Properties of nuclear force – What are Nuclear forces – eSaral

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## What are Nuclear forces?

The strong forces of attraction which firmly hold the nucleons in the small nucleus and account for the stability of the nucleus are called nuclear forces.

## Properties of Nuclear Forces

### The nuclear force is a short-range force.

1. They are appreciable when the distance between nucleons is of the order of $2.2 \times 10^{-15} \mathrm{~m}$
2. They become negligible when the distance between nucleons is greater than $4.2 \times 10^{-15} \mathrm{~m}$
3. When the distance between two nucleons is less than $1 \times 10^{-15}$ m the nuclear forces become strongly repulsive

### Nuclear forces are charge independent

The force between a pair of protons, a pair of neutrons, and a pair of neutrons and protons are equal.

$F(n-n)=F(p-p)=F(n-p)$

The net force between pair of neutrons and a pair of neutron and proton is equal. This is slightly greater than the force between pair of protons because force between protons is reduced due to electrostatic repulsion

Net force F (n — n) = Net force F(n — p) > Net force F (p — p)

### Nuclear forces are spin dependent

1. The nuclear force depends on the relative orientation of spins between two interacting nucleons
2. The force of attraction between two nucleons with parallel spin is greater than the force between nucleons with antiparallel spin.
3. Deutron is formed in a bound state only if spins of neutron and proton are parallel.

### Nuclear forces show saturation property

1. The nucleon in the nucleus interacts with its nearest neighbor only.
2. It remains unaffected by the presence of other surrounding nucleons.
3. The nuclear force between a pair of nucleons in light and a heavy nucleus is equal.

### Nuclear forces are non-central forces

1. They do not act along line joining the center of two nucleons
2. The non-central component depends on the orientation of spins relative to the line joining the center of two nucleons.

### Nuclear forces are exchange forces

1. The nuclear forces originate by exchange of mesons $\left(\pi^{+}, \pi^{\circ}, \pi^{-}\right)$ between the nucleons

2. mass of meson = 0.15 amu = 140 MeV = 280 × mass of electron

3. $p-p$ force $\mathrm{p}+\pi^{\circ} \longleftrightarrow \mathrm{p}$

n – n force $\mathrm{n}+\pi^{\circ} \longleftrightarrow \mathrm{n}$

n – p force $\mathrm{p}+\pi^{-} \longleftrightarrow \mathrm{n}$

$\mathrm{n}+\pi^{+} \longleftrightarrow \mathrm{p}$

4. The theory of exchange forces was given by Yukawa

5. The potential energy of a particle in this force field is given by Yukawa potential $U(r)=U_{0} e^{-r / r_{0}}$ where

$\mathrm{r}_{0}$ & $\mathrm{U}_{0}$ are constants.

so, that’s all from this article. I hope you get the idea about the Properties of nuclear force. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

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Types of nuclei – Nuclear Physics class 12 – eSaral
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## Types of nuclei

### Isotopes

1. These are nuclei of the same element having the same Z but different Ae.g.
${ }_{8} \mathrm{O}^{16},{ }_{8} \mathrm{O}^{17},{ }_{8} \mathrm{O}^{18}$

${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2},{ }_{1} \mathrm{H}^{3}$

${ }_{92} \mathrm{U}^{234},{ }_{92} \mathrm{U}^{235}, \underset{92}{\mathrm{U}^{238}}$
2. All isotopes of an element have the same chemical properties
3. They occupy the same place in the periodic table
4. They cannot be separated by chemical analysis
5. They can be separated by mass spectrometers or mass spectrographs

### Isotones

1. These are nuclei of different elements having the same N but different A.
${ }_{6} \mathrm{C}_{7}^{13}$ & ${ }_{7} \mathrm{~N}_{7}^{14}$

${ }_{1} \mathrm{H}_{2}^{3}$ & ${ }_{2} \mathrm{He}_{2}^{4}$${ }_{2} \mathrm{Be}_{5}^{9} & { }_{5} \mathrm{~B}_{5}^{10} 2. Isotones are different elements with different chemical properties 3. They occupy different positions in the periodic table 4. They can be separated by chemical analysis and mass spectrometers ### Isobars 1. These are nuclei of different elements having the same A but different N and Z. e.g. { }_{6} \mathrm{C}^{14} and { }_{7} \mathrm{~N}^{14}$${ }_{18} \mathrm{Ar}^{40}$

and ${ }_{20} \mathrm{Ca}^{40}$
2. Isobars are different elements with different chemical properties
3. They occupy different positions in the periodic table
4. They can be separated by chemical analysis but cannot be separated by mass spectrometers.

### Mirror nuclei

These are nuclei with the same A but in which neutron and proton numbers are interchanged.

e.g. ${ }_{4} \mathrm{Be}_{3}^{7}(\mathrm{Z}=4, \mathrm{~N}=3)$

and

$3 \mathrm{Li}_{4}^{7}(Z=3, \mathrm{~N}=4)$

### Isomer nuclei

1. These are nuclei with the same A and same Z but differ in their nuclear energy states
2. They have different lifetimes and internal structure
3. These nuclei have different radioactive properties.e.g. $\mathrm{Co}^{60}$ & $\mathrm{Co}^{60 *}$

so, that’s all from this article. I hope you get the idea about How to state and explain ohm’s law. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Nuclei. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.
State and explain ohm’s law – Ohm’s law and resistance – eSaral
Hey, do you want to learn about How to state and explain ohm’s law? If yes. Then keep reading.

## Ohm’s Law:

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

$V \propto I$

Or

$\mathrm{V}=\mathrm{I} \mathrm{R}$

where $R=\frac{V}{I}$ is resistance of conductor.

$I=n$ e $A v_{d}$

$=n$ e $A \frac{e E}{m} \tau=\left(\frac{n e^{2} \tau}{m}\right)$

$\mathrm{AE}=\left(\frac{\mathrm{ne}^{2} \tau}{\mathrm{m}}\right) \mathrm{A} \frac{\mathrm{V}}{\mathrm{L}}$

So

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\left(\frac{\mathrm{m}}{\mathrm{ne}^{2} \tau}\right) \frac{\mathrm{L}}{\mathrm{A}}$

R is resistance of conductor

### Important Points

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

2. It is a scalar quantity with a unit volt/ampere called ohm $(\Omega)$dimensions of $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$$=\frac{\mathrm{W}}{\mathrm{qI}}=\frac{\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}}{\mathrm{ATA}} =\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2} 3. The reciprocal of resistance is called conductance G=\frac{1}{R}The SI unit is \mathrm{ohm}^{-1} or siemen (s) and its dimensions are \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2} 4. The substances which obey ohm’s law are called ohmic or linear conductor. The resistance of such conductors is independent of the magnitude and polarity of applied potential differences. Here the graph between I and V is a straight line passing throughthe origin. The reciprocal of slope of straight line gives resistance R=\frac{V}{I}=\frac{1}{\tan \theta}= constant e.g. silver, copper, mercury, carbon, mica, etc. 5. The substances which do not obey ohm’s law are called nonohmic or nonlinear conductors. The I–V curve is not a straight line.i.e. p n diode, transistors, thermionic valves, rectifiers, etc. 6. V = IR defines resistance and hence it is applicable to all ohmic and nonohmic conductors. If R is constant or V \propto I then it represents ohm’s law. 7. The relation R=\frac{V}{I} is macroscopic form while \rho=\frac{E}{J} is microscopic form of ohms law. 8. The resistance of a conductor depends on temperature, nature of material, length and area of cross-section. 9. The temperature dependence of resistance is given by R=R_{0}(1+\alpha \Delta \theta) where \alpha is temperature coefficient of resistance and \Delta \theta is change in temperature. The variation is graphically represented as.The temperature coefficient of resistance \alpha=\frac{R-R_{0}}{R_{0} \Delta \theta} is defined as change in resistance per unit resistance at 0^{\circ} \mathrm{C} per degree rise of temperature. 10. For maximum metals \alpha=\frac{1}{273} per { }^{\circ} \mathrm{C} so \mathrm{R}=\mathrm{R}_{0}\left(1+\frac{\Delta \theta}{273}\right)$$=\mathrm{R}_{0}\left(\frac{273+\Delta \theta}{273}\right)$

$=\mathrm{R}_{0} \frac{\mathrm{T}}{273}$

so $R \propto T$

i.e. The resistance of pure metallic conductor is proportional to its absolute temperature.
11. $R \propto L$ and $\mathrm{R} \propto \frac{1}{\mathrm{~A}}$so$R \propto \frac{L}{A}$ or
$R=\rho \frac{L}{A}$

where $\rho$ is called resistivity or specific resistance.

12. The fractional change in resistance without a change in volume or mass are:

(a) When change in length is small $(\leq 5 \%)$ fractional change in $\mathrm{R}$ is$\frac{\Delta R}{R}=\frac{2 \Delta L}{L}$

(b) When change in radius is small $(\leq 5 \%)$ fractional change in R is

$\frac{\Delta R}{R}=\frac{-4 \Delta r}{r}$

(c) When change in area is small $(\leq 5 \%)$ fractional change in R is

$\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{-2 \Delta \mathrm{A}}{\mathrm{A}}$

1. In terms of microscopic quantities, $E=\rho J$ so resistivity is numerically equal to the ratio of the magnitude of the electric field to current density.

2. $\rho=\frac{R A}{L}$ so if L = 1 m, $A=1 \mathrm{~m}^{2}$ then $\rho=R$. Specific resistivity is numerically equal to the resistance of substance having a unit area of cross-section and unit length.

3. It is a scalar with unit ohm-meter $(\Omega-\mathrm{m})$ and dimensions $M^{1} L^{3} T^{-3} A^{-2}$

4. The reciprocal of resistivity is called conductivity or specific conductance with units mho/m dimensions$\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}$ $\sigma=\frac{1}{\rho}=\frac{n e^{2} \tau}{m}=$ ne $\mu$

5. The resistivity is independent of the shape and size of the body and depends on the nature of the material of the body. The resistivity is the property of material while resistance is the property of the object.

6. The temperature dependence of resistivity is given by relation $\rho=\rho_{0}(1+\propto \Delta \theta)$ where $\propto$is the temperature coefficient of resistivity and $\Delta \theta$ is change in temperature. For metals, $\alpha$ is positive so resistivity increases with temperature while for non-metals $\alpha$ is negative so resistivity decreases with temperature.

### Specific use 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 a steady state when red hot at 800 ${ }^{\circ} \mathrm{C}$.

2. Fuse wire is made of tin-lead alloy because it has a low melting point and low resistivity. The fuse is used in series and melts to produce an open circuit when the current exceeds the safety limit.

3. Resistances of resistance boxes are made of manganin or constantan because they have moderate resistivity and a very small temperature coefficient of resistance. The resistivity is nearly independent of temperature.

4. The filament of the bulb is made up of tungsten because it has low resistivity, a 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.

so, that’s all from this article. I hope you get the idea about How to state and explain ohm’s law. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Current Electricity. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.
Charging and discharging of capacitor – Physics – eSaral
Hey, do you want to learn about the Charging and discharging of capacitors? If yes. Then you are at the right place.

## Charging and discharging of capacitors

### Charging

When a capacitor C is connected to a battery through R then charging of capacitor takes place.

The eqn. of emf at any time t is

$R I+\frac{q}{C}=E$

$\frac{R d q}{d t}+\frac{q}{C}=E$

$\int_{0}^{Q} \frac{d q}{(C E-q)}$

$=-\int_{0}^{t} \frac{d t}{R C}$

This on solving gives $Q=Q_{0}\left(1-e^{-t / R C}\right)$

where $Q_{0}=C E$

#### Important Points

1. The charge on a capacitor increases exponentially with time
2. The current during charging process is$I=\frac{d Q}{d t}$$=-\frac{Q_{0}}{R C} e^{-t / R C}$$=-I_{0} e^{-t / R C}$

The current decreases exponentially with time.

3. $\tau=R C$ is the capacitive time constant. It is the time in which the charge on the capacitor reaches 0.632 times of its maximum value during charging.
4. The time constant is the time in which current reduces to $\frac{1}{\mathrm{e}}$ times or 0.368 times of its initial value
5. At initial time $\mathrm{t}=0$ and $I=I_{\max }$ so circuit acts as short circuit or conducting wire.At $t=\infty \quad I=0$ so circuit acts as an open circuit or as a broken wire.
6. The voltage increases exponentially with time as $V=V_{0}\left(1-e^{-t / R C}\right)$ during charging.

### Discharging

If a completely charged capacitor C having charge $\mathrm{Q}_{0}$ is discharged through a resistance R then equation of emf at any time t is

$\mathrm{RI}+\frac{\mathrm{q}}{\mathrm{C}}=0$

Or

$R \frac{d q}{d t}+\frac{q}{C}=0$

Or

$\int_{Q_{0}}^{Q} \frac{d q}{q}=-\int_{0}^{t} \frac{d t}{R C}$

Or

$Q=Q_{0} e^{-t / R C}$

#### Important Points

1. During discharging charge on a capacitor decreases exponentially with time.
2. The current during discharging process$I=\frac{d Q}{d t}=-\frac{Q_{0}}{R C} e^{-t / R C}$$=-I_{0} e^{-t / R C} The current decreases exponentially with time. 3. Time constant is the time in which charge on capacitor become \frac{1}{\mathrm{e}} or 0.368 times of its initial value \mathrm{Q}_{0} 4. Time constant is the time in which current reduces to \frac{1}{\mathrm{e}} or 0.368 times of its initial value \mathrm{I}_{0}.The direction discharging current is opposite to that of charging. 5. During discharging voltage decreases with time as V=V_{0} e^{-t / R C} so, that’s all from this article. I hope you get the idea about the Charging and discharging of capacitors. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below. For a better understanding of this chapter, please check the detailed notes of Electrostatics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App. Combination of capacitors – Series, Parallel Combinations – eSaral Hey, do you want to learn about the combination of capacitors? If so. Then keep reading. ## Combination of capacitors The process of replacing a combination of capacitors by a single equivalent capacitor is called the Combination of capacitors or grouping of capacitors. ## Capacitors in parallel 1. Capacitors are said to be connected in parallel between two points if it is possible to proceed from one point to another point along different paths. 2. Capacitors are said to be in parallel if the potential across each individual capacitor is the same and equal to the applied potential. 3. Charge on each capacitor is different and is proportional to capacity of capacitor \mathrm{Q} \propto \mathrm{C} So \mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}, \mathrm{Q}_{2}=\mathrm{C}_{2} \mathrm{~V} \mathrm{Q}_{3}=\mathrm{C}_{3} \mathrm{~V} 4. The parallel combination obeys the law of conservation of charge So Q=Q_{1}+Q_{2}+Q_{3} =\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}+\mathrm{C}_{3} \mathrm{~V} =\left(C_{1}+C_{2}+C_{3}\right) V equivalent capacitance C_{p}=\frac{Q}{V}=C_{1}+C_{2}+C_{3} 5. The equivalent capacitance may be defined as the capacitance of a single capacitor that would acquire the same total charge Q with the same potential difference V. 6. The equivalent capacitance in parallel is equal to the sum of individual capacitances. 7. The equivalent capacitance is greater than the largest of individual capacitances. 8. The capacitors are connected in parallel (a) to increase the capacitance (b) when larger capacitance is required at low potential. 9. If n identical capacitors are connected in parallel then equivalent parallel capacitance C_{p}=n C 10. The total energy stored in parallel combination U=\frac{1}{2} C_{p} V^{2} =\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}+\ldots .\right) \mathrm{V}^{2} U=\frac{1}{2} C_{1} V^{2}+\frac{1}{2} C_{2} V^{2}+\ldots =U_{1}+U_{2}$$+U_{3}+\ldots \ldots$

The total energy stored in parallel combination is equal to the sum of energies stored in individual capacitors.
11. If n plates are arranged as shown they constitute (n–1)capacitors in parallel each of capacitance $\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$

Equivalent capacitance $C_{P}=(n-1) \frac{\varepsilon_{0} A}{d}$

## Series Combination

1. Capacitors are said to be connected in series between two points if it is possible to proceed from one point to the other point along only one path. 1
2. Capacitors are said to be in series if the charge on each individual capacitor is the same
3. The potential difference across each capacitor is different and is inversely proportional to its capacitance $V \propto \frac{1}{C}$

$\mathrm{V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}$

$\mathrm{V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}}$

and $\quad V_{3}=\frac{Q}{C_{3}}$

4. The series combination obeys the law of conservation of energy

So $\quad V=V_{1}+V_{2}+V_{3}$

$=\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\frac{Q}{C_{3}}$

$=Q\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\right)$

reciprocal of equivalent capacitance

$\frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{\mathrm{V}}{\mathrm{Q}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}$

5. The equivalent capacitance in series combination is the capacitance of a single capacitor that would become charged with the same charge Q when a potential difference is V.
6. The reciprocal of the equivalent capacitance is equal to the sum of the reciprocal of individual capacitance.
7. The equivalent capacitance is smaller than the smallest of individual capacitances.
8. The capacitors are connected in series (a) to decrease the capacitance (b) to divide large potential differences across many capacitors.
9. In n identical capacitors are connected in series than equivalent series capacitance $C_{s}=\frac{C}{n}$
10. The total energy stored in a series combination $U=\frac{Q^{2}}{2 C s}$

$=\frac{Q^{2}}{2}\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}+\ldots \ldots\right)$

or

$\mathrm{U}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{1}}+\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{2}}+\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{3}}+\ldots \ldots$

$=\mathrm{U}_{1}+\mathrm{U}_{2}+\mathrm{U}_{3}+\ldots \ldots$

The total energy stored in series combinations is equal to the sum of energies stored in individual capacitors.
11. If n plates are arranged as shown then they constitute (n–1) capacitors in series each of

value $\frac{\varepsilon_{0} A}{d}$

so

$C_{S}=\frac{\varepsilon_{0} A}{(n-1) d}$

so, that’s all from this article. I hope you get the idea about the combination of capacitors. If you found this article informative then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of Electrostatics Class 12. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.
Motion of charged particle in electric field – Class 12 – eSaral

Hey, do you want to learn about the motion of charged particle in electric field? If yes. Then keep reading

## 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 $\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}$
2. The direction of force is parallel to the field if the charge is positive and opposite to the field if the charge is negative.
3. The electric field is conservative so work is 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,$W_{A B}=-U_{A B}=q\left(V_{B}-V_{A}\right)=q V$
4. Work is done in moving a charged particle in an electric field unless the points are at the same potential
5. When a charged particle is accelerated by a uniform or non-uniform electric field then by work-energy theorem

$\Delta K E=W$so $\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=q V$

Or final velocity$v=\sqrt{\left[u^{2}+\frac{2 q V}{m}\right]}$

If the charged particle is initially at rest, then $v=\sqrt{\frac{2 q V}{m}}$

If the field is uniform $E=(\mathrm{V} / \mathrm{d})$ so $v=\sqrt{\frac{2 q E d}{m}}$

6. In motion of a charged particle in a uniform electric field if the force of gravity does not exist or is balanced by some other force say reaction or neglected then$\vec{a}=\frac{\vec{F}}{m}=\frac{\overrightarrow{q E}}{m}=$ constant
[as $\vec{F}=\overrightarrow{q E}]$Here equations of motion are valid.

• If the particle is initially at rest then from v = u + at, we get $v=a t=\frac{q E}{m} t$ And from Eqn.$s=u t+\frac{1}{2} a t^{2}$

we get $\quad s=\frac{1}{2} a t^{2}=\frac{1}{2} \frac{q E}{m} t^{2}$

The motion is accelerated translatory with $\mathrm{a} \propto \mathrm{t}^{\circ}$ ; $v \propto t$ and $s \propto t^{2}$

Here $W=\Delta \mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{~m}\left[\frac{\mathrm{q} \mathrm{E}}{\mathrm{m}} \mathrm{t}\right]^{2}$

also

$W=q E d=q V$
• If the particle is projected perpendicular to the field with an initial velocity $\mathrm{V}_{0}$From Eqn. v = u + at

and

$s=u t+\frac{1}{2} a t^{2}$

$\mathrm{u}_{\mathrm{x}}=\mathrm{v}_{0}$

and $\quad a_{x}=0$

$\mathrm{u}_{\mathrm{x}}=\mathrm{v}_{0}=\mathrm{const.}$

and $\quad x=v_{0} t$

for motion along y-axis as $u_{y}=0$

and $a_{y}=\frac{q E}{m}$ $v_{y}=\left[\frac{q E}{m}\right] t$

and $\quad y=\frac{1}{2}\left[\frac{q E}{m}\right] t^{2}$

So eliminating t between equation for x and y, we have

$\mathrm{y}=\frac{\mathrm{q} \mathrm{E}}{2 \mathrm{~m}}\left[\frac{\mathrm{x}}{\mathrm{v}_{0}}\right]^{2}=\frac{\mathrm{q} \mathrm{E}}{2 \mathrm{mv}_{0}^{2}} \mathrm{x}^{2}$

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

So, that’s all from this article. I hope you get the idea about the motion of charged particles in electric field. If you liked this article then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of the Electrostatics . To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

What are electric lines of force – Properties of lines of forces.

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## Electric Lines of Force

The idea of lines of force was introduced by Michel 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.

### Important points

1. Electric lines of force usually start or diverge out from a positive charge and end or converge on a negative charge.
2. The number of lines originating or terminating on a charge is proportional to the magnitude of the charge. In SI units $1 / \varepsilon_{0}$ 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 $\mathrm{q} / \varepsilon_{\mathrm{p}}$. If the body is cubical and the charge is situated at its center the lines of force through each face will be $\mathrm{q} / 6 \varepsilon_{0}$.
3. Lines of force never cross each other because if they cross, then intensity at that point will have two directions which are 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 the electric field will not remain conservative.
5. Lines of force have a tendency to contract longitudinally like a stretched elastic string producing attraction between opposite charges and repel each other laterally resulting in, the repulsion between similar charges and ‘edge-effects (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 a non-uniform field while the last shows a 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, the 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 the magnitude of electric field intensity. The crowded lines represent a 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 the direction of the 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 the direction of velocity and acceleration will be the 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 $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{at}}$

## 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. Electric lines of force always emerge or terminate normally on the surface of a charged conductor, while magnetic lines of force start or terminate on the surface of 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 monopolies do not exist).

So, that’s all from this article. I hope you get the idea about What are electric lines of force. If you liked this article then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of the Electric charge and Fields. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Principle of parallel plate capacitors – Definition, Capacitance – eSaral
Hey, do you want to learn about the Principle of Parallel plate capacitor? If yes. Then keep reading.

## Principle of Parallel plate capacitor

Let an insulated metal plate A be given a positive charge till its potential becomes maximum. When another insulated plate B is brought near A. Then by induction inner face of B becomes negatively charged and the outer face becomes positively charged. The negative charge tries to reduce the potential of A and the positive charge tries to increase it. When the outer surface of B is earthed positive charge flows to the earth while the negative charge stays on causing a reduction in the potential of A. Thus, a larger amount of charge can be given to A to raise it to maximum potential.

### Important Points

1. The capacitance of an insulated conductor is increased by bringing an uncharged earthed conductor near it.
2. An arrangement of two conductors carrying equal and opposite charge separated by a dielectric medium are said to form a capacitor.
3. The capacitor is an arrangement for storing a large amount of charge and hence electrical energy in a small space.
4. The capacity of a capacitor is defined as the ratio of charge Q on the plates to the potential difference between the plates i.e. $C=\frac{Q}{V}$
5. Capacitors are used in various electrical circuits like oscillators, in tuning circuits, filter circuits, electric fan, and motor, etc.
6. The shape of conductors can be a plane, spherical or cylindrical make a parallel plate, spherical or cylindrical capacitor.

## Capacitors in parallel

1. Capacitors are said to be connected in parallel between two points if it is possible to proceed from one point to another point along different paths.
2. Capacitors are said to be in parallel if the potential across each individual capacitor is the same and equal to the applied potential.
3. Charge on each capacitor is different and is proportional to capacity of capacitor $Q \propto C$ so $\mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}$ , $\mathrm{Q}_{2}=\mathrm{C}_{2} \mathrm{~V}$ , $Q_{3}=C_{3} V$
4. The parallel combination obeys law of conservation of charge So

$\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}+\mathrm{Q}_{3}$

$=C_{1} V+C_{2} V+C_{3} V$

$=\left(C_{1}+C_{2}+C_{3}\right) V$equivalent capacitance $C_{p}=\frac{Q}{V}$

$=C_{1}+C_{2}+C_{3}$
5. The equivalent capacitance may be defined as the capacitance of a single capacitor that would acquire the same total charge Q with the same potential difference V.
6. The equivalent capacitance in parallel is equal to the sum of individual capacitances.
7. The equivalent capacitance is greater than the largest of individual capacitances.
8. The capacitors are connected in parallel (a) to increase the capacitance (b) when larger capacitance is required at low potential.
9. If n identical capacitors are connected in parallel then equivalent parallel capacitance $C_{p}=n C$
10. The total energy stored in parallel combination

$U=\frac{1}{2} C_{p} V^{2}$

$=\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}+\ldots .\right) \mathrm{V}^{2}$

or$\mathrm{U}=\frac{1}{2} \mathrm{C}_{1} \mathrm{~V}^{2}+\frac{1}{2} \mathrm{C}_{2} \mathrm{~V}^{2}+\ldots \ldots=\mathrm{U}_{1}+\mathrm{U}_{2}$

$+U_{3}+\ldots \ldots$

The total energy stored in parallel combination is equal to the sum of energies stored in individual capacitors.
11. If n plates are arranged as shown they constitute (n–1) capacitors in parallel each of capacitance $\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$
Equivalent capacitance $C_{P}=(n-1) \frac{\varepsilon_{0} A}{d}$

## Capacitance of parallel plate capacitor with conducting slab

The original uniform field $E_{0}$ exists in distance d-t so potential difference between the plates

$V=E_{0}(d-t)=\frac{\sigma}{\varepsilon_{0}}(d-t)$

$=\frac{Q}{\varepsilon_{0} A}(d-t)$

Capacitance $C=\frac{Q}{V}$

$=\frac{\varepsilon_{0} A}{d(1-t / d)}=\frac{C_{0}}{1-t / d}$

$c>c_{0}$ so capacitance increases on introducing a metallic slab between the plates.

## The capacitance of parallel plate capacitor with dielectric slab

When a dielectric is introduced between plates then $\mathrm{E}_{0}$ field is present outside the dielectric and field E exists inside the dielectric. The potential difference between the plates

$V=E_{0}(d-t)+E t=E_{0}(d-t)$

$+\frac{E_{0} t}{K}=E_{0}\left[d-t\left(1-\frac{1}{K}\right)\right]$

$\mathrm{V}=\frac{\sigma}{\varepsilon_{0}}\left[\mathrm{~d}-\mathrm{t}\left(1-\frac{1}{\mathrm{~K}}\right)\right]$

$=\frac{\mathrm{Qd}}{\varepsilon_{0} \mathrm{~A}}\left[1-\frac{\mathrm{t}}{\mathrm{d}}\left(1-\frac{1}{\mathrm{~K}}\right)\right]$

Capacitance $C=\frac{Q}{V} \frac{\varepsilon_{0} A}{d\left[1-\frac{t}{d}\left(1-\frac{1}{K}\right)\right]}$

$=\frac{C_{0}}{1-\frac{t}{d}\left(1-\frac{1}{K}\right)}$

1. $C>C_{0}$ so capacitance increases on introducing a dielectric slab between plates of capacitor.
2. The capacitance is independent of position of dielectric slab between the plates.
3. If whole space is filled with dielectric than t = d and $C=K C_{0}$

## Energy stored in Capacitor

The charging of a capacitor involves transfer of electrons from one plate to another. The battery transfers a positive charge from negative to positive plate. Some work is done in transferring this charge which is stored as electrostatic energy in the field.

If dq charge is given to capacitor at potential V

then dW = V dq

or

$W=\int_{0}^{Q} \frac{q}{C} d q$

$\frac{Q^{2}}{2 C}=\frac{1}{2} C V^{2}=\frac{1}{2} Q V$

### Important Points

1. The energy is stored in the electric field between plates of capacitors.
2. The energy stored depends on capacitance, charge, and potential difference. This does not depend on the shape of the capacitor.
3. The energy is obtained at cost of the chemical energy of the battery

So, that’s all from this article. If you liked this article on the Principle of Parallel plate capacitors then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of the Electric charge and field. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.
Gauss law in electrostatics – Application, Important Points – eSaral

Hey, do you want to learn about Gauss Law in electrostatics? If yes. Then keep reading.

## Gauss law

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 $1 / \varepsilon_{0}$ times the charge enclosed by the closed surface,

Mathematically

$\oint_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} s}=\frac{\mathrm{q}}{\varepsilon_{0}}$

### Important points

[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 centre. By Coulomb’s

law intensity at a point P on the surface will be

$\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{3}} \overrightarrow{\mathrm{r}}$

electric flux linked with area $\overrightarrow{\mathrm{d} s}$

$\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} s}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{3}} \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{d}}$

direction of $\vec{r}$ and $\overrightarrow{\mathrm{d}} \mathrm{s}$ are same i.e.,

$\vec{r} \cdot \overrightarrow{d s}=r d s \cos 0^{\circ}=r d s$

So,

$\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d}} \mathrm{s}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \mathrm{ds}$

Or

$\oint_{s} \vec{E} \cdot \overrightarrow{d s}$

$=\oint_{s} \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} d s$

for all points on the sphere r = constant

$\oint \vec{E} \cdot \overrightarrow{d s}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$

$\oint_{s} d s=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}\left(4 \pi r^{2}\right)$

[as $\left.\oint \mathrm{ds}=4 \pi \mathrm{r}^{2}\right]$

$=\frac{q}{\varepsilon_{0}}$

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:

(i) 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.

(ii) 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 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 $\lambda$.

According to Gauss law

$\oint \vec{E} \cdot \overrightarrow{d s}=\frac{q_{i n}}{\varepsilon_{0}}$

$E(2 \pi r h)=\frac{\lambda h}{\varepsilon_{0}}$

So

$E=\frac{\lambda}{2 \pi \varepsilon_{0} r} \quad\left(E \propto \frac{1}{r}\right)$

### 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 = $\sigma$

According to gauss law

$\oint \vec{E} \cdot \overrightarrow{d S}=q_{i n} / \varepsilon_{0}$

Or

$E A+E A+0=\frac{\sigma A}{2 \varepsilon_{0}}$

Or

$E=\frac{\sigma}{2 \varepsilon_{0}}$

### Important Points

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.

[3] 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

$\oint_{S_{1}} \overrightarrow{E_{0}} \cdot \overrightarrow{d s}=\frac{Q}{\varepsilon_{0}}$

Or

$\mathrm{E}_{0}\left(4 \pi \mathrm{r}^{2}\right)=\frac{\mathrm{Q}}{\varepsilon_{0}}$

$\mathrm{E}_{0}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}$

$=\frac{\sigma}{\varepsilon_{0}} \frac{R^{2}}{r^{2}}$

where the surface charge density

$\sigma=\frac{\text { total charge }}{\text { surface area }}$

$=\frac{Q}{4 \pi R^{2}}$

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 $P_{s}$ on surface of shell $(r=R)$

$E_{S}=\frac{Q}{4 \pi \varepsilon_{0} R^{2}}=\frac{\sigma}{\varepsilon_{0}}$

3. At a point $P_{\text {in }}$ inside the shell $(r<R)$

According to gauss law

$\oint_{S_{2}} \vec{E} \cdot \overrightarrow{d s}=\frac{q_{i n}}{\varepsilon_{0}}$

As enclosed charge $q_{\text {in }}=0$

So $\quad E_{\text {in }}=0$

The electric field inside the spherical shell is always zero.

[4] 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 $\rho=\frac{\text { total charge }}{\text { total volume }}$

$=\frac{Q}{\frac{4}{3} \pi R^{3}}=\frac{3 Q}{4 \pi R^{3}}$

1. At a point $P_{0}$ outside the sphere $(r>R)$

according to gauss law $\oint \vec{E}_{0} \cdot \overrightarrow{d s}=\frac{Q}{\varepsilon_{0}}$

$E_{0}\left(4 \pi r^{2}\right)=\frac{Q}{\varepsilon_{0}}$

Or $\mathrm{E}_{0}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}$

$=\frac{\rho}{3 \varepsilon_{0}}\left(\frac{\mathrm{R}^{3}}{\mathrm{r}^{2}}\right)$

2. At a point $P_{s}$ on surface of sphere $(r=R)$

$\mathrm{E}_{\mathrm{s}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}$

$=\frac{\rho}{3 \varepsilon_{0}} R$

3. At a point $P_{\text {in }}$ inside the sphere $(r<R)$

According to gauss law

$\oint \vec{E}_{i n} \cdot \overrightarrow{d s}=\frac{q_{i n}}{\varepsilon_{0}}$

$=\frac{1}{\varepsilon_{0}} \rho \cdot \frac{4}{3} \pi r^{3}$

$=\frac{Q r^{3}}{\varepsilon_{0} R^{3}}$

$E_{\text {in }}\left(4 \pi r^{2}\right)=\frac{Q r^{3}}{\varepsilon_{0} R^{3}}$

or $E_{i n}=\frac{Q r}{4 \pi \varepsilon_{0} R^{3}}$

$=\frac{\rho}{3 \varepsilon_{0}} \mathrm{r}$

$\left(E_{\text {in }} \propto r\right)$

For a better understanding of this chapter, please check the detailed notes of the Electric charge and Fields. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

State and explain coulomb law – Class 12 Physics – eSaral

Hey, do you want to learn how to State and explain Coulomb law? If yes. Then keep reading.

## Coulomb’s law

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 the distance between them. This force acts along the line joining the center of two charges.

If $q_{1} \& q_{2}$ are charges, r is the distance between them and F is the force acting between them

Then, $F \alpha q_{1} q_{2}$

$\mathrm{F} \alpha 1 / \mathrm{r}^{2}$

$\mathrm{F} \alpha \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}$

Or $\mathrm{F}=\mathrm{c} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}$

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

$C=\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}$

(In SI unit)

C = 1 in electrostatic unit (esu)

$\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$

= 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.

$\mathrm{F}_{\text {air }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}$

And

$F_{\text {medium }}=\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{\mathrm{r}}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}$

$\frac{F_{\text {medium }}}{\mathrm{F}_{\text {air }}}=\frac{1}{\varepsilon_{\mathrm{r}}}=\mathrm{K}$

$\varepsilon_{r}$ or $\mathrm{K}$ = 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 $\left(\varepsilon_{r}\right.$ or $\left.\mathrm{K}\right)$ of a medium is defined as the ratio of the permittivity $\varepsilon$ of the medium to the permittivity$\varepsilon_{0}$ of free space i.e.

$\varepsilon_{\mathrm{r}}$

Or

$\mathrm{K}=\frac{\varepsilon}{\varepsilon_{0}}$

Dimensions of permittivity

$\varepsilon_{0}=\frac{Q^{2}}{F \times \text { length }^{2}}$

$=\frac{T^{2} A^{2}}{M L T^{-2} L^{2}}$

$=M^{-1} L^{-3} T^{4} A^{2}$

The dielectric constants of different mediums are

## Coulomb’s law in vector form

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

$\vec{F}_{21}=$ force on $q_{2}$ due to $q_{1}$

$\overrightarrow{F_{21}}=\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \frac{q_{1} q_{2}}{r_{12}^{2}} \hat{r}_{12}$

$\overrightarrow{\mathrm{F}}_{12}=$ Force on $\mathrm{q}_{1}$ due to $\mathrm{q}_{2}$

$\overrightarrow{F_{12}}=\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \frac{q_{1} q_{2}}{r_{21}^{2}} \hat{r}_{21}$

$\vec{F}_{12}=-\vec{F}_{21} \quad$ (as $\left.\hat{r}_{12}=-\hat{r}_{21}\right)$

Or

$\overrightarrow{\mathrm{F}}_{12}+\overrightarrow{\mathrm{F}}_{21}=0$

### Important points

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 the 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 $\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}$

If $q_{1}=q_{2}=1$ coulomb,

$r=1 m$

then $F=\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9}$ 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 repels it and is repelled by it with a force of

$9 \times 10^{9}$ newton.

So, that’s all from this article. I hope you get the idea about how to State and explain coulomb law. If you liked this article then please share it with your friends. If you have any confusion related to this topic then feel free to ask in the comments section down below.

For a better understanding of this chapter, please check the detailed notes of the Electric charge and Fields. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Nuclear fission and fusion – Physics Class 12 – eSaral

Hey, do you want to learn about Nuclear fission and fusion? if so. Then you are at the right place.

## Nuclear fission

1. The process of splitting of a heavy nucleus into two nuclei of comparable size and release of large energy is called fission.
${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{2}{ }_{2} \mathrm{U}^{236} \longrightarrow{ }_{56} \mathrm{Ba}^{141}+{ }_{36} \mathrm{Kr}^{92}+3{ }_{0} \mathrm{n}^{1}+$ energy $\mathrm{y}$
• U235 nucleus captures a thermal neutron
• This forms a compound nucleus U236 in excited state
• The shape of nucleus is distorted and nucleus splits into two fragments emitting several neutrons.

2. Nuclear fission was discovered by Otto Hahn and Strassman.
3. The energy of a thermal neutron ~ kT = 0.025 eV
4. The binding energy per nucleon of products is greater than the reactants.
5. The energy released in fission of Uranium is about 200 MeV. The fission energy released per nucleon is about 0.84 MeV
6. The fission of U235 may take place by different routes but amount of energy released per fission is nearly equal
7. The fission fragments are highly radioactive.
8. Nuclear fission can be explained on basis of liquid drop model.
9. The natural Uranium has following isotopes
${ }_{92} \mathrm{U}^{234}(0.006 \%) ; \quad{ }_{92} \mathrm{U}^{235}(0.72 \%) ; \quad{ }_{92} \mathrm{U}^{238}(99.27 \%)$
10. 92U238 is not fissionable. This can be converted to plutonium which is fissionable by neutrons.
${ }_{92} \mathrm{U}^{238}+{ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{93} \mathrm{~Np}^{239}+{ }_{-1} \mathrm{e}^{0}+\bar{v}$
${ }_{93} \mathrm{~Np}^{239} \longrightarrow{ }_{94} \mathrm{Pu}^{239}+{ }_{-1} \mathrm{e}^{0}+\bar{v}$
11. The Uranium in which fraction of U235 is increased from 0.7% to 2.3% is called enriched Uranium.
12. The neutrons emitted in fission are fast neutrons. Their energy is about 2MeV. On an average 2.5 neutrons are emitted per fission.
13. Energy released per gm of Uranium

Energy released per gm of Uranium  $=\frac{\text { Avogadro number }}{\text { mass number }} \times$ energy released per fission

$=\frac{6.023 \times 10^{23}}{235} \times 200=5.12 \times 10^{23} \mathrm{MeV}$

energy released by 1gm of U235 = 5.12 × 1023 MeV = 8.2 × 1010 J = 2.28 × 104 kWh = 2 × 1010 calorie
This energy is equivalent to
(i) energy obtained by burning 2560 kg of coal
(ii) energy obtained by burning 20 tonne of explosive TNT
14. The energy is released in form of kinetic energy of fission fragments, g-rays, heat, sound and light energy.
15. The fission process can take place at normal pressure and temperature.
16. Mass distribution in fission fragments

1. The mass distribution of fission fragments is highly assymetric.
2. The curve shows two peaks. at A = 95 and
A = 139. These represent the most probable mass numbers obtained in fission fragments.
3. The mass number of lighter fragment is between 85 to 104.
4. The mass number of heavier fragment is between 130 to 149.

## Nuclear fusion

1. The process in which two or more lighter nuclei combine to form a heavy nucleus is known as nuclear fusion
$4_{1} \mathrm{H}^{1} \longrightarrow{ }_{2} \mathrm{He}^{4}+2_{+1} \mathrm{e}^{0}+2 \mathrm{v}+\mathrm{Q}$
2. The binding energy per nucleon of product is greater than the reactants.
3. The energy released per nucleon is large ~ 6.75 MeV.
4. Fusion is possible at high pressure (~ 106 atom) and high temperature (~ 108 ºC)
5. The proton-proton cycle happens at lower temperature as compared to carbon-nitrogen cycle.
6. Nuclear fusion in possible at a place which has reactants in large quantity.
7. Hydrogen bomb works on principle of nuclear fusion
8. The explosion of a hydrogen bomb needs an explosion of atom bomb to generate required temperature.
9. No harmful radiations are produced in fusion.

## Difference Between Nuclear Fission and Nuclear Fusion

For a better understanding of this chapter, please check the detailed notes of Nuclear Physics. If you want more Free Learning Videos and Study Material Then don’t forget to download the eSaral App.

Mind Maps for Electrostatics Revision: Class XII, JEE, NEET
Here is the Mind Maps for Class 12Th Electrostatics Comprising all the important Formulae and Key Points. Therea re very important for quick revision during Exam.

### Mind Maps Electric Conductors

Mind Maps for Ray Optics Revision – Class XII, JEE, NEET
Ray Optics in Class 12th comprises variety of cases with important formula and key points. So here is the mind map to help you in remembering all the formula and important key concepts on finger tips.

### Mind Map of Optical Instruments

Mind Maps for Modern Physics Revision – Class XII, JEE, NEET
Atomic Structure in Class 12 comprises variety of cases with important formulae and key points. So here are the mind maps to help you in remembering all the formulas and important key concepts on finger tips.

### Mind Map of X-Rays

Mind Maps for Thermodynamics Revision – Class 11, JEE, NEET

True Dip and Apparent Dip – Magnetism and Matter Class 12
Here we will study about True Dip and Apparent Dip.

## Apparent Dip

The dip at a place is determined by a dip circle. It consists of a magnetized needle capable of rotation in a vertical plane about a horizontal axis. The needle moves over a vertical scale graduated in degrees.

If the plane of the scale of the dip circle is not in the magnetic meridian then the needle will not indicate the correct direction of the earth’s magnetic field. The angle made by the needle with the horizontal is called Apparent Dip.

## True Dip

When the plane of scale of the dip circle is in the magnetic meridian the needle comes to rest in direction of the earth’s magnetic field. The angle made by the needle with the horizontal is called True Dip.

Suppose dip circle is set at angle $\alpha$ to the magnetic meridian.

Horizontal component $${B_H}’ = {B_H}\cos \alpha$$

Vertical component ${B_V}’ = {B_V}$ (remains unchanged)

Apparent dip is ${\theta ^\prime }\tan {\theta ^\prime } = {{{B_V}’} \over {{B_H}’}} = {{{B_V}} \over {{B_H}\cos \alpha }} = {{\tan \theta } \over {\cos \alpha }}\left( {\tan \theta = {{{B_V}} \over {{B_H}}} = {\rm{ true dip }}} \right)$

1. For a vertical plane other than magnetic meridian $\alpha>0$ or $\cos \alpha<1$ so $\theta^{\prime}>\theta$ In a vertical plane other than magnetic meridian angle of dip is more than in magnetic meridian.
2. For a plane perpendicular to magnetic meridian $\alpha=\frac{\pi}{2}$ $\therefore \tan \theta^{\prime}=\infty \quad$ so $\quad \theta^{\prime}=\frac{\pi}{2}$ So in a plane perpendicular to magnetic meridian dip needle will become vertical.

### At magnetic equator :

1. The angle of dip is zero.
2. Vertical component of earths magnetic field becomes zero $B_{V}=B \sin \theta=B \sin 0=0$
3. A freely suspended magnet will become horizontal at the magnetic equator.
4. At the equator earth’s magnetic field is parallel to the earth’s surface i.e., horizontal.

### At magnetic poles :

1. The angle of dip is $90^{\circ}$
2. Horizontal component of earth’s magnetic field becomes zero. $B_{H}=B \cos \theta=B \cos 90=0$
3. A freely suspended magnet will become vertical at magnetic poles.
4. At poles, the earth’s magnetic field is perpendicular to the surface of the earth i.e. vertical.

Ex. If $\theta_{1}$ and $\theta_{2}$ are angles of dip in two vertical planes at right angle to each other and $\theta$ is true dip then prove $\cot ^{2} \theta=\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}$.

Sol. If the vertical plane in which dip is $\theta_{1}$ subtends an angle $\alpha$ with meridian than other vertical plane in which dip is $\theta_{2}$ and is perpendicular to first will make an angle of $90-\alpha$ with magnetic meridian. If $\theta_{1}$ and $\theta_{2}$ are apparent dips than

$\tan \theta_{1}=\frac{B_{V}}{B_{H} \cos \alpha}$

$\tan \theta_{2}=\frac{B_{V}}{B_{H} \cos (90-\alpha)}=\frac{B_{V}}{B_{H} \sin \alpha}$

$\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\frac{1}{\left(\tan \theta_{1}\right)^{2}}+\frac{1}{\left(\tan \theta_{2}\right)^{2}}$

$=\frac{B_{H}^{2} \cos ^{2} \alpha+B_{H}^{2} \sin ^{2} \alpha}{B_{V}^{2}}=\frac{B_{H}^{2}}{B_{V}^{2}}$

$=\left(\frac{B \cos \theta}{B \sin \theta}\right)^{2}=\cot ^{2} \theta$

So $\quad \cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\cot ^{2} \theta$
Properties of Paramagnetic & Diamagnetic Materials

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Neutral Point of Magnet – Magnetism and Matter Class 12 Physics Notes
A neutral point of Magnet is a point at which the resultant magnetic field is zero. In general, the neutral point is obtained when a horizontal component of the earth’s field is balanced by the produced by the magnet. When the N pole of the magnet points South and the magnet in the magnetic meridian.   When we plot the magnetic field of a bar magnet the curves obtained represent the superposition of magnetic fields due to bar magnet and earth.

## DEFINITION

A neutral point in the magnetic field of a bar magnet is that point where the field due to the magnet is completely neutralized by the horizontal component of the earth’s magnetic field. At neutral point field due to bar magnet (B) is equal and opposite to horizontal component of earth’s magnetic field $\left( B _{ H }\right) or \quad B = B _{ H }$
• #### The neutral point when the north pole of a magnet is towards the geographical north of the earth.

The neutral points $N_{1}$ and $N_{2}$ lie on the equatorial line. The magnetic field due to magnet at a neutral point is

$B =\frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$

Where M is the magnetic dipole moment of the magnet, $2 l$is its length, and r is a distance of the neutral point.

At neutral point $B = B _{ H } \cdot \quad$ so $\quad \frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}}= B _{ H }$

• #### Neutral point when south pole of a magnet is towards geographical north of earth.

The neutral points $N_{1}$ and $N_{2}$ lie on the axial line of magnet. The magnetic field due to magnet

at neutral point is $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$

At neutral point $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$ so $\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{2 M }{ r ^{3}}= B _{ H }$

#### Special Point

When a magnet is placed with its S pole towards north of earth neutral points lie on its axial line. If magnet is placed with its N pole towards north of earth neutral points lie on its equatorial line. So neutral points are displaced by $90^{\circ}$ on rotating magnet through $180^{\circ}$ In general if magnet is rotated by angle $\theta$ neutral point turn through an angle $\frac{\theta}{2}$

#### Neutral Point in Special Cases

(a) If two bar magnets are placed with their axis parallel to each other and their opposite poles face each other then there is only one neutral point (x) on the perpendicular bisector of the axis equidistant from the two magnets.

(b) If two bar magnets are placed with their axis parallel to each other and their like poles face each other then there are two neutral points on a line equidistant from the axis of the magnets.

Also Read: Properties of Paramagnetic & Diamagnetic Materials
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Bar Magnet as an Equivalent Solenoid – Magnetism || Class 12 Physics Notes

### Bar Magnet as an Equivalent Solenoid

In solenoid each turn behaves as a small magnetic dipole having dipole moment $\mathrm{I} \mathrm{A}$. A solenoid is treated as arrangement of small magnetic dipoles placed in line with each other. The number of dipoles is equal to number of turns in a solenoid. The south and north poles of each turn cancel each other except the ends. So solenoid can be replaced by single south and north pole separated by distance equal to length of solenoid. The magnetic field produced by a bar magnet is identical to that produced by a current carrying solenoid. Derivation of Bar Magnet as an Equivalent Solenoid To demonstrate the similarity of a current carrying solenoid to a bar magnet, let us calculate axial field of a finite solenoid carrying current.

Consider: $a=$ radius of solenoid

$2 l=$ length of solenoid with centre O

$n=$ number of turns per unit length $I=$ current passing through solenoid

$O P=r$

Consider a small element of thickness $d x$ of solenoid at distance $x$ from O. and number of turns in element $=n d x$

We know magnetic field due to n turns coil at axis of solenoid is given by

$d B=\frac{\mu_{0} n d x I a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{\frac{3}{2}}}$

The magnitude of the total field is obtained by summing over all the elements $-$ in other words by integrating from $x=-1$ to $x=+1 .$ Thus,

$B=\frac{\mu_{0} n I a^{2}}{2} \int_{-1}^{l} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}$

This integration can be done by trigonometric substitutions. This exercise, however, is not necessary for our purpose. Note that the range of $x$ is from $-1$ to $+1 .$ Consider the far axial field of the solenoid, i.e., $r>>$ a and $r>>1 .$ Then the denominator is approximated by

\begin{aligned}\left[(r-x)^{2}+a^{2}\right]^{3 / 2} &=r^{3} \\ \text { and } B &=\frac{\mu_{0} n I a^{2}}{2 r^{3}} \int_{-1}^{1} d x \\ &=\frac{\mu_{0} n I}{2} \frac{2 l a^{2}}{r^{3}} \end{aligned}

Note that the magnitude of the magnetic moment of the solenoid is, (total number of turns $\times$ current $\times$ cross-sectional area). Thus,

$B=\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}$

It is clear from the above expression that magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Terrestrial Magnetism – Earth’s Magnetism || Class 12 Physics Notes
Do you know that the Earth is also a magnet? Yes!!! How do you think then that the suspended bar magnet always points in the north-south direction? Adware about the concept of the Terrestrial Magnetism that we are going to discuss in this chapter. It is really interesting to study and analyze this concept of earth’s magnetism. The branch of Physics which deals with the study of earth’s magnetic field is called terrestrial magnetism.
1. William Gilbert suggested that earth itself behaves like a huge magnet. This magnet is so oriented that its S pole is towards geographic north and N pole is towards the geographic south.
2. The earth behaves as a magnetic dipole inclined at small angle $11.5^{\circ}$ to the earth’s axis of rotation with its south pole pointing geographic north.
3. The idea of earth having magnetism is supported by following facts.
4. A freely suspended magnet always comes to rest in N-S direction.
5. A piece of soft iron buried in N-S direction inside the earth acquires magnetism.
6. Existence of neutral points. When we draw field lines of bar magnet we get neutral points where magnetic field due to magnet is neutralized by earth’s magnetic field.
7. The magnetic field at the surface of earth ranges from nearly 30 $\mu T$ near equator to about 60$\mu T$ near the poles. The magnetic field on the axis is nearly twice the magnetic field on the equatorial line.
Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Magnetic Elements – Magnetism Class 12 || Physics Notes
The physical quantities which determine the intensity of earth’s total magnetic field completely both in magnitude and direction are called magnetic elements.

### Angle of Declination $(\phi)$:

The angle between the magnetic meridian and geographic meridian at a place is called angle of declination.

(a) Isogonic Lines : Lines drawn on a map through places that have same declination are called isogonic lines.

(b) Agonic Line : The line drawn on a map through places that have zero declination is known as an agonic line.

### Angle of Dip or Inclination

The angle through which the N pole dips down with reference to horizontal is called the angle of dip. At magnetic north and south pole angle of dip is $90^{\circ}$. At magnetic equator the angle of dip is zero.

OR The angle which the direction of resultant field of earth makes with the horizontal line of magnetic meridian is called angle of dip.

(a) Isoclinic Lines : Lines drawn up on a map through the places that have same dip are called isoclinic lines.

(b) Aclinic Line : The line drawn through places that have zero dip is known as aclinic line. This is the magnetic equator.

### Horizontal component of earth’s magnetic field

The total intensity of the earth’s magnetic field makes an angle  with horizontal. It has

(i) Component in horizontal plane called horizontal component $B_{H}$.

(ii) Component in vertical plane called vertical component $B_{V}$

$B _{ V }= B \sin \theta \quad B _{ H }= B \cos \theta$

So $\frac{ B _{ V }}{ B _{ H }}=\tan \theta \quad$ and $\quad B =\sqrt{ B _{ H ^{2}}+ B _{ V ^{2}}}$

#### IMPORTANT POINTS

1. If $\theta$ and $\phi$ are known we can find direction of B.
2. If $\theta$ and $B_{H}$ are known we can find magnitude of B.
3. So if $\theta$, $\phi$ and $B_{H}$ are known we can find total field at a place. So these are called as Elements of earth’s magnetic field.
4. The declination gives the plane, dip gives the direction and horizontal component gives magnitude of earth’s magnetic field.
5. If declination is ignored, then the horizontal component of earth’s magnetic field is from geogrophic south to geographic north.
6. Angle of dip is measured by instrument called dip circle.
Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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