NCERT Class 12 Physics Chapter 2 Notes - Electrostatic Potential and Capacitance

Have you ever wondered how a small capacitor in the phone in your pocket yellow jacket is capable of storing and delivering energy in such an efficient way? That is the strength of Electrostatic Potential and Capacitance which is the subject of Physics Class 12 Chapter 2. The chapter enables you to learn how to make electric charges that generate potential, how the storage of energy in electric fields works and the role of capacitors in electronics in the present times. This chapter provides the foundation of both theory and practice since we have seen the concept of electric potential, the behavior of capacitors in various combinations.

This Electric potential and capacitance chapter is highly significant to students who are going to take CBSE board exams and other competitive exams such as JEE, NEET because they learn the fundamental principles of electrostatics and then obtain important applications of the principles in electric circuits and electrical arrangements. Some of the topics discussed are the electric potential energy, equipotential surfaces, capacitance, combination of capacitors, and the effect of dielectrics. So to make your preparation easy, professional teachers in Careers360 have prepared elaborate Class 12 Physics notes of Chapter 2 that comprises key concepts that can be grasped in a short manner, all the useful formulas that are necessary during exams, clear and labeled diagrams to feel the visual picture better.

Also, students can refer,

• NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance
• NCERT Exemplar Class 12 Physics Chapter 2 Solutions Electrostatic Potential and Capacitance

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NCERT Notes for Class 12 Chapter 2

The NCERT Notes for Physics Class 12 Chapter Electrostatic Potential and Capacitance is crucial for revision and last-minute study. Using these notes, students can view all the topics and prepare for their exams.

Electric Potential:

The electrostatic potential in a region of the electric field is equal to the amount of work done in bringing a unit-positive test charge from infinity to that point against the electrostatic force.

$V=\frac{W}{q_o}$

Where,

W - work done and $q_o$ - unit charge

• Electric potential is a scalar quantity and the SI unit is Volt (V).
• CGS unit is stat-volt.
• Dimension - $M L^2 T^{-3} A^{-1}$.
• 1 volt $=\frac{1}{300}$ stat volt.

Potential due to system of point charges:

$V=\sum_{i=1}^n \frac{K Q_i}{r_i}$

Potential difference:

The potential difference between two points A and B in an electric field is equal to the amount of work done (by an external agent) in moving a unit positive charge from point A to another point B.

$V_A$=Electric potential at point A

$V_B$=Electric potential at point B

$V_B-V_A=\frac{W}{q}$

$r_B \rightarrow$ the distance of charge at $B$
$r_A \rightarrow$ distance of charge at $A$
$\Delta V=$ The Electric potential difference in bringing charge q from point A to point B in the Electric field produced by Q .

$
\begin{aligned}
& \Delta V=V_B-V_A=\frac{W_{A \rightarrow B}}{q} \\
& \Delta V=K Q\left[\frac{1}{r_B}-\frac{1}{r_A}\right]
\end{aligned}
$

Where,

W is the amount of work done and q is the unit positive charge.

Electric potential due to a point charge:

$\begin{aligned} & V= \frac{Kq}{r} \\ & K=\frac{1}{4 \pi \epsilon_0}\end{aligned}$

Relation between electric field and potential

$
\vec{E}=-\frac{d V}{d r}
$

Where E is Electric field
And V is Electric potential
And $r$ is the position vector
And Negative sign indicates that in the direction of intensity the potential decreases.
If $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$

Then $E_x=\frac{\delta V}{d x}, E_y=\frac{\delta V}{d y}, E_z=\frac{\delta V}{d z}$
where

$
\begin{aligned}
E_x & =-\frac{\partial V}{d x} \quad \text { (partial derivative of } \mathrm{V} \text { w.r.t. x) } \\
E_y & =-\frac{\partial V}{d y} \quad \text { (partial derivative of } \mathrm{V} \text { w.r.t. y) } \\
E_z & =-\frac{\partial V}{d z} \quad \text { (partial derivative of } \mathrm{V} \text { w.r.t. z) }
\end{aligned}
$

Potential Due to an Electric Dipole

A. On the Axial Line:

As shown in the above figure We want to find out Electric Potential due to an Electric Dipole at a Point M which is on axial line and at a distance r from the center of a dipole.

Where $V_1$ and $V_2$ is the Electric Potential at M due to $-q$ and $+q$ charges respectively.

$
\begin{aligned}
& V_1=\frac{k q}{(r+a)} \\
& \begin{aligned}
& V_2=\frac{k q}{(r-a)} \\
& \begin{aligned}
V_{\text {net }} & =V_2-V_1 \\
V_{\text {net }} & =V_1+V_2 \\
& =\frac{-k q}{(r+a)}+\frac{k_q}{(r-a)} \\
& =k_q\left\{\frac{1}{r-a}-\frac{1}{r+a}\right\} \\
& =k q\left\{\frac{(r+a)-(r-a)}{(r-a)(r+a)}\right\}
\end{aligned}
\end{aligned}
\end{aligned}
$

So $V_{\text {net }}=\frac{2 k q a}{r^2-a^2}$
Using $P=q(2 a)$
So $V_{\text {net }}=\frac{k P}{r^2-a^2}$
- if $r \gg a$
then $V_{n e t}=\frac{K P}{r^2}=\frac{P}{4 \pi \epsilon_0 r^2}$

B. On the Equitorial line:

As shown in the above figure We want to find out Electric potential due to an Electric Dipole at a Point M which is on the Equitorial line and at a distance r from the center of a dipole.

Where $V_1$ and $V_2$ is the Electric Field Intensity at M due to $-q$ and $+q$ charges respectively.

$
\begin{aligned}
& V_1=-\frac{1}{4 \pi \epsilon_0} * \frac{q}{\sqrt{r^2+a^2}} \\
& V_2=\frac{1}{4 \pi \epsilon_0} * \frac{q}{\sqrt{r^2+a^2}} \\
& V_{\text {net }}=V_2-V_1=0
\end{aligned}
$

C. At any general point:

As shown in the above figure We want to find out Electric potential due to an Electric Dipole at a Point M which at a distance r from the center of a dipole and making an angle with the axial line.

From the figure, M is at the axial line of dipole having dipole moment as $P \cos \theta$ and M is at the Equitorial line of dipole having dipole moment as $P \sin \theta$.
So $P \sin \theta$ has no contribution in electric potential at point M.
if $r>>a$
then
$V_a=\frac{1}{4 \pi \varepsilon_0} \times \frac{2 P \cos \theta}{r^2}$ and $V_{\perp}=0$
$\mathrm{So} V_{\text {net }}=V_a=\frac{K P \cos \theta}{r^2}$

Electric Potential Energy:

Consider a system with two charges, q₁ and q₂ fixed at points A and B, respectively, and separated by AB =r₂. If q₂ is moved from B to a new point C along AB and AC =r₂, and the charge is displaced from r to r + dr, then the work done (dW) is as follows:

dW=F.dr $=\frac{K q_1 q_2}{r^2} \cdot d r$

• Total work done

$\begin{aligned} & W=\int_{r_1}^{r_2} \frac{K q_1 q_2}{r^2} \cdot d r \\ & =\frac{q_1 q_2}{4 \pi \epsilon_0}\left[\frac{1}{r_1}-\frac{1}{r_2}\right]\end{aligned}$

• Change in potential energy

$\begin{aligned} & U\left(r_2\right)-U\left(r_1\right)=-W \\ & =\frac{q_1 q_2}{4 \pi \epsilon_0}\left[\frac{1}{r_2}-\frac{1}{r_1}\right]\end{aligned}$

• When the system of two charges have infinite separation then potential energy

$U(\infty)=0$

• The potential energy when separation is r is

$\begin{aligned} & U(r)_0=U(r)-U(\infty) \\ & =\frac{q_1 q_2}{4 \pi \epsilon_0 r}\end{aligned}$

Equipotential Surface

An equipotential surface is a surface where the electric potential remains constant at every point. This means that no work is required to move a charge along this surface because the potential difference between any two points is zero.

• The electric field is always perpendicular to an equipotential surface.
• No work is done when moving a charge along an equipotential surface.
• In a uniform electric field, equipotential surfaces are parallel planes.
• For a point charge, equipotential surfaces are concentric spheres centered around the charge.

Work done in rotation of dipole and equilibrium of dipole:

$\begin{aligned} & W=p E\left(\cos \Theta_1-\cos \Theta_2\right) \\ & W=p E\left(\cos \Theta_1-\cos \Theta_2\right) \\ & \text { if } \theta_1=90 \text { and } \theta_2=\theta \\ & W=-p E \cos \theta=-\vec{p} . \vec{E}\end{aligned}$

This work done is stored as potential energy.

Condition for the stable equilibrium of a dipole:

1. Angle (θ): The system is stable when the angle between the dipole moment (p) and the electric field (E) is 0° (aligned).
2. Torque: In this position, the net torque acting on the dipole is zero. Any slight deviation from this position causes a restoring torque, bringing the dipole back into alignment with the field.
3. Potential Energy: When the dipole is aligned with the electric field, its potential energy is at its lowest.

Condition for the unstable equilibrium of a dipole:

1. Angle (θ): The system is in unstable equilibrium when the angle between p and E is 180° (anti-aligned).
2. Torque: In this position, the net torque acting on the dipole is zero. Any small displacement from this position, however, produces a torque that increases the angle between p and E, pushing the dipole out of alignment.
3. Potential Energy: The dipole's potential energy is greatest when it is anti-aligned with the electric field.

Electrostatics of Conductors

• At electrostatic equilibrium, the electric field inside a charged conductor is zero.
• The electric field at every point on the surface of a charged conductor is normal (perpendicular) to the surface.
• In a static situation, the excess charge on a charged conductor exists only on its surface.
• There is no electric field inside the cavity of a conductor, providing electrostatic shielding.
• At electrostatic equilibrium, the electrostatic potential remains constant throughout the conductor.

Dielectrics and Polarization

• Dielectrics are non-conducting materials, and they do not have free charge carriers that can move easily within the material.

Non-Polar Molecules:

The centres of negative and positive charges coincide in a non-polar molecule. The non-polar molecule lacks a permanent dipole moment.

Example: O₂, H₂

Polar Molecule:

Polar molecules have negative and positive charge centres that are separated and have a permanent dipole moment.

Example: H₂O, HCl

NOTE : Both polar and non-polar dielectrics acquire a net dipole moment in the presence of an external electric field.

Polarization:

It is the dipole moment per unit volume

$
P=\chi_e E
$

$\chi_e$ is the electric susceptibility of the dielectric medium

• Polarized dielectrics are similar to two charge surfaces with an induced charge density of opposite polarity.

Capacitor

A capacitor is a system of two conductors, which are separated by an insulator. A capacitor is used to store a large amount of charge.

The charge stored in a capacitor:

$Q=C V$

where, C is capacitance and V is voltage

Capacitance (C):

The capacitance of a capacitor

$C=Q / V$

Dielectric Strength:

Dielectric strength is the maximum amount of electric field that a dielectric medium can withstand.

Parallel Plate Capacitor:

Two conducting plates of area A separated by a distance d. If the dielectric medium between the capacitor plate is vacuum or air, then

$C=\frac{\epsilon_0 A}{d}$

• When a dielectric of dielectric constant k is inserted between the above capacitor, the new capacitance

$\begin{aligned} & C^{\prime}=\frac{k \epsilon_0 A}{d} \\ & C^{\prime}=k \stackrel{C}{C}\end{aligned}$

Combination of capacitors:

Series Combination :

In series: $\frac{1}{C_{e q}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+$. $\qquad$
Note : In series combination equivalent capacitance is always less the smallest capacitor of combination.

Parallel Combination :

Equivalent capacitance of parallel combination $\mathrm{C}_{e q}=\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$

Note : Equivalent capacitance is always greater than the largest capacitor of combination.

• For n capacitors connected in parallel, the net value of capacitance is

$C=C_1+C_2+C_3+\ldots \ldots \ldots C_n$

• For n capacitors connected in series, the net value of capacitance is

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \ldots \ldots \ldots \ldots \frac{1}{C_n}$

Energy Stored in a Capacitor:

The energy U stored in a capacitor of capacitance C, charge Q and voltage V is

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

The electric energy density

In a region with an electric field, the electric energy density,

Energy per unit volume $=\frac{1}{2} \epsilon_0 E^2$

Electric Potential and Capacitance Previous year Question and Answer

Q1: A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge

A. remains a constant because the electric field is uniform

B. increases because the charge moves along the electric field

C. decreases because the charge moves along the electric field

D. decreases because the charge moves opposite to the electric field

Answer:

A positively charged particle in a uniform electric field experiences a force in the direction of the electric field.
When the charge moves along the direction of the field, it moves to a region of lower electric potential.
As it moves, the electric potential energy decreases because the work done by the field on the charge reduces the potential energy and converts it into kinetic energy.

Hence, the correct answer is option (C).

Q2: Two point charges $4 \mu C$ and $+1 \mu C$ are separated by a distance of 2 m in air. Find the point on the line-joining charges at which the net electric field of the system is zero.

Answer:

$\text { Let the electric field is zero at } \mathrm{p} \text { at distance } \mathrm{x} \text { from } 1 \mu c \text { charge, then }\\$
$\frac{k \times 4 \mu c}{(2-x)^2}=\frac{k \times 1 \mu c}{x^2} \\$
$\frac{2}{2-x}=\frac{1}{x} \\$
$2 x=2-x \\$
$x=\frac{2}{3} m \\$
$\therefore 2-x=2-\frac{2}{3}=\frac{4}{3} m$

Q3: Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.

Answer:

We know that the electric field strength is proportional to the distance between the surface, so, the surface will be evenly spaced.

The electric field is always perpendicular to the equipotential surface.

Hence, the equipotential surfaces are going to be infinite plane sheets in the z-direction parallel to the x-y plane.

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