Reflection by plane mirror – Ray Optics, Physics – eSaral

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## Reflection by plane mirror:

1. The image formed by the plane mirror is always erect, of the same size, and at the same distance as the object is.
2. To see the full image in a plane mirror, its length is just half the height of the man and it has to be kept in a specific position.
3. When the plane mirror or any reflecting surface is turned through an angle $\theta$ and if the incident ray remains stationary, then the reflected ray will turn through $2 \theta$.
4. The image of an object formed by a plane mirror is perverted & if the object is real then a virtual image formed.
5. Image due to a plane mirror is as far behind the mirror as is the object in Infront of it.
6. The magnification produced by the plane mirror is 1 i.e. the size of the image is equal to the size of the object.
7. When the two plane mirrors are parallel to each other, the number of images is infinity.
8. When two plane mirrors is held at angle $\theta$ with their reflecting surfaces facing each other and an object is placed between them, images are formed by successive reflections.

First of all, we will calculate

$n=\frac{360}{\theta}$

Then
9. If the angle between the two mirrors is $\theta$, the deviation produced by successive reflections is

$\delta=\delta_{1}+\delta_{2}=2 \pi-2 \theta$

Note : When reflection takes place by a smooth surface, it is called regular reflection, but when reflection takes place by a rough surface, it is called diffused reflection.

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Young double slit experiment class 12 – Physics – eSaral

Hey, do you want to learn about the Young double-slit experiment? If yes. Then keep reading.

## The experiment of young’s Dual-slit

• This experiment shows interference of light
• $s_{1} \& s_{2}$ slit behaves like two coherent sources.
• On-Screen, Bright & Dark portions are alternatively found.
• The bright portion is called Bright fringe & the dark portion is called dark fringe.
• The central fringe is always bright.
• Energy conserves in interference of light.
• It is explained on the basis of hygen principle.
• This experiment verifies the wave nature of light.

1. At a point on screen to find dark or bright fringe, it depends upon the path difference between $\mathrm{s}_{1} \mathrm{P}$ & $\mathrm{s}_{2} \mathrm{P}$ light waves.

2. Two types of path difference between light waves:
(i) Geometrical path difference.
(ii) Optical path difference.

In above experiment, optical difference s1P & s2P has geometrical path difference so, Total path difference = Geometrical difference

$\mathrm{s}_{1} \mathrm{P}-\mathrm{s}_{2} \mathrm{P}=\frac{\mathrm{xd}}{\mathrm{D}}$ (note $-\sin \theta \approx \tan \theta \approx Q)$

• Bright Fringes: If nth bright fringe forms at point P, then for Bright fringe:

$\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=\frac{\mathrm{X}_{\mathrm{n}} \mathrm{d}}{\mathrm{D}}=\mathrm{n} \lambda$

$X_{n}=\frac{n \lambda D}{d}$

So distance between central fringe & nth bright fringe :

$X_{n}=\frac{n \lambda D}{d}$

n = 1 first bright fringe
n = 2 Second bright fringe

• Dark Fringe: If nth dark fringe forms at a point P, then for dark fringe.

$\frac{X_{n} d}{D}=\frac{(2 n-1)}{2} \lambda \mid$

distance between central fringe & nth fringe.

$X_{n}=\frac{(2 n-1) \lambda D}{2 d}$
n = 1 First bright fringe.
n = 2 Second bright fringe.

Distance between dark & bright fringe which are incoming orders is called fringe width. In Interference Fringe width of dark & bright fringes are the same $\beta=\frac{\lambda D}{d}$

• The angular width of the fringe $\alpha \quad D=\beta$ $\alpha=\beta / D=\frac{\lambda}{d}$

## Comments on young’s interference experiment:

1. Energy is conserved in interference. This indicated that energy is redistributed from the destructive interference region to the constructive interference region.

2. If the entire arrangement of young’s double-slit experiment is immersed in water then fringe width decreases $\frac{\beta_{\text {water }}}{\beta_{\text {air }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}=\frac{1}{a_{\mathrm{w}}}=\frac{1}{(4 / 3)}$

3. If white light is used in place of monochromatic light in young’s double-slit experiment.

(a) central fringe is white

(b) Colored fringe around the central white fringe

(c) Inner edge of the dark fringe is red. While the outer edge is violet (or blue)

(d) Inner edge of the bright fringe is violet (or blue) and the outer edge is red.

4. If a filter allowing only $\lambda_{\text {red }}$ (of $\lambda_{1}$) is placed in front of slit $\mathrm{s}_{1}$ and filter allowing only $\lambda_{\text {blue }}\left(\right.$ or $\left.\lambda_{1}\right)$ is placed is front of slit $\mathrm{s}_{2}$. Then there is no interference pattern. (refer to point no.3 of conditions)

5. If a thin glass plate or mica sheet is placed in front of one of the slit, then the central fringe shifts towards that slit, the refractive index of glass is $\mu$ and the thickness of sheet t, then the optical path = $\mu \mathrm{t}$ so extra path difference $(\mu-1) t$

If the central fringe now appears at the location of previously formed nth bright fringe then $(\mu-1) t$ $=n \lambda$ if the central fringe appears at the position of previously formed nth dark fringe then $(\mu-1) t$

$=(2 n-1) \frac{\lambda}{2}$

6. If the width of the slit S increased then the degree of spatial coherence decreases. As a result the interference pattern gradually disappears similar occurs if the distance between $\mathrm{s}_{1}$ and $\mathrm{s}_{2}$ is increased.

7. The fringe visibility :

$V=\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$

$V_{\max }:$ if $I_{1}=I_{2}=I_{0}$

or $\mathrm{I}_{\min }=0$

If widths of slits $\mathrm{s}_{1}$ and $\mathrm{s}_{2}$ are unequal the brightness of the bright fringe and the darkness of the dark fringe decreases.

If $\left.\right|_{1}>>\left.\right|_{2}$ then $\left.\right|_{\max }=\left.\right|_{\min }$

8. When waves from two coherent sources $S_{1}$ are $S_{2}$ interfere in space the shape of the fringe is hyperbolic with foci at $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$.

So, that’s all from this article. I hope you get the idea about the Young double-slit experiment. 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 the Wave Optics. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Interference by Thin Films – Physics – eSaral

Hey, do you want to learn about the interference by thin films? If so. Then you are at the right place.

## Interference by Thin Films

The interference caused by thin films is due to the interference between the waves reflected from the upper and lower surfaces. These two reflected coherent waves are obtained from the same incident wave by division of amplitude.

A film of thickness t and refractive index $\mu$ produces a path difference of $2 \mu \mathrm{cos} \mathrm {r}$ between the two reflected waves and an additional path difference $\frac{\lambda}{2}$

or phase difference of $\pi$ is produced due to the reflection of one wave from a denser medium. Thus, the total path difference in reflected waves is $\left(2 \mu t \cos r+\frac{\lambda}{2}\right)$

For maxima $2 \mu t \quad \cos r+\frac{\lambda}{2}=n \lambda$

or $2 \mu t \cos r=(2 n-1) \frac{\lambda}{2}$

and for minima $2 \mu t \cos r+\frac{\lambda}{2}=(2 n+1) \frac{\lambda}{2}$

or $2 \mu t \cos r=(2 n-1) \frac{\lambda}{2}$

Here r is the angle of refraction. For normal incidence or near-normal incidence $r \approx 0$ ,so that

for maxima $2 \mu t=(2 n \pm 1) \frac{\lambda}{2}$

for minima $2 \mu t=n \lambda$ If a film is very thin $t \approx 0$, then the condition of minima is satisfied and the film appears dark in reflected light.

In transmitted waves, similar interference is observed, but now the additional path difference of $\frac{\lambda}{2}$ is absent. So, in transmitted waves.

$2 \mu t \cos r=n \lambda$ for maxima

and $2 \mu t \cos r=(2 n \pm 1) \frac{\lambda}{2}$ for minima.

## Points to remember

• The fringe width increases with the increase in distance between the source & the screen.
• Fringe width decreases by increasing distance between two slits $s_{1} \& s_{2}$.
• If the experiment is repeated in water instead of air, then $\beta$ decreases.
• When one of the slits of $s_{1} \& s_{2}$ is close then interference does not take place.
• When the two slits are illuminated by two independent sources then interference fringes are not obtained.
• When one of the slits is closed & the width of another is made of the order of $\lambda$, then diffraction fringes are observed
• When the slit is illuminated with different colors then fringes are obtained of the same color but their fringes width is different.
• In young’s double-slit experiment light waves undergo diffraction at both the slits and the diffracted waves superimpose to produce interference.
• If biprism experiment is a liquid instead of air, then the fringe width increases (whereas in young’s double-slit experiment it decreases)
• The wavelength undergoing destructive interference, the color of that wavelength will be absent.
• The wavelength for which the condition of constructive interference is fulfilled that color will be visible maximum consequently the fringes will be colored.

so, that’s all from this article. I hope you get the idea about the Interference by Thin Films. 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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Fresnel biprism experiment – Physics – eSaral

Hey, do you want to learn about the fresnel biprism experiment? If yes. Then keep reading.

## Fresnel’s Biprism Experiment

1. It is an optical device to obtain two coherent sources by the refraction of lights.
2. The angle of biprism is $179^{\circ}$ & refracting angle is $\alpha=1 / 2^{\circ}$.

3. Distance between source & screen D = a + b. Distance between two coherent sources

$=\mathrm{d}=2 \mathrm{a}(\mu-1) \alpha$

Where a = distance between source & Biprism

b = distance between screen & Biprism

$\mu$ = refractive index of the material of the prism.

$\lambda=\frac{d \beta}{D}=\frac{2 a(\mu-1) \alpha \beta}{(a+b)}$

$=\frac{\sqrt{d_{1} d_{2}} \cdot \beta}{(a+b)}$

Note-

$\alpha$ is in radian $\alpha^{0}=\alpha \times \frac{3.14}{180}$ Suppose refracting angle & refractive index is not known then d can be calculated by a convex lens.

One convex lens whose focal length (f) and 4f < D. First convex lens is kept near biprism & $\mathrm{d}_{1}$ is calculated then it is kept near eyepiece & $\mathrm{d}_{2}$ is calculated. $\mathrm{d}=\sqrt{\mathrm{d}_{1} \mathrm{~d}_{2}}$

Application :

With the help of this experiment the wavelength of monochromatic light, the thickness of thin films, and their refractive index & distance between apparent coherent sources can be determined. When Fresnel’s arrangement is immersed in water

(a) Effect on d

$\mathrm{d}_{\text {water }}<\mathrm{d}_{\mathrm{air}}$. Thus when the Fresnel’s biprism experiment is immersed in water, then the separation between the two virtual sources decreases but in young’s double-slit experiment it does not change.

(b) In young’s double-slit experiment $\beta$ decreases and in Fresnel’s biprism experiment $\beta$ increases.

so, that’s all from this article. I hope you get the idea about the fresnel biprism experiment. 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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What are coherent sources of light – Physics – eSaral

Hey, do you want to know what are coherent sources of light? If yes. Then keep reading.

## Coherent source

The two sources of light, whose frequencies are the same and the phase difference between the waves emitted by which remains constant with respect to time are defined as coherent sources. There are two independent concepts of coherence namely

1. Temporal coherence
2. spatial coherence

## Temporal coherence:

In a typical light source (like sodium lamp) a light wave (photon) is produced when an excited atom goes to the ground state and emits light the duration of this $\sim 10^{-9}$ to $10^{-10} \mathrm{sec}$. Thus the electromagnetic (light) wave remains sinusoidal for this much time. This time is known as coherence time it is denoted by $\tau_{c}$.

## Spatial coherence:

Two waves at a different points in space are said to be space coherent if they preserve a constant phase difference over any time t. Note: Laser is a source of monochromatic light waves of a high degree of coherence. The entire wavefront of the laser beam is spatially coherent. Important points: –

1. They are obtained from the same single source.
2. Their state of polarization is the same

Note:

1. Laser light is highly coherent & monochromatic
2. The light emitted by two independent sources (candles, bulbs, etc.) is non-coherent and interference phenomenon cannot be produced by such two sources

(a) Division of wavefront

(b) Division of Amplitude

so, that’s all from this article. I hope you get the idea about What are the coherent sources of light. 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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Einstein equation of the photoelectric effect – Physics – eSaral

Hey, do you want to learn about the Einstein equation of the photoelectric effect? If yes. Then keep reading.

## Einstein’s photoelectric equation

1. Einstein explained the photoelectric effect using the quantum nature of radiation.
2. The emission of a photoelectron is a result of the interaction of one photon with a loosely bound electron in which the photon is completely absorbed by the electron.
3. Some part of incident energy equal to work function is used to remove an electron from metal and remaining is given to electron as its kinetic energy.

$\mathrm{hv}=\mathrm{W}+\mathrm{E}_{\max }=\mathrm{W}+\frac{1}{2} \mathrm{mv}_{\max }^{2}$

$=h v_{0}+\frac{1}{2} m v_{\max }^{2}$

$\frac{1}{2} \mathrm{mv}_{\max }^{2}=\mathrm{h}\left(\mathrm{v}-\mathrm{v}_{0}\right)$

$=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\mathrm{eV}_{\mathrm{S}} \ldots(1)$

equation 1 is called Einstein photoelectric equation. maximum velocity of the emitted electron

$v_{\max }=\sqrt{\frac{2 h\left(v-v_{0}\right)}{m}}=\sqrt{\frac{2 h c\left(\lambda-\lambda_{0}\right)}{m \lambda \lambda_{0}}}$ $=\sqrt{\frac{2 \mathrm{eV}_{\mathrm{S}}}{\mathrm{m}}}$

The stopping potential

$V_{S}=\frac{h\left(v-v_{0}\right)}{e}=\frac{h c\left(\lambda_{0}-\lambda\right)}{e \lambda \lambda_{0}}$

4. The Einstein’s photoelectric equation is in accordance with the conservation of energy. Here light energy is converted into electric energy.
5. The equation explains the laws of photoelectric emission.

(a) The increase in intensity increases the number of photons with the same energy hv. So the number of photoelectrons will proportionally increase.

(b) If $v<v_{0}$ then KE will become negative which is not possible so in this condition photoemission is not possible.

(c) If $v>v_{0}$ Then $\mathrm{KE} \propto\left(v-v_{0}\right)$ so maximum kinetic energy or stopping potential increases linearly with frequency of incident radiation.

## Important points

1. In the photoelectric effect, all photoelectrons do not have the same kinetic energy. Their KE ranges from zero to $E_{\max }$ which depends on the frequency of incident radiation and nature of cathode.
2. The photoelectric effect takes place only when photons strikebound electrons because for free electrons energy and momentum conservations do not hold together.
3. Cesium is the best photo-sensitive material.
4. The efficiency of a photoemission

$\eta=\frac{\text { Number of photoelectrons emitted per unit area per unit time }}{\text { Number of photons incident per unit area per unit time }}$

$=\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{p}}}$ $\eta=\frac{\text { Intensity of emitted electrons }}{\text { Intensity of incident radiation }}$

$=\frac{I_{e}}{I_{p}}$ Therefore $\eta=\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{p}}}=\frac{\mathrm{I}_{\mathrm{e}}}{\mathrm{I}_{\mathrm{p}}}$

so, that’s all from this article. I hope you get the idea about the Einstein equation of the photoelectric effect. 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 the Dual Nature of Radiation and matter. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

What is photoelectric effect class 12 – Modern Physics – eSaral

Hey, do you want to know What is photoelectric effect in class 12? If yes. Then keep reading.

## What is Photoelectric effect?

The phenomenon of emission of electrons from the surface of the metal when the light of suitable frequency falls on it is called the photoelectric effect.

1. The ejected electrons are called photoelectrons
2. The current produced due to emitted electrons is called photocurrent.
3. The photoelectric effect proves the quantum nature of radiation.
4. The classical electromagnetic theory fails to explain photoelectric effect.
5. Einstein explained the photoelectric effect using quantum nature of radiation.
6. Hallwach is credited with discovery of the photoelectric effect.

## Experimental study of photoelectric effect

### Effect of intensity of incident radiation

1. The number of incident photons per second on a metal plate is called intensity of incident radiation.
2. For a fixed incident frequency, the saturation photocurrent is directly proportional to the intensity of incident radiation e.g. when intensity of radiation is doubled at constant frequency saturation photocurrent is also doubled.
3. Saturation current: When all photo electrons produced reach anode the photocurrent becomes maximum and is independent of applied potential difference. This current is called saturation or maximum current.

### Effect of potential

1. When polarity of electrodes is reversed with commutator the current is reduced but does not become zero. This shows that emitted photoelectrons have kinetic energy.
2. The negative potential of the anode at which the photo current becomes zero is called stopping potential $\left(\mathrm{V}_{\mathrm{s}}\right)$. At this potential the electrons with maximum kinetic energy are stopped from reaching the anode.
3. No photo current is produced even on increasing the intensity of incident radiation when the anode is at stopping potential. Thus, stopping potential is independent of the intensity of incident radiation.
4. The stopping potential is a measure of maximum kinetic energy of photo electrons. $E_{\max }$ $=\mathrm{eV}_{\mathrm{s}}$

### Effect of frequency of incident radiation

1. Threshold frequency the minimum frequency of incident radiation that can eject photo electrons from a metal is known as threshold frequency $\left(v_{0}\right)$.
2. At stopping potential if the frequency of incident radiation is increased then current starts flowing again. This can be made zero by increasing stopping potential.
3. Thus, the maximum kinetic energy of photoelectron or stopping potential increases with an increase in the frequency of incident radiation. The maximum kinetic energy of photoelectron increases linearly with increase in the frequency of incident light.

### Effect of material of cathode

The stopping potential, work function, and threshold frequency depend on the nature of material of cathode. Work function : The minimum energy required for emission of electrons from metal is called work function work function $\phi=\mathrm{hv}_{0}$ where $v_{0}$ is threshold frequency.

so, that’s all from this article. I hope you get the idea about What is photoelectric effect in class 12. 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 the Dual Nature of Radiation and matter. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Types of radioactive decay – Alpha Decay, Beta Decay – eSaral

Hey, do you want to learn about the Types of Radioactive decay or Types of radioactive processes? If yes. Then you are at the right place.

## Types of radioactive decay

### (a) Alpha decay

$_{z} X^{A}$ (parent nucleus) $\longrightarrow_{Z-2} Y^{\mathrm{A}-4}$ (daughter nucleus)

$+{ }_{2} \mathrm{He}^{4}$ (alpha particle)

e. g. ${ }_{92} \mathrm{U}^{238} \longrightarrow{ }_{90} \mathrm{Th}^{234}+{ }_{2} \mathrm{He}^{4}$

1. Alpha particle consists of 2 neutrons, 2 protons and carries a positive charge in magnitude 2 electrons. It is doubly ionized helium nuclei.

2. $\alpha$ emission takes place when the size of the nucleus becomes too large. The decay reduces the size of the nucleus.

3. $\alpha$ emission is explained on basis of the quantum mechanical tunnel effect.

4. The energy released in $\alpha$ decay $\mathrm{Q}=\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}-\mathrm{M}_{\alpha}\right) \mathrm{c}^{2}$

5. The kinetic energy of $\alpha$ particle $\mathrm{E}_{\alpha}=\left(\frac{\mathrm{A}-4}{\mathrm{~A}}\right) \mathrm{Q}$ where A is mass number and Q is disintegration energy

### (b) Beta decay

• Electron emission

$\left(\beta^{-}\right)$ ${ }_{z} X^{A} \longrightarrow_{z+1} Y^{A}$

$+_{-1} \mathrm{e}^{0}\left(\beta^{-}\right.$ particle $)+\bar{v}$

e.g. ${ }_{6} \mathrm{C}^{14} \longrightarrow{ }_{7} \mathrm{~N}^{14}$

$+_{-1} \mathrm{e}^{0}+\bar{v}$ (antineutrino)

1. $\beta^{-}$ Particles are fast-moving electrons carrying a negative charge

2. $\beta^{-}$ Particles are emitted when the nucleus has too many neutrons relative to a number of protons i.e. N/Z ratio is larger than required.

3. The emission of electrons takes place when a neutron is converted to a proton inside the nucleus. This helps in the correction of the N/Z ratio. ${ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{1} \mathrm{p}^{1}+{ }_{-1} \mathrm{e}^{0}+\overline{\mathrm{v}}$

4. The interaction responsible for $\beta$ decay is weak interaction.

• Positron emission

$\mathrm{z} \mathrm{X}^{\mathrm{A}} \longrightarrow \mathrm{z}_{-1} \mathrm{X}^{\mathrm{A}}$

$+_{+1} e^{0}\left(\beta^{+}\right.$ particle $)+v$

$\mathrm{eg} \cdot{ }_{29} \mathrm{Cu}^{64} \longrightarrow{ }_{28} \mathrm{Ni}^{64}$

$+_{+1} \mathrm{e}^{0}+v$ (neutrino)

1. $\beta^{+}$ Particles are positrons with a mass equal to an electron but carry a unit positive charge.

2. $\beta^{+}$ Particles are emitted when the nucleus has too many protons relative to a number of neutrons i.e. N/Z ratio is smaller than required.

3. The emission of positron takes place when a proton is converted to a neutron inside the nucleus. This increases the N/Z ratio.

${ }_{1} \mathrm{p}^{1}={ }_{0} \mathrm{n}^{1}+{ }_{+1} \mathrm{e}^{0}+\boldsymbol{v}$

### (c) Gamma decay

${ }_{Z} \mathrm{X}^{\mathrm{A}^{*}} \longrightarrow{ }_{Z} \mathrm{X}^{\mathrm{A}}+\gamma$

e.g. $\quad{ }_{5} \mathrm{~B}^{12} \longrightarrow{ }_{6} \mathrm{C}^{12^{\star}}+{ }_{-1} \mathrm{e}^{0}+\overline{\mathrm{v}}$

${ }_{6} \mathrm{C}^{12^{*}} \longrightarrow{ }_{6} \mathrm{C}^{12}+\gamma$

1. $\gamma$ rays are electromagnetic radiations that are chargeless and massless

2. $\gamma$ rays are emitted when the nucleus has excess energy

3. $\gamma$ rays are emitted when the nucleus jumps from the excited state to a lower level or ground state. This reduces the energy of the nucleus.

4. $\gamma$ rays are electromagnetic radiations of short wavelength $\left(\sim 10^{-12} \mathrm{~m}\right)$ which travel with speed of light.

### (d) Electron capture

${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}+{ }_{+1} \mathrm{e}^{0} \longrightarrow+\underset{\mathrm{z}-1}{\mathrm{Y}}^{\mathrm{A}}+\mathrm{v}$

${ }_{4} \mathrm{Be}^{7}+{ }_{-1} \mathrm{e}^{0} \longrightarrow{ }_{3} \mathrm{Li}^{7}+\mathrm{v}$

1. This process takes place when the nucleus has too many protons relative to a number of neutrons. i.e. N/Z ratio is larger than required.

2. This process occurs when a parent nucleus captures one of its own orbital atomic electrons and emits a neutrino.

so, that’s all from this article. I hope you get the idea about the Types of Radioactive decay. 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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What is binding energy in physics – Binding energy per nucleon – eSaral

Hey, do you want to learn about What is binding energy in physics? If yes. Then keep reading.

## Binding energy

1. The energy required to break a nucleus into its constituent nucleons and place them at an infinite distance is called binding energy.

2. The energy equivalent to mass defect is called binding energy.

3. This is the energy with which the nucleons are held together.

4. The binding energies $(\sim \mathrm{MeV})$ are very large as compared to molecular binding energies

$(\sim \mathrm{eV})$

Binding energy

$B E=(\Delta m) c^{2}$

$=c^{2}\left[Z m_{p}+(A-Z) m_{n}-M\left(z X^{A}\right)\right]$

the rest mass of protons + rest mass of neutrons = rest mass of nucleus + BE

## Binding energy per nucleon

1. The binding energy per nucleon of a nucleus is the average energy required to extract a nucleon from the nucleus. Binding energy per nucleon

$\overline{\mathrm{B}}=\frac{\text { Total binding energy }}{\text { Total number of nucleons }}$

$=\frac{\mathrm{BE}}{\mathrm{A}}=\frac{\Delta \mathrm{mc}^{2}}{\mathrm{~A}}$

$=\frac{c^{2}}{A}$ x
$\left[Z m_{p}+(A-Z) m_{n}-M\left(z X^{A}\right)\right]$

2. The plot of binding energy per nucleon with mass number A is shown as
3. Binding energy per nucleon gives a measure of the stability of the nucleus. More is the binding energy per nucleon more is the stability of the nucleus.

4. Binding energy per nucleon gives a measure of the stability of the nucleus. More is the binding energy per nucleon more is the stability of the nucleus.

5. Binding energy per nucleon is small for lighter nuclei i.e. ${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2}$ etc.

6. For A < 28 at A = 4n the curve shows some peaks at

${ }_{2} \mathrm{He}^{4},{ }_{4} \mathrm{Be}^{8},{ }_{6} \mathrm{C}^{16},{ }_{8} \mathrm{O}^{16},{ }_{10} \mathrm{Ne}^{20},{ }_{12} \mathrm{Mg}^{24}$.

7. The binding energy per nucleon is almost constant about 8.5 MeV in range 40 < A < 120. The binding energy per nucleon is maximum about 8.8 MeV for $\mathrm{F} \mathrm{e}^{56}$.

8. The binding energy per nucleon decreases for A > 200 They become less stable and exhibit radioactivity.

9. In fusion lighter nuclei fuse to form heavier nuclei. The process in accompanied by an increase in binding energy per nucleon.

10. In fission a heavy nucleus splits into two lighter nuclei. Here also increase in binding energy per nucleon takes place.

11. The heaviest stable nuclide is ${ }_{83} \mathrm{Bi}^{209}$.

so, that’s all from this article. I hope you get the idea about What is binding energy in physics. 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 Chapter Nuclei. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.

Properties of nuclear force – What are Nuclear forces – eSaral

Hey, do you want to learn about the Properties of nuclear force? If yes. Then keep reading.

## 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.

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

Types of nuclei – Nuclear Physics class 12 – eSaral
Hey, do you want to learn about the Types of nuclei? If yes. Then keep reading

## 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

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

### Important points about resistivity:

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.

Hey, do you want to learn about What are electric lines of force? If yes. Then keep reading.

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

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

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

So, that’s all from this article. I hope you get the idea about Gauss law in electrostatics. 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.

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.

Principle of superposition of waves – Types of Interference – eSaral
Hey do you learn about the Principle of superposition of waves? If so. Then keep reading.

## Principle of Superposition

Two, or more, progressive waves can travel simultaneously in a medium without affecting the motion of one another. Therefore, the resultant displacement of each particle of the medium m any instant is equal to the vector sum of the displacements produced by the two waves separately, this principle is called the ‘principle of superposition.

$y=y_{1} \pm y_{2} \pm$

$\vec{y}=\vec{y}_{1}+\vec{y}_{2}+\ldots \ldots \ldots \ldots$

Special Point –
1. It holds for all types of waves, provided the waves are not of very large amplitude. If waves are of very large amplitude, as laser waves, then this principle does not hold.
2. When we listen to an orchestra, we receive a complex sound due to the superposition of sound waves of different characteristics produced by different musical instruments. Still, we can recognize separately the sounds of different instruments.
3. Similarly, our radio antenna is open to the waves of different frequencies transmitted simultaneously by different radio stations. But when we tune the radio of a particular station, we receive the program of that station only as if the other stations were silent. Thus, the principle of superposition holds not only for the mechanical waves but also for the electromagnetic waves.

## Interference :

when two waves of the same frequency travel in a medium simultaneously in the same direction then, due to their superposition, the resultant intensity at any point of the medium is different from the sum of intensities of the two waves. At some points, the intensity of the resultant waves is very large while at some other points it is very small or zero. This phenomenon is called the ‘interference’ of waves.

Mathematically,

Ist wave $\Rightarrow \mathrm{Y}_{1}=\mathrm{a}_{1} \sin (\omega \mathrm{t})$

Ilnd wave $\Rightarrow Y_{2}=a_{2} \sin (\omega t+\phi)$

By principle of Superposition

$y=y_{1}+y_{2}$

$y=a_{1} \sin \omega t+a_{2} \sin (\omega t+\phi)$

$y=a_{1} \sin \omega t+\left[a_{2} \sin \omega t \cos \phi\right.$
$\left.+a_{2} \cos \omega t \sin \phi\right]$

$y=\sin \omega t\left(a_{1}+a_{2} \cos \phi\right)$
$+a_{2} \cos \omega t \sin \phi$

Suppose that

$a_{1}+a_{2} \cos \phi=A \cos \theta$ ……….(1)

$\mathrm{a}_{2} \sin \phi=\quad \mathrm{A} \sin \theta$ ……….(2)

some the square of Eq. (1) & (2)

$A^{2}=a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi$

This is amplitude of resultant wave.

Divide Eq. (2) By Eq. (1)

$\tan \theta=\frac{a_{2} \sin \phi}{a_{1}+a_{2} \cos \phi}$ ………………..(3)

$\phi$ Phase difference between two waves

Eq. of Resultant wave

$y=A \sin (w t+\theta)$

$\theta$ Initial phase of resultant wave

For interference I $\propto a^{2}$

$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{l}_{1}} \sqrt{\mathrm{I}_{2}} \cos \phi$

### (1) Constructive interference:

When waves are meat in same phase

phase difference

$\Delta \phi \quad=\quad 0,2 \pi, 4 \pi \quad \ldots \ldots 2 n \pi$

Path difference

$\Delta \mathrm{x} \quad=\quad 0, \lambda, 2 \lambda \ldots \ldots . . \mathrm{n} \lambda$

$\mathrm{n} \quad=\quad 0,1,2, \ldots \ldots \ldots \ldots$

$\cos \phi=+1$

$\mathrm{I}_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$

$A_{\max }=a_{1}+a_{2}$

If

$a_{1}=a_{2}=a$

$\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}$

$\mathrm{I}_{\max }=4 \mathrm{I}$

$A_{\max }=2 a$

### (2) Destructive Interference

Phase difference $\Delta \phi \quad=\quad \pi, 3 \pi \quad \ldots \ldots(2 n+1) \pi$

Path difference $\Delta \mathrm{x} \quad=\quad \frac{\lambda}{2}, \frac{3 \lambda}{2} \quad \ldots \ldots(2 n+1) \frac{\lambda}{2}$

$\begin{array}{lll}n & = & 0,1,2, \ldots \ldots \ldots \ldots \ldots . .\end{array}$

$\mathrm{A}^{2}=\mathrm{a}_{1}{ }^{2}+\mathrm{a}_{2}{ }^{2}+2 \mathrm{a}_{1} \mathrm{a}_{2} \cos \phi$

$\cos \phi=-1$

$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{l}_{1}} \quad \sqrt{\mathrm{I}_{2}} \cos \phi$

$\cos \phi=-1$

$A_{\min }=a_{1}-a_{2}$

$\mathrm{I}_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$

If

$a_{1}=a_{2}=a$

$\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}$

$A_{\min }=0$

$I_{\min }=0$

## Special results about interference pattern

Results
1. Maximum and minimum intensities in any interference wave form. $\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\operatorname{Min}}}=\left(\frac{\sqrt{\mathrm{l}_{1}}+\sqrt{\mathrm{I}_{2}}}{\sqrt{\mathrm{l}_{1}}-\sqrt{\mathrm{I}_{2}}}\right)^{2}=\left(\frac{\mathrm{a}_{1}+\mathrm{a}_{2}}{\mathrm{a}_{1}-\mathrm{a}_{2}}\right)^{2}$
2. Average intensity of interference wave form <I> or $I_{\text {av }}=\frac{\text { I }_{\max }+I_{\min }}{2}$
Put the value of $I_{\max } \& I_{\min }$

Or

$\left.\right|_{\mathrm{av}}=\mathrm{I}_{1}+\mathrm{I}_{2}$

$a=a_{1}=a_{2}$

And

$I_{1}=I_{2}=I$

Then

$I_{\max }=4$I

Then

$I_{\max }=4$I

$I_{\min }=0$

And

$I_{A V}=2 I$

So, that’s all from this article. If you liked this article on the principle of superposition of waves 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 Waves. To watch Free Learning Videos on physics by Saransh Gupta sir Install the eSaral App.