductor, number density of charge carriers, current in conductor, area of cross-section of conductor.

Mobility $\mu = {{{v_d}} \over E} = {{e\tau } \over m} = {e \over m}{{m\sigma } \over {n{e^2}}} = {\sigma \over {ne}}$

The unit is ${{\rm{m}}^2}{{\rm{V}}^{ – 1}}{{\rm{s}}^{ – 1}}$ and dimensions are ${{\rm{M}}^{ – 1}}{{\rm{T}}^2}{{\rm{A}}^1}$

The mobility depends on applied potential difference, length of conductor, number density of charge carriers, current in conductor, area of cross-section of conductor.

Q. Though the drift velocity for electrons is small, an electric bulb lights up immediately as we turn the switch on. Why?

Sol. When switch is made on the electric field $\vec E$ responsible for setting up current propagates through wires at speed of light $3 \times {10^8}{\rm{m}}/{\rm{s}}$. So field is set up immediately in time ${\rm{L}}/{\rm{c}}$ causing electrons to drift and hence bulb lights up immediately.

Types & Effects of Electric Current | Current Density

In electrostatics, we studied charges at rest. In real life the most important application without which we can’t think of our lives i.e. electricity is produced by charges in motion. Charges in motion constitute electric current. Here we will see the basic things like Unit of Electric Current, Types of Electric Current, Effects of Electric Current and Current density.

Let’s start:

## Introduction to Electric Current

Electricity, phenomenon associated with stationary or moving electric charges. Electric charge is a fundamental property of matter and is borne by elementary particles. In electricity the particle involved is the electron, which carries a charge designated, by convention, as negative. Thus, the various manifestations of electricity are the result of the accumulation or motion of numbers of electrons. [source]

Electric Current is defined as rate of flow of charge through any cross-section.
If $\Delta q$ charge passes through a cross-section in time $\Delta {\rm{t}}$ then average current ${{\rm{I}}_{{\rm{av}}}} = {{\Delta {\rm{q}}} \over {\Delta {\rm{t}}}}\quad$and $\quad$ instantaneous current ${\rm{I}} = \mathop {\lim }\limits_{\Delta {\rm{t}} \to 0} {{\Delta {\rm{q}}} \over {\Delta {\rm{t}}}} = {{{\rm{dq}}} \over {{\rm{dt}}}}$

## Unit of Electric Current

Current is a fundamental quantity with dimension $\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{\circ} \mathrm{A}^{1}$

Current is a scalar quantity with its SI unit Ampere.

Ampere: The current through a conductor is said to be one ampere if one coulomb of charge is flowing per second through a cross-section of wire.

CGS unit of current is Biot (Bi)

1 ampere $= {{1{\rm{ coulomb }}} \over {{\rm{ sec ond }}}} = {{(1/10){\rm{ emu of charge }}} \over {{\rm{ sec ond }}}} = {1 \over {10}}$ Biot

### Important Points

1. The conventional direction of current is the direction of flow of positive charge or applied field.
It is opposite to direction of flow of negatively charged electrons.

2. The current may be constituted by motion of different type of charge carries in different situations.

3. If n electrons pass through a point in a conductor in time $t$ then current through the conductor is ${\rm{I}} = {{\rm{q}} \over {\rm{t}}} = {{{\rm{ne}}} \over {\rm{t}}}$. Number of electrons flowing through conductor in ${\rm{t}}$ second is${\rm{n}} = {{{\rm{I}} \times {\rm{t}}} \over {\rm{e}}}$. For 1 ampere of current ${\rm{n}} = {1 \over {1.6 \times {{10}^{ – 19}}}} = 6.25 \times {10^{18}}$ electrons/second
4. The conductor remains uncharged when current flows through it because the charge entering at one end per second is equal to charge leaving the other end per second.
5. For a given conductor current does not change with change in its cross-section because current is simply rate of flow of charge.

6. If n particles each having a charge q pass per second per unit area then current associated with cross-sectional area A is
${\rm{I}} = {{\Delta {\rm{q}}} \over {\Delta {\rm{t}}}} = {\rm{nqA}}$
7. If there are n particles per unit volume each having a charge q and moving with velocity v then current through cross-sectional area A is ${\rm{I}} = {{\Delta {\rm{q}}} \over {\Delta {\rm{t}}}} = {\rm{nqvA}}$
8. If a charge q is moving in a circle of radius r with speed v then its time period is
T = 2πr/v. The equivalent current ${\rm{I}} = {{\rm{q}} \over {\rm{T}}} = {{{\rm{qv}}} \over {2\pi {\rm{r}}}}$

## Current Density

This is defined as current flowing per unit area held normal to direction of flow of current.

Current density $J = {I \over A} = {{nq} \over {At}}$ and current $I = \int {\vec J} \cdot \overrightarrow {da}$

The direction of current density $\vec \jmath$ is same as that of electric field $\vec E.$ J is a vector with unit ${\rm{amp}}/{{\rm{m}}^2}$and dimension $\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{\circ} \mathrm{A}^{1}$

## Effects of Electric Current

### (a) Heating Effects of Electric Current

The phenomenon of heating of a conductor by the flow of an electric current through it is called heating effect of electric current.

### (b) Magnetic Effects of Electric Current

The phenomenon of production of a magnetic field in space around a current carrying conductor is called magnetic effect of electric current.

### (c) Chemical Effects of Electric Current

The phenomenon of decomposition of an electrolyte into ions on passing an electric current through it is called chemical effect of electric current.

## Types of Electric Current

### (a) Direct Current

The current whose magnitude and direction does not vary with time is called direct current (dc).
The various sources are cells, battery, dc dynamo etc.

### (b) Alternating Current

The current whose magnitude continuously changes with time and periodically changes its direction is called alternating current.
It has constant amplitude and has alternate positive and negative halves. It is produced by ac dynamo.

Gauss’s Law : Introduction | Electric Flux | Applications | Examples

Electric lines of Force are used to measure Electric Flux. It is a property of Electric Field which tells us the number of field lines crossing a particular area. It is rate of flow of electric field through a surface which can be open or closed. We will also discuss Gauss’s Law in detail which is an application of Electric Flux which helps to calculate Electric Field for a given charge distribution enclosed by a closed surface. We will study its application in detail here.

Here you will get to know in detail about:

## Introduction to Electric Flux

Electric flux through an elementary area, ds is defined as the scalar product of area and field, i.e. ${\varphi _{\rm{E}}} = \int {\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to }$ .

It represents the total lines of force passing through the given area. Here area is treated as a vector. The direction of area vector is given by direction of normal to the surface.

## Important Points on Electric Flux

1. It is a real scalar physical quantity with unit volt × m and dimensions
${\varphi _E} = Eds = {F \over q}ds = {{ML{T^{ – 2}}} \over {AT}}{L^2} = M{L^3}{T^{ – 3}}{A^{ – 1}}$

2. It will be maximum when cos θ is max = 1, i.e., θ = 0°, i.e., electric field is normal to the area with (dΦE)max = E ds.
3. It will be minimum when cos θ is min = 0, i.e., θ = 90°, i.e. field is parallel to the area with (dΦE)min = 0.
4. For a closed body outward flux is taken as positive while inward flux is taken as negative.

## Gauss’s Law

Gauss’s law for electricity states that the electric flux across any closed surface is proportional to the net electric charge enclosed by the surface. The law implies that isolated electric charges exist and that like charges repel one another while unlike charges attract. [source]

It relates the total flux of an electric field through a closed surface to the net charge enclosed by that surface. According to it, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface,

Mathematically$\oint\limits_s {\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to } = {q \over {{\varepsilon _0}}}$

## Important Points on Gauss’s Law

1. Gauss’s law and Coulomb’s law are equivalent. If we assume Coulomb’s law, we can prove Gauss’s law and vice versa. To prove Gauss’s law from Coulomb’s law, consider a hypothetical spherical surface called Gaussian surface of radius r with point charge q at its center. By Coulomb’s law intensity at a point P on the surface will be $\mathop E\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^3}}}\mathop r\limits^ \to$ electric flux linked with area $\mathop {ds}\limits^ \to$ $\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^3}}}\mathop r\limits^ \to \bullet \mathop {ds}\limits^ \to$ direction of $\mathop r\limits^ \to$ and $\mathop {ds}\limits^ \to$ are same i.e., $\mathop r\limits^ \to \bullet \mathop {ds}\limits^ \to$ = r ds cos 0° = r ds. So,$\mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}ds$
or $\oint\limits_s {} \mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = \oint\limits_s {} {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}ds$

for all points on the sphere r = constant $\oint {} \mathop E\limits^ \to \bullet \mathop {ds}\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}\oint\limits_s {} ds = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}(4\pi {r^2})$ = ${q \over {{\varepsilon _0}}}$ [as$\oint {ds = 4\pi {r^2}}$] The results are true for any arbitrary surface provided the surface is closed.
2. It relates the total flux linked with a closed surface to the charge enclosed by the closed surface:
a) If a closed body not enclosing any charge is placed in either uniform on non-uniform electric field total flux linked with it will be zero.

b) If a closed body encloses a charge q, total flux linked with the body is independent of the shape and size of the body and position of charge inside it.

## Applications of Gauss’s Law

### Electric Field due to a line charge:

Gauss law is useful in calculating electric field intensity due to symmetrical charge distributions.

We consider a gaussian surface which is a cylinder of radius r which encloses a line charge of length h with line charge density λ.

According to Gauss law $\oint {\mathop E\limits^ \to .\mathop {ds}\limits^ \to }$=${{{q_{in}}} \over {{\varepsilon _0}}}$

\begin{align}& \int{\overset{\to }{\mathop{E}}\,.\overset{\to }{\mathop{ds}}\,} \\ & Cylindrical \\ & surface \\ \end{align}+\begin{align}& \int{\overset{\to }{\mathop{E}}\,.\overset{\to }{\mathop{ds}}\,} \\ & \,circular \\ & surface \\\end{align}+\begin{align}& \int{\overset{\to }{\mathop{E}}\,.\overset{\to }{\mathop{ds}}\,} \\ & \,circular \\ & surface \\\end{align}=$\frac{\lambda h}{{{\varepsilon }_{0}}}$

\begin{align}&\int{Eds\,\,\cos \,\,0{}^\text{o}} \\& Cylindrical \\& suface \\\end{align}+\begin{align}& \int{Eds\,\,\cos } \\& I\,circular \\& surface \\\end{align}\frac{\pi }{2}+\begin{align}& \int{Eds\,\,\cos } \\& II\,circular \\& surface \\\end{align}\frac{\pi }{2}=$\frac{\lambda h}{{{\varepsilon }_{0}}}$

E(2 πr h) =$\frac{\lambda h}{{{\varepsilon }_{0}}}$

So,   E =  $\frac{\lambda }{2\ \pi \ {{\varepsilon }_{0}}\ r}$$(E\propto \frac{1}{r}) ### Electric Field due to an infinite plane thin sheet of charge: To find electric field due to the plane sheet of charge at any point P distant r from it, choose a cylinder of area of cross-section A through the point P as the Gaussian surface. The flux due to the electric field of the plane sheet of charge passes only through the two circular caps of the cylinder. Let surface charge density = σ According to gauss law \oint {\mathop E\limits^ \to .\mathop {dS}\limits^ \to = {q_{in}}/{\varepsilon _0}} \int\limits_{\begin{smallmatrix}I\,circular \\surface\end{smallmatrix}}{E\,\,ds\cos \theta+}\int\limits_{\begin{smallmatrix}II\,\,circular \\surface\end{smallmatrix}}{E\,\,ds\cos \theta+}\int\limits_{\begin{smallmatrix}cylindrical \\surface\end{smallmatrix}}{E\,\,ds\cos \theta =}\frac{\sigma A}{{{\varepsilon }_{0}}} or EA + EA + 0 = {{\sigma A} \over {2{\varepsilon _0}}} or E ={\sigma \over {2{\varepsilon _0}}} #### Points to remember 1. The magnitude of the electric field due to the infinite plane sheet of charge is independent of the distance from the sheet. 2. If the sheet is positively charged, the direction of the field is normal to the sheet and directed outward on both sides. But for a negatively charged sheet, the field is directed normally inwards on both sides of the sheet. ### Electric Field Intensity due to uniformly charged spherical shell: We consider a thin shell of radius R carrying a charge Q on its surface 1. At a point P0 outside the shell (r > R) According to gauss law \oint\limits_{{S_1}} {\mathop {{E_0}}\limits^ \to .\mathop {ds}\limits^ \to } = {Q \over {{\varepsilon _0}}} or E0 (4πr2) ={Q \over {{\varepsilon _0}}} E0 = {Q \over {4\;\pi \;{\varepsilon _0}{r^2}}} = {\sigma \over {{\varepsilon _0}}}$${{{R^2}} \over {{r^2}}}$
where the surface charge density σ = ${{total\,\,ch\arg e} \over {surface\,\,area}}$ = ${Q \over {4\pi {R^2}}}$
The electric field at any point outside the shell is same as if the entire charge is concentrated at center of shell.

2. At a point Ps on surface of shell (r = R)
ES = ${Q \over {4\pi {\varepsilon _0}{R^2}}}$ = ${\sigma \over {{\varepsilon _0}}}$
3. At a point Pin inside the shell (r < R)

According to gauss law $\oint\limits_{{S_2}} {\mathop E\limits^ \to .\mathop {ds}\limits^ \to }$ = ${{{q_{in}}} \over {{\varepsilon _0}}}$. As enclosed charge qin = 0.
So, Ein = 0.
The electric field inside the spherical shell is always zero.

### Electric field intensity due to a spherical uniformly charge distribution:

We consider a spherical uniformly charge distribution of radius R in which total charge Q is uniformly distributed throughout the volume.
The charge density ρ= ${{total\,\,ch\arg e} \over {total\,\,volume}}$ = ${Q \over {{4 \over 3}\pi {R^3}}}$ = ${{3Q} \over {4\pi {R^3}}}$

1. At a point P0 outside the sphere (r > R)
According to gauss law $\oint {{{\mathop E\limits^ \to }_0}.\mathop {ds}\limits^ \to }$ = ${Q \over {{\varepsilon _0}}}$ or E0 (4πr2) = ${Q \over {{\varepsilon _0}}}$
or E0 = ${Q \over {4\;\pi \;{\varepsilon _0}{r^2}}}$ = ${\rho \over {3\;{\varepsilon _0}}}$$\left( {{{{R^3}} \over {{r^2}}}} \right) 2. At a point Ps on surface of sphere (r = R) Es ={Q \over {4\pi \;{\varepsilon _0}{R^2}}} = {\rho \over {3{\varepsilon _0}}} R 3. At a point Pin inside the sphere (r < R) According to Gauss law, \oint {{{\mathop E\limits^ \to }_{in}}.\mathop {ds}\limits^ \to } ={{{q_{in}}} \over {{\varepsilon _0}}}={1 \over {{\varepsilon _0}}} ρ. {4 \over 3}πr3 = {{Q{r^3}} \over {{\varepsilon _0}{R^3}}} Ein(4πr2) ={{Q{r^3}} \over {{\varepsilon _0}{R^3}}} or Ein = {{Qr} \over {4\pi \;{\varepsilon _0}{R^3}}} = {\rho \over {3{\varepsilon _0}}} r (Ein ∝ r) ## Example based on Gauss’s Law Q. It has been experimentally observed that the electric field in a large region of earth’s atmosphere is directed vertically down. At an altitude of 300 m, the electric field is 60 Vm–1. At an altitude of 200m, the field is 100 Vm–1. Calculate the net amount of charge contained in the cube of 100 m edge, located between 200m and 300m altitude. Sol. According to Gauss’s theorem, electric flux: {\varphi _E} = {q \over {{\varepsilon _0}}} = \int {\mathop E\limits^ \to .\mathop {dS}\limits^ \to } The surface integral in the above equations contains six terms – the surface integral over the bottom surface, the surface integral over the top surface and surface integral over the four vertical faces. For the bottom surface, both the vectors and are in the same direction. For the top surface, they act in opposite directions while for the vertical faces, they are perpendicular to each other. Hence,{\varphi _E} = \int\limits_{bottom} {\mathop {{E_1}}\limits^ \to .\mathop {dS}\limits^ \to } + \int\limits_{top} {\mathop {{E_2}}\limits^ \to .\mathop {dS}\limits^ \to } + 4\int\limits_{vertical} {\mathop E\limits^ \to .\mathop {dS}\limits^ \to } = \int\limits_{{\rm{bottom}}} {{{\rm{E}}_{\rm{1}}}{\rm{ dS cos 0}}^\circ } + \int\limits_{{\rm{top}}} {{{\rm{E}}_{\rm{2}}}{\rm{ dS cos 180}}^\circ } + {\rm{4}}\int\limits_{{\rm{vertical}}} {{\rm{E dS cos 90}}^\circ } = \int\limits_{{\rm{bottom}}} {{{\rm{E}}_{\rm{1}}}{\rm{ dS}}} – \int\limits_{{\rm{top}}} {{{\rm{E}}_{\rm{2}}}{\rm{ dS }}} = {E_1}S – {E_2}S = \left( {{E_1} – {E_2}} \right)S Also, ΦE = q/ε0 ∴q = ε0(E1 – E2)S = (8.85 × 10–12 C2 N–1 m–2) (100 V/m – 60 V/m) (100 m2) = 3.54 × 10–6 C Q. The charge density of a uniformly charged metallic plate is σ = – 2μC/m2. Find the distance from which a 50eV electron must be projected so that it does not strike the plate. Sol. The electric field due to metallic sheet \mathop E\limits^ \to ={\sigma \over {{\varepsilon _0}}}$$\hat n$= K(4)$\hat n$

where $\hat n$ is unit vector perpendicular to plate

$\mathop E\limits^ \to$ = 9 × 109 × 4 × π(–2 × 10–6) $\hat n$ = 72π× 103 (–$\hat n$) N/C

Force on electron due to this field $\mathop F\limits^ \to$ = $\mathop {qE}\limits^ \to$ = 72π× 10+3 (– 1.6 × 10–19) (–$\hat n$)

= 72× 1.6 × π× 10–16$\hat n$

Let electron be left from a distance d then F.d = energy of electron

or 72 × 1.6 × π× 10–16 × d = 50 × 1.6 × 10–19 or d = 0.221 × 10–3 m

Q. If a point charge q is placed at the center of a cube what is the flux linked

(a) with the cube and

(b) with each face of the cube?

Sol. (a) According to Gauss’s law flux linked with a closed body is 1/ε0 times the charge enclosed and here the closed body cube is enclosing a charge q so, ${\varphi _T} = {q \over {{\varepsilon _0}}}$ .

(b) Now as cube is a symmetrical body with 6-faces and the point charge is at its center, so electric flux linked with each face will be${\varphi _F} = {1 \over 6}\left( {{\varphi _T}} \right) = {q \over {6{\varepsilon _0}}}$ .

Q. If a point charge q is placed at one corner of a cube, what is the flux linked with the cube?

Sol. In this case by placing three cubes at three sides of given cube and four cubes above, the charge will be in the center, So, the flux linked with each cube will be one-eighth of the flux q/ε0.

Flux associated with given cube =${q \over {8{\varepsilon _0}}}$ .

Q. A point charge q is placed at one corner of a cube of edge a. What is the flux through each face of the cube?

Sol. At a corner, 8 cubes can be placed symmetrically, flux linked with each cube due to a charge q at the corner will be q/8ε0

For the faces passing through the edge A, electric field E will be parallel to area of face and so flux through these three faces will be zero.

As the cube has six faces and flux linked with three faces (through A) is zero, so flux linked with remaining three face will be (q/8ε0). The remaining three faces are symmetrical so flux linked with each of the three faces passing through B will be,$= {1 \over 3} \times \left[ {{1 \over 8}\left( {{q \over {{\varepsilon _0}}}} \right)} \right] = {1 \over {24}}{q \over {{\varepsilon _0}}}$

Electric lines of Force – Important points | Solved Examples

If we bring an electric charge into an electric field, how many places can we keep it? Infinitely many points, Right. As we know, it will experience a force which is different at different points. The problem is how can we represent the strength of the force at any point without calculating it mathematically, Michael Faraday came up with concept of imaginary lines, which are called as Electric lines of Force. They are also known as Electric Field Lines.

Here you will get to know in detail about:

## Introduction

The idea of lines of force was introduced by Michael Faraday. A line of force is an imaginary curve the tangent to which at a point gives the direction of intensity at that point and the number of lines of force per unit area normal to the surface surrounding that point gives the magnitude of intensity at that point.

The electric lines of force that represent the field of a positive electric charge in space consist of a family of straight lines radiating uniformly in all directions from the charge where they originate. A second positive charge placed in the field would travel radially away from the first charge. [source]

## Important Points on Electric lines of Force

1. Electric lines of force usually start or diverge out from positive charge and end or converge on negative charge.

2. The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In SI units 1/ε0 shows electric lines associated with unit (i.e., 1 coulomb) charge. So, if a body encloses a charge q, total lines of force or flux associated with it is q/ε0. If the body is cubical and charge is situated at its center the lines of force through each face will be q/6ε0.

3. Lines of force never cross each other because if they cross then intensity at that point will have two directions which is not possible.
4. In electrostatics, the electric lines of force can never be closed loops, as a line can never start and end on the same charge. If a line of force is a closed curve, work done round a closed path will not be zero and electric field will not remain conservative.
5. Lines of force have tendency to contract longitudinally like a stretched elastic string producing attraction between opposite charges and repel each other laterally resulting in, repulsion between similar charges and ‘edge-effect‘ (curving of lines of force near the edges of a charged conductor).

6. If the lines of force are equidistant straight lines the field is uniform and if lines of force are not equidistant or straight lines or both, the field will be non-uniform. The first three represent non-uniform field while last shows uniform field.

7. Electric lines of force end or start normally on the surface of a conductor. If a line of force is not normal to the surface of a conductor, electric intensity will have a component along the surface of the conductor and hence conductor will not remain equipotential which is not possible as in electrostatics conductor is an equipotential surface.

8. If in a region of space, there is no electric field there will be no lines of force. This is why inside a conductor or at a neutral point where resultant intensity is zero there is no line of force.
9. The number of lines of force per unit normal area at a point represents magnitude of electric field intensity. The crowded lines represent strong field while distant lines show a weak field.
10. The tangent to the line of force at a point in an electric field gives the direction of intensity. It gives direction of force and hence acceleration which a positive charge will experience there (and not the direction of motion). A positive point charge free to move may or may not follow the line of force. It will follow the line of force if it is a straight line (as direction of velocity and acceleration will be same) and will not follow the line if it is curved as the direction of motion will be different from that of acceleration. The particle will not move in the direction of motion or acceleration (line of force) but other than these which will vary with time as $\mathop v\limits^ \to = \mathop u\limits^ \to + \mathop {at}\limits^ \to$

## Comparison of Electric and Magnetic lines of Force

1. Electric lines of force never form closed loops while magnetic lines of force are always closed loops.
2. They always emerge or terminate normally on the surface of a charged conductor, while magnetic lines of force start or terminate on the surface of a magnetic material at any angle.
3. Electric lines of force do not exist inside a conductor but magnetic lines of force may exist inside magnetic material.
4. Total electric lines of force linked with a closed surface may or may not be zero but total magnetic lines of force linked with a closed surface is always zero (as mono-poles do not exist).

## Solved Examples

Q. A solid metallic sphere is placed in a uniform electric field. Which of the lines A, B, C and D shows the correct path and why?

Sol. Path (A) is wrong as lines of force do not start or end normally on the surface of a conductor.

Path (B) and (C) are wrong as lines of force should not exist inside a conductor. Also, lines of force are not normal to the surface of conductor.

Path (D) represents the correct situation as here line of force does not exist inside the conductor and starts and ends normally on its surface.

Q. A metallic slab is introduced between the two plates of a charged parallel plate capacitor. Sketch the electric lines of force between the plates.

Sol. Keeping in mind the following properties:

1. Lines of force start from positive charge and end on negative charge.
2. Lines of force start and end normally on the surface of a conductor.
3. Lines of force do not exist inside a conductor (as field inside a conductor is zero) The field between the plates is shown as

What is Electric Potential & Potential difference?

As we know that electric charge forms an electric field around itself. But what if some other charge tries to come into that electric field? Is it that easy to enter into someone’s electric field? The answer is No. As the charge trying to enter also has an electric field. The amount of work needed to move a unit positive charge from one point to another (in the electric field of a charge) is called Electric Potential. The difference of Electric Potential between two points is called Potential difference.

Here you will get to know in detail about:

## Electric Potential & Potential Difference

An electric potential (also called the electric field potential, potential drop or the electrostatic potential) is the amount of work needed to move a unit of positive charge from a reference point to a specific point inside the field without producing an acceleration. Typically, the reference point is the Earth or a point at infinity. [source]

In an electric field, potential at a point is defined as a scalar function of position whose negative gradient at that point gives electric intensity $\mathop E\limits^ \to$

$\mathop E\limits^ \to = – \nabla \,V$

Or $\mathop E\limits^ \to = – grad\,V\,\,or\,\,\mathop E\limits^ \to = – {{dV} \over {dr}}\mathop n\limits^ \to$

So, $dV = – \mathop E\limits^ \to \bullet d\mathop r\limits^ \to$

i.e. $\int_{{V_0}}^V {dV = – } \int_{{r_0}}^r {\mathop E\limits^ \to \bullet \mathop {dr}\limits^ \to }$

or$V – {V_0} = – \int_{{r_0}}^r {\mathop E\limits^ \to \bullet \mathop {dr}\limits^ \to }$

Whenever and wherever possible we take the reference point at infinity and assume the potential to be zero there, so if r0 = ∞, V0 = 0

$V = – \int_\infty ^r {\mathop E\limits^ \to \bullet d\mathop r\limits^ \to }$

or $V = – \int_\infty ^r {{{\mathop F\limits^ \to } \over {{q_0}}} \bullet d\mathop r\limits^ \to }$ $\left[ {as\,\,\mathop E\limits^ \to = {{\mathop F\limits^ \to } \over {{q_0}}}} \right]$

or $V = – {W \over {{q_0}}}$ $\left[ {as\,\,\int {\mathop F\limits^ \to \bullet \mathop {dr}\limits^ \to } = W} \right]$

So potential at a point can be physically interpreted as the negative of the work done by the field in displacing a unit positive charge from some reference point to the given point.

Furthermore, as electric field is conservative, so U = –W,

$V = {U \over {{q_0}}}$ i.e., U = q0V

i.e., potential at a point is numerically equal to the potential energy per unit charge at that point.

## Unit & Dimensions of Electric Potential

1. It is a scalar having dimensions: $V = {W \over {{q_0}}} = {{M{L^2}{T^{ – 2}}} \over {AT}} = M{L^2}{T^{ – 3}}{A^{ – 1}}$ and SI or practical unit is (J/C) called volt (V).
2. In CGS system as 1 J = 107 erg and 1 coulomb = 3 × 109 esu of charge = (1/10) emu of charge
So, $1V = 1{J \over C} = {{{{10}^7}erg} \over {3 \times {{10}^9}esu\,\,of\,\,ch\arg e}} = {1 \over {300}}esu\,\,of\,\,potential$
and, $1V = 1{J \over C} = {{{{10}^7}erg} \over {(1/10)\,\,emu\,\,of\,\,ch\arg e}} = {10^8}emu\,\,of\,\,potential$

## Important Points on Electric Potential & Potential Difference

1. As $V = – \int {\mathop E\limits^ \to \bullet \mathop {dr}\limits^ \to }$
so,$E = – {{dV} \over {dr}}$
(a) if E is given, V can be calculated and vice-versa.
(b) Negative of the slope of V versus r curve gives electric intensity E.
2. If the field is produced by a point charge then, $\mathop E\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^3}}}\mathop r\limits^ \to$ and $V = – \int_\infty ^r {\mathop E\limits^ \to \bullet \mathop {dr}\limits^ \to }$
So, $V = – \int_\infty ^r {{1 \over {4\pi {\varepsilon _0}}}} {q \over {{r^3}}}\mathop r\limits^ \to \bullet \mathop {dr}\limits^ \to = – {1 \over {4\pi {\varepsilon _0}}}\int_\infty ^r {{q \over {{r^2}}}dr}$
$V = {1 \over {4\pi {\varepsilon _0}}}{q \over r}$
3. For discrete-distribution of charges V = V1 + V2 + V3 + ……… =${1 \over {4\pi {\varepsilon _0}}}\sum {{{q_i}} \over {{r_i}}}$

4. For continuous charge distribution, treating the element as point:
$dV = {1 \over {4\pi {\varepsilon _0}}}{{dq} \over r}$ , or $V = {1 \over {4\pi {\varepsilon _0}}}\int {{{dq} \over r}}$
5. Potential in free space due to a point charge is given by ${V_0} = {1 \over {4\pi {\varepsilon _0}}}{q \over r}$
In a medium of permittivity ε, $V = {1 \over {4\pi \varepsilon }}{q \over r}$
So, ${V \over {{V_0}}} = {{{\varepsilon _0}} \over \varepsilon }$
i.e., $V = {{{V_0}} \over K}$ $\left[ {as{\varepsilon \over {{\varepsilon _0}}} = K} \right]$
i.e., by presence of a medium potential decreases and becomes (1/K) times of its free space value.
6. In case of a spherical charge distribution, for external point (r > R), the charge distribution behaves as whole of its charge is concentrated at the center, i.e.,${V_{out}} = {1 \over {4\pi {\varepsilon _0}}}{q \over r}$

7. In case of spherical charged conductor hollow or solid for an internal point (i.e., r < R), potential everywhere inside is same constant, maximum and equal to its value at the surface, i.e.,
Vin = Vcentre = Vsurface = ${1 \over {4\pi {\varepsilon _0}}}{q \over R}$

8. In case of spherical volume distribution of charge (e.g., nucleus or uniformly charged dielectric sphere) for an internal point (i.e., r < R) potential varies non-linearly with r as, ${V_{in}} = {1 \over {4\pi {\varepsilon _0}}}{{q\,[3{R^2} – {r^2}]} \over {2{R^3}}}$

V will be maximum at the center, at r = 0 with ${V_C} = {3 \over 2}\left( {{q \over {4\;\pi \;{\varepsilon _0}\;R}}} \right)$ while at the surface at r = R, VS =${q \over {4\;\pi \;{\varepsilon _0}\;R}}$
so that, ${V_C} = {3 \over 2}{V_S}$

9. The electric field is conservative, so work done and hence potential difference between two points is path independent and depends only on the position of points between which the charge is moved. If a charged particle is moved in an electric field between two fixed points A and B whatever be the path, potential difference or work done will be same.

## Equipotential Surfaces

For a given charge distribution, locus of all points having same potential is called ‘equipotential surface‘.

#### Important Points about Equipotential Surfaces:

1. Equipotential surfaces can never cross each other otherwise potential at a point will have two values which is not possible.
2. Equipotential surfaces are always perpendicular to lines of force.
3. If a charge is moved from one point to the other over an equipotential surface, work done will be zero as WAB = – UAB
= q(VB – VA) = 0 [as VB = VA]
4. The different shapes of equipotential surfaces (EPS) are shown below

## Examples based on Electric Potential & Potential Difference

Q. Infinite number of same charges q are placed at x = 1, 2, 4, 8…… What is the potential at x = 0?

Sol. $V = {1 \over {4\pi {\varepsilon _0}}}\left( {{q \over 1} + {q \over 2} + {q \over 4} + {q \over 8} + ….} \right)$

$= {q \over {4\pi {\varepsilon _0}}}{1 \over {\left( {1 – {1 \over 2}} \right)}} = {{2q} \over {4\pi {\varepsilon _0}}} = {q \over {2\pi {\varepsilon _0}}}$ [∴ a + ar + …. ∞= r < 1 of r <1]

Q. If the alternative charges are unlike, then what will be the potential?

Sol. Then,$V = {1 \over {4\pi {\varepsilon _0}}}\left[ {{q \over 1} – {q \over 2} + {q \over 4} – {q \over 8} + ……\infty } \right]$

$= {1 \over {4\pi {\varepsilon _0}}}\left[ {\left( {{q \over 1} + {q \over 4} + …..\infty } \right) – \left( {{q \over 2} + {q \over 8} + …..\infty } \right)} \right]$

$= {1 \over {4\pi {\varepsilon _0}}}\left[ {{1 \over {1 – {1 \over 4}}} – {1 \over 2}\left( {{1 \over {1 – {1 \over 4}}}} \right)} \right] = {1 \over {4\pi {\varepsilon _0}}}{{2q} \over 3}$ [∴ a + ar + ar² ……. ∞ = ${a \over {1 – r}}$ r < 1]

Q. Electric potential for a point (x, y, z) is given by V = 4x² volt. Electric field at point (1, 0, 2) is-

Sol. $E = – {{dV} \over {dx}}$ = –8x E at (1, 0, 2) = –8V/m

∴ Magnitude of E = 8 V/m and direction is along x-axis.

Q. Electric field is given by $E = {{100} \over {{x^2}}}$. Find potential difference between x = 10 and x = 20 m.

Sol. $E = – {{dV} \over {dx}}$

or dV = – Edx

or $\int_A^B {\;dV} = – \int_A^B {\;E.dx}$

or ${V_B} – {V_A} = – \int_{10}^{20} {} {{100} \over {{x^2}}}dx = – 5$ volt

Potential difference = 5 volt.

Q. Two circular loops of radius 0.05 m and 0.09 m respectively are put such that their axes coincide and their centers are 0.12 m apart. Charge of 10–6 coulomb is spread uniformly on each loop. Find the potential difference between the centers of loops.

Sol.  The potential at the center of a ring will be due to charge on both the rings and as every element of a ring is at a constant distance from the center,

so, ${V_1} = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{{q_1}} \over {{R_1}}} + {{{q_2}} \over {\sqrt {R_2^2 + {x^2}} }}} \right]$

${V_1} = 9 \times {10^9}\left[ {{{{{10}^{ – 4}}} \over 5} + {{{{10}^{ – 4}}} \over {\sqrt {{9^2} + {{12}^2}} }}} \right]$

${V_1} = 9 \times {10^5}\left[ {{1 \over 5} + {1 \over {15}}} \right] = 2.40 \times {10^5}V$

Similarly, ${V_2} = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{{q_2}} \over {{R_2}}} + {{{q_1}} \over {\sqrt {R_1^2 + {x^2}} }}} \right]$

or, ${V_2} = 9 \times {10^5}\left[ {{1 \over 9} + {1 \over {13}}} \right] = {{198} \over {117}} \times {10^5}$ = 1.69 × 105 V

So, V1 – V2 = (2.40 – 1.69) × 105 = 71 kV

Q. A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such that the surface densities are equal. Find the potential at the common center.

Sol. If q1 and q2 are the charges on spheres of radii r and R respectively, then in accordance with ‘conservation of charge’ q1 + q2 = Q

According to given problem σ1 = σ2

Or ${{{q_1}} \over {4\pi {r^2}}} = {{{q_2}} \over {4\pi {R^2}}}$

or ${{{q_1}} \over {{q_2}}} = {{{r^2}} \over {{R^2}}}$

So ${q_1} = {{Q{r^2}} \over {({r^2} + {R^2})}}$ and ${q_2} = {{Q{R^2}} \over {({r^2} + {R^2})}}$

Now as potential inside a conducting sphere is equal to its surface, so potential at the common center:$V = {V_1} + {V_2} = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{{q_1}} \over r} + {{{q_2}} \over R}} \right]$

$V = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{Qr} \over {({R^2} + {r^2})}} + {{QR} \over {({R^2} + {r^2})}}} \right] = {1 \over {4\pi {\varepsilon _0}}}{{Q(R + r)} \over {({R^2} + {r^2})}}$

## Potential Energy of a System

1. The electric potential energy of a system of charges is the work done in bringing these charges from infinity near each other to form the system.
2. If a system is given negative of its potential energy, then all charges will move to infinity. This negative value of total energy is called the Binding energy.
3. Energy of a system of two charges $PE = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over d}$
4. Energy of a system of three charges PE = ${1 \over {4\pi {\varepsilon _0}}}\left[ {{{{q_1}{q_2}} \over {{r_{12}}}} + {{{q_2}{q_3}} \over {{r_{23}}}} + {{{q_3}{q_1}} \over {{r_{31}}}}} \right]$
5. Energy of a system of n charges, PE = $\frac{1}{2}.\frac{1}{4\pi {{\varepsilon }_{0}}}\sum\limits_{i=1}^{n}{{}}\sum\limits_{\begin{smallmatrix}j=1 \\i\,\ne\end{smallmatrix}}^{n}{\,\,\frac{{{q}_{i\,\,}}{{q}_{j}}}{{{r}_{ij}}}}$

## Work Done in an Electric Field

1. If electric potential at a point is V then potential energy of a charge placed at that point will be qV.
2. Work done in moving a charge from A to B is equal to change in PE of that charge
WAB = work done from A to B = PEB – PEA = q (VB – VA)
3. Work done in moving a charge along a closed path in an electric field is zero.
4. Total energy remains constant in an electric field i.e. KEA + PEA = KEB + PEB
KE = Kinetic Energy
PE = Potential Energy
5. A free charge moves from higher PE to lower PE state in an electric field.
(i) A positive charge moves from higher potential to lower potential
(ii) a negative charge moves from lower potential to higher potential
(iii) Work done in displacement of a charge through $\mathop r\limits^ \to$ by a force $\mathop F\limits^ \to \,\,\,is\,\,W = \mathop F\limits^ \to .\mathop r\limits^ \to$

## Motion of a charged particle in an Electric Field

In case of motion of a charged particle in an electric field:

1. A point charge experiences a force whether it is at rest or in motion $\mathop F\limits^ \to = q\mathop E\limits^ \to$.
2. The direction of force is parallel to the field if the charge is positive and opposite to the field if charge is negative.
3. Electric field is conservative so work done is independent of path and work done in moving a point charge q between two fixed points having a potential difference V is equal to,
WAB = – UAB = q(VB – VA) = qV
4. Work is done in moving a charged particle in an electric field unless the points are at same potential

5. When a charged particle is accelerated by a uniform or non-uniform electric field then by work energy theorem ΔKE = W,
So, ${1 \over 2}m{v^2} – {1 \over 2}m{u^2} = qV$ or final velocity $v = \sqrt {\left[ {{u^2} + {{2qV} \over m}} \right]}$
If the charged particle is initially at rest, then $v = \sqrt {{{2qV} \over m}}$ . If the field is uniform,
E = (V/d)
so $v = \sqrt {{{2qEd} \over m}}$
6. In motion of a charged particle in a uniform electric field if force of gravity does not exist or is balanced by some other force say reaction or neglected then $\mathop a\limits^ \to = {{\mathop F\limits^ \to } \over m} = {{\mathop {qE}\limits^ \to } \over m} = cons\tan t$ [as $\mathop F\limits^ \to = \mathop {qE}\limits^ \to$]
Here equations of motion are valid.

(i) If the particle is initially at rest then from v = u + at, we get v = at =${{qE} \over m}t$

And from Eqn. $s = ut + {1 \over 2}a{t^2}$

we get $s = {1 \over 2}a{t^2} = {1 \over 2}{{qE} \over m}{t^2}$

The motion is accelerated translatory with a $\propto$ t°; v∝ t and s∝ t²

Here $W = \Delta KE = {1 \over 2}m{v^2} = {1 \over 2}m\;{\left[ {{{qE} \over m}t} \right]^2}\left[ {as\,\,from\,\,Eqn.\left( 6 \right)v = {{qEt} \over m}} \right]$

also, W = qEd = qV

(ii) If the particle is projected perpendicular to the field with an initial velocity v0

From Eqn. v = u + at and s = ut + ${1 \over 2}$at² ux = v0 and ax = 0,

ux = v0 = const. and x = v0t

for motion along y-axis as uy = 0 and ay =${{qE} \over m}$,

${v_y} = \left[ {{{qE} \over m}} \right]\;t$ and $y = {1 \over 2}\left[ {{{qE} \over m}} \right]\,\;{t^2}$

So, eliminating t between equation for x and y, we have $y = {{qE} \over {2m}}{\left[ {{x \over {{v_0}}}} \right]^2} = {{qE} \over {2mv_0^2}}{x^2}$

If particle is projected perpendicular to field the path is a parabola.

## Example based on motion of a charged particle in an Electric Field

Q. An oil drop ‘B’ has charge 1.6 × 10–19 C and mass 1.6 × 10–14 kg. If the drop is in equilibrium position, then what will be the potential diff. between the plates. [The distance between the plates is 10mm]

Sol. For equilibrium, electric force = weight of drop

⇒ qE = mg or $q.{V \over d} = mg$

⇒ $V = {{mgd} \over q}$ = 1.6 × 10–14 × 9.8 × 10 × 10–3 / 1.6 × 10–19 = 104 V

Electric Field Definition | Electric Field Intensity | Examples

As we have studied Coulomb’s Law, we know that charges either attract or repel each other even when they are a distance apart. Have you ever wondered why they experience such a force? It is somewhat like a Tiger. Every charge forms a territory around itself. If some other charge or charges comes in its territory then they experience a force according to their nature. This space a charge creates around itself is called as Electric field.

So, here we are going to know more about Electric Field, its definition and also about electric field intensity in the following order:

## Electric Field Definition

To explain ‘action at a distance‘, i.e. ‘force without contact‘ between charges we assume that a charge or charge distribution produces a field in space surrounding it. The region surrounding a charge or charge distribution in which its electrical effects are perceptible is called the electric field of the given charge.

Electric field at a point is characterized either by a vector function of position $\mathop E\limits^ \to$ called Electric Field Intensity or by a scalar function of position V called Electric Potential. The electric field in a certain space is also visualized graphically in terms of lines of force. So electric intensity, potential and lines of force are different ways of describing the same field.

## Electric Field Intensity $\mathop E\limits^ \to$

The Electric Field Intensity at a point in an Electric Field is defined as the force experienced by a unit positive point charge called test charge supposed to be placed at that point. The test charge does not affect the source charge or charge distribution producing the field. If a test charge q0 at a point P in an electric field experiences a force$\mathop F\limits^ \to$, then electric field$\mathop E\limits^ \to = (\mathop F\limits^ \to /{q_0})$

If the electric field is produced by a point charge q, then from Coulomb’s law $\mathop F\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{{q{q_0}} \over {{r^3}}}\mathop r\limits^ \to$ field due to point-charge q at position $\mathop r\limits^ \to$ in free space.

$\mathop E\limits^ \to = {{\mathop F\limits^ \to } \over {{q_0}}} = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^3}}}\mathop r\limits^ \to \,\,\,\,or\,\,\,\,E = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$

If electric field is produced by a charge distribution, then by ‘principle of superposition‘ field is given as

$\mathop E\limits^ \to = \mathop {{E_1}}\limits^ \to + \mathop {{E_2}}\limits^ \to + ……. = \sum\limits_{i = 1}^n {\,\mathop {{E_i}}\limits^ \to }$ with $\mathop {{E_i}}\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{{{q_i}} \over {r_i^3}}\mathop {{r_i}}\limits^ \to$

while for continuous charge distribution (treating small charge element as a point charge),

$d\mathop E\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}{{dq} \over {r_{}^3}}\mathop r\limits^ \to ,$ i.e., $\mathop E\limits^ \to = {1 \over {4\pi {\varepsilon _0}}}\int {{{dq} \over {{r^3}}}} \mathop r\limits^ \to$

## Important Points on Electric Field

1. It is a vector quantity having dimensions
$E = {F \over q} = {{ML{T^{ – 2}}} \over {AT}} = ML{T^{ – 3}}{A^{ – 1}}$
The SI unit is N/C or V/m as ${N \over C} = {{N \times m} \over {C \times m}}\buildrel {\left( {N \times m = J} \right)} \over \longrightarrow {J \over {C \times m}}\buildrel {\left( {J/C = V} \right)} \over\longrightarrow {V \over m}$
2. By definition $\mathop E\limits^ \to = \mathop F\limits^ \to /{q_0},$ or $\mathop F\limits^ \to = {q_0}\mathop E\limits^ \to$
A charged particle in an electric field experiences a force whether it is at rest or in motion. The direction of force is along the field if it is positive and opposite to the field if it is negative.
3. In free space Electric field is ${E_0} = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$
In a medium of permittivity ε field is $E = {1 \over {4\pi \varepsilon }}{q \over {{r^2}}}$
So, ${E \over {{E_0}}} = {{{\varepsilon _0}} \over \varepsilon } = {1 \over K}$ [as ε = ε0 K]
Or, E = E0/K
4. In presence of a dielectric, electric field decreases and becomes 1/K times of its value in free space.The electrostatic field i.e. field produced by charge at rest is conservative.
$F = – {{dU} \over {dr}}$        or         ${F \over {{q_0}}} = – {1 \over {{q_0}}}{{dU} \over {dr}}$
or  $E = – {{dV} \over {dr}}$  $\left[ {as{F \over {{q_0}}} = E\,\,and\,\,{U \over {{q_0}}} = V} \right]$
So, ${E_x} = – {{\partial V} \over {\partial x}}$ ${E_y} = – {{\partial V} \over {\partial y}}$ and ${E_z} = – {{\partial V} \over {\partial z}}$ with $\mathop E\limits^ \to = \hat i{E_x} + \hat j{E_y} + \hat k{E_z}$ i.e. if potential V is given $\mathop E\limits^ \to$ can be calculated.
5. The electric field between two parallel plates separated by a distance d and having a potential difference V, will therefore be given by:$E = – {{dV} \over {dr}} = – {{\left( {{V_2} – {V_1}} \right)} \over {\left( {{r_2} – {r_1}} \right)}} = – {{\left( {0 – V} \right)} \over {\left( {d – 0} \right)}} = {V \over d}$

6. The value of the electric field has dimensions of force per unit charge. In the meter-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per meter. In the centimeter-gram-second system, the electric field is expressed in units of dynes per electrostatic unit (esu), equivalent to stat-volts per centimeter. [source]

## Solved Examples

Q. A thin spherical conducting shell of radius ‘a’ carries a charge q. Concentric with it is another thin metallic spherical shell of radius b > a. Calculate electric field at radial distance r when

(a) r < a

(b) a < r < b and

(c) r > b.

What will be the change in the field in above cases if the outer shell

(i) is given a charge Q

(ii) is earthed?

Sol.  We know that intensity inside a charged conductor is zero while for external point a charged sphere behaves as if whole of its charge were concentrated at the center, i.e.,

Ein = 0 and ${E_{out}} = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$

(a) If r < a, the point A is internal to both the shells, so EA = 0 + 0 = 0

(b) If a < r < b, the point B is external to shell of radius a and internal to shell of radius b, so

${E_B} = {1 \over {4\pi {\varepsilon _0}}}{q \over {r_B^2}} + 0 = {1 \over {4\pi {\varepsilon _0}}}{q \over {r_B^2}}$

(c) If r > b, the point C is external to both the shells and as due to induction –q and +q charges will be induced on the inner and outer surface of spherical shell of radius b,

${E_C} = {1 \over {4\pi {\varepsilon _0}}}{{\left[ {q – q + q} \right]} \over {r_C^2}} = {1 \over {4\pi {\varepsilon _0}}}{q \over {r_C^2}}$ [< EB as rC > rB]

This all is shown in fig. (A)

Now if the outer sphere is given a charge Q or is earthed, the result of case (a) and (b) will not change as external charge on a conductor does not affect the field inside it. However, in case (c), i.e., r > b.

(i) If the outer sphere is given a charge Q, the charge Q will also produce electric field so that now:

$E_{C}^{‘}={{E}_{C}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{r_{C}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(q+Q)}{r_{C}^{2}}$   [> EC]

This is shown in fig. (B).

(ii) If the outer sphere is earthed, the outer sphere will acquire induced charge say q’ such that its potential becomes zero as potential of earthed conductor is assumed to be zero then

$V = {1 \over {4\pi {\varepsilon _0}}}{{\left( {q + q’} \right)} \over b} = 0$ i.e.,       q’ = –q

i.e., the induced charge on the outer shell will now be (–q) [and not zero] so now

$E_{C}^{‘}={{E}_{C}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-q}{r_{C}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( q-q \right)}{r_{C}^{2}}=0$   [< EC]

i.e., field outside the shell is zero. In this situation field exists only in the space between the shells as shown in fig. (C).

Q. A conducting hollow spherical shell having an inner radius ‘a’ and outer radius ‘b’ carries a net charge Q. If a point charge q is placed at the center of this sphere, determine the surface charge density at

(a) the inner surface and

(b) the outer surface.

Sol. If qa and qb are the charges on the inner and outer surfaces of the given shell

${\sigma _a} = {{{q_a}} \over {4\pi {a^2}}}$ and ${\sigma _b} = {{{q_b}} \over {4\pi {b^2}}}$       —1

The isolated sphere is given a charge Q, so in accordance with conservation of charge

qa + qb = Q      —2

Intensity at a point inside a conductor is zero and external charge does not produce field inside a conductor, so for point P (a < r < b), we have

${E_P} = {1 \over {4\pi {\varepsilon _0}}}{{\left( {q + {q_a}} \right)} \over {{r^2}}} = 0$, i.e., qa = –q

So that Eqn. (2) yields qb = Q – qa = Q – (–q) = Q + q

And hence in the light of eqn. (1)

(a) ${\sigma _a} = – {q \over {4\pi {a^2}}}$

and

(b) ${\sigma _b} = {{Q + q} \over {4\pi {b^2}}}$

Q. A block having mass m and charge q is resting on a friction-less plane at a distance L from the wall as shown in fig. Discuss the motion of the block when a uniform electric field E is applied horizontally towards the wall assuming that collision of the block with the wall is perfectly elastic.

Sol. Electric force $\mathop F\limits^ \to = q\mathop E\limits^ \to$ will accelerate the block towards the wall producing acceleration $a = {F \over m} = {{qE} \over m}$.

As initially the block is at rest and acceleration is constant, so time taken by the block to reach the wall $L = {1 \over 2}a{t^2}$ i.e. $t = \sqrt {{{2L} \over a}} = \sqrt {{{2mL} \over {qE}}}$,

As collision with the wall is perfectly elastic, the block will rebound with same speed and will move in opposite direction. It will come to rest after travelling same distance L in same time t. After stopping it will be again accelerated towards the wall and so the block will execute oscillatory motion with ‘span’ L and time period $T = 2t = 2\sqrt {{{2mL} \over {qE}}}$

The restoring force F = qE is constant and is not proportional to displacement x, so the motion is not simple harmonic.

Electric field at a point is characterized either by a vector function of position called Electric Field Intensity or by a scalar function of position V called Electric Potential.

Vector form of Coulomb’s Law – Definition | Examples

Now we have the basic idea of what is Electric Charge. So, the next step is How two charges interact with each other? How do they interact with each other? What all parameters affect the forces they exert on each other? What is the resultant force and in what is its direction? Here we are going to know more about Vector form of Coulomb’s Law to get answers of all these questions.

So, let’s begin!!

## Coulomb’s Law Definition

The force of attraction or repulsion between two stationary point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them. This force acts along the line joining the center of two charges.

If q1 & q2 are charges, r is the distance between them and F is the force acting between them

Then, F ∝ q1q2

F ∝ 1/r²

∴ F ∝ ${{{q_1}{q_2}} \over {{r^2}}}$

Or $F = C{{{q_1}{q_2}} \over {{r^2}}}$

C is const. which depends upon system of units and also on medium between two charges.

In SI unit,

C = 1 in electrostatic unit (esu)

ε0 = 8.85 × 10–12 C²/Nm² = permittivity of free space or vacuum

## Effect of medium

The dielectric constant of a medium is the ratio of the electrostatic force between two charges separated by a given distance in air to electrostatic force between same two charges separated by same distance in that medium.

Fair =${1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}}$ and Fmedium = ${1 \over {4\pi {\varepsilon _0}{\varepsilon _r}}}{{{q_1}{q_2}} \over {{r^2}}}$

${{{F_{medium}}} \over {{F_{air}}}} = {1 \over {{\varepsilon _r}}}$= K

εr or K = Dielectric constant or Relative permittivity or specific inductive capacity of medium.

1. Permittivity: Permittivity is a measure of the ability of the medium surrounding electric charges to allow electric lines of force to pass through it. It determines the forces between the charges.
2. Relative Permittivity: The relative permittivity or the dielectric constant (εr or K) of a medium is defined as the ratio of the permittivity ε of the medium to the permittivity ε0 of free space i.e. εr or $K = {\varepsilon \over {{\varepsilon _0}}}$

Dimensions of permittivity, ${\varepsilon _0} = {{{Q^2}} \over {F\,\, \times lengt{h^2}}}$$= {{{T^2}{A^2}} \over {ML{T^{ – 2}}{L^2}}}$= M–1 L–3 T4A2

The dielectric constants of different mediums are:

 Medium Vacuum Air Water Mica Teflon Glass PVC Metal εr 1 1.00059 80 6 2 5-10 4.5 ∞

## The direction of the force acting between two charges also depends on their nature and it is along the line joining the center of two charges.

$\mathop {{F_{21}}}\limits^ \to$= force on q2 due to q1

$\mathop {{F_{21}}}\limits^ \to = {1 \over {4\pi {\varepsilon _0}{\varepsilon _r}}}\,\,\,{{{q_1}{q_2}} \over {r_{12}^2}}\,\,{\hat r_{12}}$

$\mathop {{F_{12}}}\limits^ \to$= Force on q1 due to q2

$\mathop {{F_{12}}}\limits^ \to = {1 \over {4\pi {\varepsilon _0}{\varepsilon _r}}}\,\,\,{{{q_1}{q_2}} \over {r_{21}^2}}\,\,{\hat r_{21}}$

$\mathop {{F_{12}}}\limits^ \to = – \mathop {{F_{21}}}\limits^ \to$ (∵ ${\hat r_{12}} = – {\hat r_{21}}$ )

Or ${\mathop F\limits^ \to _{12}} + {\mathop F\limits^ \to _{21}} = 0$

This is known as Vector form of Coulomb’s Law.

## Important Points on Coulomb’s Law

1. The electrostatic force is a medium dependent force.
2. The electrostatic force is an action-reaction pair, i.e., the force exerted by one charge on the other is equal and opposite to the force exerted by the other on the first.
3. The force is conservative, i.e., work done in moving a point charge around a closed path under the action of Coulomb’s force is zero.
4. Coulomb’s law is applicable to point charges only. But it can be applied for distributed charges also.
5. This law is valid only for stationary point charges and cannot be applied for moving charges.
6. The law expresses the force between two-point charges at rest. In applying it to the case of extended bodies of finite size care should be taken in assuming the whole charge of a body to be concentrated at its ‘center’ as this is true only for spherically charged body, that too for a external point.
7. The equilibrium of a charged particle under the action of Coulombian forces alone can never be stable. This statement is called Earnshaw’s theorem.
8. Unit of charge:
$F = {1 \over {4\pi {\varepsilon _0}}}\,\,{{{q_1}{q_2}} \over {{r^2}}}$
If q1 = q2 = 1 coulomb, r = 1m
then $F = {1 \over {4\pi {\varepsilon _0}}}\,\,$ = 9 × 109 N
One Coulomb of charge is that charge which when placed at rest in vacuum at a distance of one meter from an equal and similar stationary charge is repelled by it with a force of 9 × 109 Newton.

## Comparison of Electrostatic and Gravitational Forces

Two charged bodies experience electrostatic force and also a gravitational force on account of their masses.

### Similarities:

(a) Both the forces are central forces, i.e., they act along the line joining the center of two charges or masses.

(b) Both the forces obey inverse square law i.e., F ∝ (1/r²).

(c) Both are conservative forces, i.e., the work done by these is independent of the nature of path.

(d) Both the forces can operate in vacuum.

### Differences:

(a) The gravitational forces are always attractive while the electrostatic forces may be attractive or repulsive.

(b) The gravitational force is independent of medium while electrostatic force depends on nature of medium.

(c) Electrostatic forces are very large as compared to gravitational forces.

For an electron-proton system, electrostatic force of attraction

${F_e} = {1 \over {4\pi {\varepsilon _0}}}{{e.e} \over {{r^2}}} = 9 \times {10^9}.{{{{(1.6 \times {{10}^{ – 19}})}^2}} \over {{r^2}}}$N

where ‘r’ meter is the separation between the electron and proton,

Force of gravitational attraction ${F_g} = G{{{m_e}{m_p}} \over {{r^2}}} = 6.67 \times {10^{ – 11}}.{{(9.1 \times {{10}^{ – 31}}) \times (1.67 \times {{10}^{ – 27}})} \over {{r^2}}}N$

Thus,${{{F_e}} \over {{F_g}}} = 2.26 \times {10^{39}}$ i.e., electrostatic force between a proton and an electron is about 1039 times stronger than the gravitational force.

## Vector form of Coulomb’s Law Examples

Q. Three equal charges Q each are placed on the vertices of an equilateral triangle of side a. What is the resultant force on any one charge due to the other two?

Sol. The charges are shown in fig.

The resultant force $F = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos 60^\circ }$

With F1 = F2 = kQ2/a2 $F = {{\sqrt 3 k{Q^2}} \over {{a^2}}}$

From symmetry the direction is along y-axis.

Q.  Two equally charged identical metal spheres A and B repel each other with a force of 2 × 10–5 N. Another identical uncharged sphere C is touched to B and then placed at the mid-point between A and B. What is the net electric force on C?

Sol. Let initially the charge on each sphere is q and separation between their centers is r:

$F = {1 \over {4\pi {\varepsilon _0}}}{{q \times q} \over {{r^2}}} = 2 \times {10^{ – 5}}N$

When sphere C touches B, the charge of B, q will distribute equally on B and C as spheres are identical now: qB = qC = (q/2)

So, sphere C will experience a force

${F_{CA}} = {1 \over {4\pi {\varepsilon _0}}}{{q\left( {q/2} \right)} \over {{{\left( {r/2} \right)}^2}}} = 2F$ along $\mathop {AB}\limits^ \to$ due to charge on A

and, ${F_{CB}} = {1 \over {4\pi {\varepsilon _0}}}{{(q/2)\,(q/2)} \over {{{(r/2)}^2}}} = F$ along $\mathop {BA}\limits^ \to$ due to charge on B

So, the net force on C due to charges on A and B,

FC = FCA – FCB = 2F – F = F = 2 × 10–5 N along $\mathop {AB}\limits^ \to$.

Q.  Force F is acting between two charges. If a sheet of glass (εr = 6) is placed between the two charges, what will be the force?

Sol.  $F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}}$

or $F’ = {1 \over {4\pi { \in _0}K}}{{{q_1}{q_2}} \over {{r^2}}}$

or $F’ = {F \over K} = {F \over 6}$

Q. How should we divide a charge ‘Q’ to get maximum force of repulsion between them?

Sol. Let q & Q – q be the two parts.

$F = {1 \over {4\pi {\varepsilon _0}}}{{q\left( {Q – q} \right)} \over {{r^2}}}$

for maximum F, ${{dF} \over {dq}} = 0$

or ${1 \over {4\pi {\varepsilon _0}}}{{Q – 2q} \over {{r^2}}} = 0$

or $q = {Q \over 2}$

Hence Q should be divided in two equal parts.

Basic Properties of Electric Charge, Definition & Solved Examples

Have you ever thought how electricity is generated? What is it made up of? Have you heard of electric charges? Let us start to know more about these bit by bit. In this article we are going to learn about Electric Charge definition and the basic properties of electric charge:

## Electric Charge Definition

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two-types of electric charges; positive and negative (commonly carried by protons and electrons respectively). [Source]

The excess or deficiency of electrons in a body gives the concept of charge. A negatively charged body has excess of electrons while a positively charged body has lost some of its electrons.

## Basic Properties of Electric Charge

We have an idea about electric charge since our childhood. There are a few basic properties of electric charge listed below, some of which you might be knowing already and some you might get to know now. Let’s get started:

1. Like charges repel while unlike charges attract each other. The true test of electrification is repulsion and not attraction as attraction may also take place between a charged and an uncharged body and also between two oppositely charged bodies.
2. Charge is a scalar.
3. Charge is transferable: When a charged body is put in contact with an uncharged body, the uncharged body becomes charged due to transfer of electrons from one body to the other.
If the charged body is positive it will draw some electrons from the uncharged body and if it is negative then it will transfer some of its excess electrons to the uncharged body.
4. Charge is always associated with mass: Charge cannot exist without mass though mass can exist without charge.
Example:

• The particles such as photon or neutrino which have zero rest mass do not have a charge.
• As charge cannot exist without mass, the presence of charge itself is a convincing proof of existence of mass. So a beam (such as β-rays, canal rays or cosmic rays) which gets deflected by electric or magnetic field will be charged and hence will consist of particles with mass. However, converse may or may not be true. If electric or magnetic field does not deflect a beam, the beam will be uncharged and so may consist of neutral particles like atoms or neutrons or electromagnetic waves like γ-rays or X-rays.
• In charging, the mass of a body changes. If electrons are removed from the body, the mass of the body will decrease and the body will become positively charged. If electrons are added to a body, the mass of the body will increase and the body will acquire a net negative charge. Due to extremely small mass of electron (= 9.1 × 10–31 kg) the change in mass of a body due to charging is negligible as compared to the mass of the body.

5. Charge is quantized: When a physical quantity has only discrete values the quantity is said to be quantized. Millikan oil drop experiment established that the smallest charge that can exist in nature is the charge of an electron. If the charge of an electron (e = 1.6 × 10–19 C) is taken as the elementary unit, i.e., quanta of charge, the charge on a body will be an integral multiple of e. i.e., q = ± ne with n = 1, 2, 3, 4…Charge on a body can never be (2/3)e, 17.2e or 10–5 e, etc. The elementary particles such as proton or neutron are made of quarks having charge (± 1/3)e and (± 2/3)e. The quarks do not exist in free state so the quantum of charge is still e.
6. Charge is conserved: In an isolated system the total charge does not change with time, though individual charge may change. Charge can neither be created nor destroyed. Equal quantities of positive and negative charges can appear or disappear simultaneously. Conservation of charge holds good in all types of reactions i.e. chemical, nuclear or decay. No exceptions to the rule have been found.
7. Charge is invariant: This means that charge is independent of frame of reference, i.e., charge on a body does not change with speed. The charge density or mass of a body depends on speed and increases with increase in speed.
8. Accelerated charges radiate energy: Electromagnetic theory shows that a charged particle at rest produces only electric field. If the charged particle is accelerated, it produces electric and magnetic fields but does not radiate energy. In accelerated motion charge produce electric and magnetic fields and also radiate energy in the form of electromagnetic waves.

9. A body can be charged by:
• Friction
• Induction
• Conduction

### Let’s see these 3 in detail below:

#### Friction

In friction when two bodies are rubbed together, electrons are transferred from one body to the other. This makes one body become positively charged while the other becomes negatively charged, e.g., when a glass rod is rubbed with silk, the rod becomes positively charged while the silk is negatively charged. Clouds are also charged by friction. Charging by friction is in accordance with conservation of charge. The positive and negative charges appear simultaneously in equal amounts due to transfer of electrons from one body to the other.

#### Induction

If a charged body is brought near a neutral body, the charged body will attract opposite charge and repel similar charge present in the neutral body. This makes one side of the neutral body becomes positively charged while the other side negative.

Important Points

1. Inducing body neither gains nor loses charge.
2. The nature of induced charge is always opposite to that of inducing charge.
3. Induced charge can be lesser or equal to inducing charge (but never greater) and its maximum value is $q’ = – q\,\,\left[ {1 – {1 \over K}} \right]$ where q is the inducing charge and K is the dielectric constant of the material of the uncharged body.
4. As K = ∞, so q’ = –q for metals, i.e., in metals induced charge is equal and opposite of inducing charge.
5. Induction takes place only in bodies (either conducting or non-conducting) and not in particles.

#### Conduction

Conduction takes place when an insulated conductor is brought in contact with a charged body and it gets the same charge as the charged body. Conduction is only possible in conductors and not in insulators.

## Unit of Electric Charge

• S.I unit of electric charge is Coulomb.
1 Coulomb = 1 ampere × 1 second
• C.G.S. unit of electric charge is Static Coulomb or Franklin
1 Coulomb = 3 × 109 static Coulomb
1 Coulomb = 3 × 109 electrostatic unit (esu) of charge = ${1 \over {10}}$ electromagnetic unit of charge
[esu = electrostatic unit]
[emu = electromagnetic unit]
• Practical units of charge are amp × hr (= 3600 Coulomb) and Faraday (= 96500 Coulomb)

## Solved Examples

Q.  Take two metallic spheres which are same in all respects. Give one of them a positive charge of Q coulomb and give the other same but negative charge. Which sphere will have a higher mass?

Sol. Negatively charged sphere will have a higher mass. This is due to increase in number of electrons to make it negatively charged.

Q. Which of the following charge is not possible?

[1] 1.6 × 10–18 C

[2] 1.6 × 10–19 C

[3] 1.6 × 10–20 C

[4] None of these

Sol. As only 1.6 × 10–20 C is 1/10th of electric charge and hence not an integral multiple.

Q. How many electrons are present in 1 coulomb charge?

Sol. q = ne

q = 1C e = 1.6 × 10–19 C

So, n = q/e = 6.25 × 1018 electrons.

Q. Identify X in the following nuclear reactions-

[a] 1H + 9Be → X + n

[b] 12C + 1H → X

[c] 15N + 1H → 4He + X

Sol. Using conservation of charge

a) Charge on nucleus of X must be (because Q (n) = 0) = Q (H) + Q (Be)

= 1e + 4e = 5e

Thus, X is boron (B)

b) Q (X) = Q (C) + Q (H)

= 6e + 1e = 7e

Thus, X is nitrogen (N)

c) Q (X) + Q (He) = Q (N) + Q (H)

Q(X) + 2e = 7e + 1e

Q(X) = 6e

Thus, X is carbon (C)