Introduction | Electrostatic Potential and Capacitance Class 12 Notes

Hello students. In Chapter 1, we spent a lot of time discussing the Electric Field. We treated the electric field as a vector quantity—it had a direction and a magnitude, and we had to be very careful with vector addition.

Now, in Chapter 2, we are going to look at the same electric phenomena but from a different perspective: Energy. Think back to Class 11. You learned that you can solve mechanics problems using Newton’s Laws (Vectors) or using Work and Energy (Scalars). Often, the scalar approach was much easier because you didn’t have to worry about angles and directions. That is exactly what we are doing here. We are introducing Electrostatic Potential, which is a scalar quantity. This will make solving complex problems involving charges much simpler. Let’s dive into the concepts of Potential and Capacitance.

1. Electrostatic Potential Energy

1.1 The Concept of Conservative Forces

Before we define potential, we must recall what a conservative force is. A force is conservative if the work done by it depends only on the initial and final positions, not on the path taken.

• Gravity is a conservative force.
• The Spring force is a conservative force.
• The Coulomb force (Electrostatic force) is also a conservative force.

Because the electrostatic force is conservative, we can define a potential energy associated with it.

1.2 Defining Electrostatic Potential Energy Difference

Imagine a source charge $Q$ fixed at the origin. We want to bring a small test charge $q$ from a point $R$ to a point $P$ against the repulsive force of $Q$. To do this, we must apply an external force ($F_{ext}$) that is equal and opposite to the electric force ($F_E$). We move the charge very slowly (without acceleration). The work done by this external force is stored in the system as Electrostatic Potential Energy.

Mathematically, the potential energy difference ($\Delta U$) between point P and point R is the work done by the external force:
$$\Delta U = U_P – U_R = W_{RP}$$

Figure-1: Work done in bringing a test charge from point R to P against the repulsive force of source charge Q.

Important Points to Remember:

• Path Independence: It does not matter if you take a straight line or a curved path from R to P; the work done is the same because the electric field is conservative.
• Only Difference Matters: Physically, the absolute value of potential energy isn’t as important as the difference in energy. However, to define a specific value at a point, we need a reference.
• The Reference Point: By convention, we assume that at infinity, the potential energy is zero ($U_{\infty} = 0$). So, if we bring a charge from infinity to point P, the work done defines the potential energy at P.

2. Electrostatic Potential (V)

While Potential Energy ($U$) depends on the size of the test charge $q$, we want a quantity that depends only on the source charge $Q$ and the location. This quantity is the Electrostatic Potential ($V$).

Definition: Electrostatic Potential at any point is the work done by an external force in bringing a unit positive test charge from infinity to that point (without acceleration).

Formula:
$$V = \frac{W}{q}$$
or
$$V_P – V_R = \frac{U_P – U_R}{q}$$

SI Unit: The unit is the Volt (V), named after Alessandro Volta.
$$1 \text{ Volt} = 1 \text{ Joule per Coulomb} (1 \text{ J/C})$$

Physical Meaning: Think of Potential as “Electrical Pressure”. Positive charges naturally want to move from High Potential to Low Potential (just like water flows downhill). Negative charges move from Low Potential to High Potential.

3. Potential Due to a Point Charge (Detailed Derivation)

This is a very common derivation in CBSE Board exams.

Goal: Find the potential $V$ at a distance $r$ from a source charge $Q$.

Setup:

• Place a source charge $Q$ at the origin.
• We want to calculate potential at point P, which is at a distance $r$ from the origin.
• Imagine bringing a unit positive test charge (+1 C) from infinity ($\infty$) to P.

Figure-2: Calculating the potential at point P due to a charge Q at the origin. The path of integration is from infinity to r.

The Process:
At any intermediate point $P’$ (at distance $r’$), the electrostatic force on a unit positive charge is given by Coulomb’s law:
$$F = \frac{1}{4\pi\epsilon_0} \frac{Q}{(r’)^2}$$
The force is repulsive (acting outwards). To move the charge inwards by a small distance $dr’$, the external force must be inwards.
Work done $dW = \vec{F}_{ext} \cdot \vec{dr}$. Since force and displacement are opposite, the work done by the external agent is positive, but mathematically, we are integrating against the field.
$$dW = – \frac{Q}{4\pi\epsilon_0 (r’)^2} dr’$$
(Note: The negative sign appears in the force integration relative to displacement).

Total Work ($W$) is the integral from $\infty$ to $r$:
$$W = – \int_{\infty}^{r} \frac{Q}{4\pi\epsilon_0 (r’)^2} dr’$$
Take constants out:
$$W = – \frac{Q}{4\pi\epsilon_0} \int_{\infty}^{r} (r’)^{-2} dr’$$
Perform integration ($\int x^{-2} dx = -x^{-1} = -1/x$):
$$W = – \frac{Q}{4\pi\epsilon_0} \left[ -\frac{1}{r’} \right]_{\infty}^{r}$$
$$W = \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{r} – \frac{1}{\infty} \right]$$
Since $1/\infty = 0$:
$$W = \frac{Q}{4\pi\epsilon_0 r}$$

Final Result:
Since Potential $V$ is work done per unit charge:
$$V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$$

Key Observation:

• For $E$ (Electric Field), the dependence is $1/r^2$.
• For $V$ (Potential), the dependence is $1/r$.
• If $Q$ is positive, $V$ is positive. If $Q$ is negative, $V$ is negative.

4. Potential Due to an Electric Dipole

Recall that an electric dipole is a pair of equal and opposite charges ($q$ and $-q$) separated by a small distance $2a$.

Setup:
Dipole moment $p = q \times 2a$, directed from $-q$ to $+q$.
We want potential at a point P at distance $r$ from the center, making an angle $\theta$ with the dipole axis.

Derivation Logic:
Potential is a scalar, so we simply add the potential due to $+q$ and the potential due to $-q$.
$$V = V_{+q} + V_{-q}$$
$$V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_1} – \frac{q}{r_2} \right)$$
where $r_1$ is distance from $+q$ and $r_2$ is distance from $-q$.

Approximation (Far Field $r \gg a$):
Using geometry and the cosine rule, for points far away, we can approximate:
$$\frac{1}{r_1} – \frac{1}{r_2} \approx \frac{2a \cos\theta}{r^2}$$
Substituting this back:
$$V = \frac{q}{4\pi\epsilon_0} \frac{2a \cos\theta}{r^2}$$
Since $p = 2aq$:
$$V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$$

Special Cases:

• Axial Line ($\theta = 0$ or $\pi$):
  $$V = \pm \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$$
  (Positive near $+q$, negative near $-q$).
• Equatorial Line ($\theta = 90^\circ$): $\cos(90^\circ) = 0$, so $V = 0$.
  
  Teacher’s Note: This is a favorite exam question. The Electric Field is not zero on the equatorial line, but the Potential is zero because the point is equidistant from the positive and negative charges.

Contrast with Single Charge:

• Dipole Potential $\propto 1/r^2$.
• Point Charge Potential $\propto 1/r$.
• Dipole potential depends on angle $\theta$; Point charge potential is spherically symmetric.

5. Equipotential Surfaces

Definition: An equipotential surface is a region where the potential is constant at all points ($V$ is same everywhere).

Figure-3: Equipotential surfaces for a positive point charge. Notice that the Electric Field lines (arrows) are always perpendicular to the surfaces (concentric circles).

Important Properties (Must Learn for Exams):

1. No Work Done: Since potential difference $\Delta V = 0$ between any two points on the surface, the work done in moving a charge along the surface is zero ($W = q\Delta V = 0$).
2. Perpendicular to Field: The Electric Field is always normal (perpendicular) to the equipotential surface at every point. Reasoning: If the field had a component parallel to the surface, work would be done in moving a charge. Since work is zero, the parallel component must be zero. Thus, the field is purely perpendicular.
3. Crowding: Surfaces are closer together where the electric field is stronger.

Examples of Shapes:

• Point Charge: Concentric Spheres.
• Uniform Electric Field: Parallel planes perpendicular to the field lines.
• Dipole: The surfaces get distorted, coming closer between the charges (see textbook diagrams).

6. Relation Between Electric Field and Potential

We know that Field ($E$) is related to Force, and Potential ($V$) is related to Work. Since Work = Force $\times$ Distance, $E$ and $V$ must be related.
Consider two equipotential surfaces separated by a tiny distance $\delta l$, with potential difference $\delta V$.
Work done to move unit charge = $E \cdot \delta l$.
Also, Work = Change in Potential = $-\delta V$ (Work against field).
So,
$$E \delta l = -\delta V$$
$$E = – \frac{\delta V}{\delta l}$$

In standard calculus notation:
$$E = – \frac{dV}{dr}$$
Teacher’s Explanation:

• The negative sign means the Electric Field points in the direction where Potential decreases most steeply.
• The magnitude of the field is the change in potential per unit displacement.

7. Potential Energy of a System of Charges

7.1 System of Two Charges

Imagine space is empty.

• Bring charge $q_1$ from infinity. Work done = 0 (no field to oppose it).
• Bring charge $q_2$ from infinity to a distance $r_{12}$ from $q_1$.

Now, $q_1$ creates a potential $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_{12}}$.
Work done $W = q_2 \times V_1$.
$$W = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$$
This work is stored as Potential Energy $U$.

7.2 System of Three Charges ($q_1, q_2, q_3$)

• Bring $q_1$: Work = 0.
• Bring $q_2$: Work against $q_1 = \frac{k q_1 q_2}{r_{12}}$.
• Bring $q_3$: It faces forces from both $q_1$ and $q_2$.

Work = (Force from 1) + (Force from 2).
$$U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right)$$

8. Potential Energy in an External Field

What if there is already an electric field $E$ (and potential $V$) created by some “external” source we don’t worry about?

1. Single Charge q:
$$U = q V(r)$$
Where $V(r)$ is the external potential at that point.
Example: An electron accelerated through 1 Volt gains 1 eV ($1.6 \times 10^{-19}$ J) of energy.

2. Two Charges in External Field:
We have to account for:
(a) Work done by external field on $q_1$.
(b) Work done by external field on $q_2$.
(c) Interaction energy between $q_1$ and $q_2$.
$$U_{total} = q_1 V(r_1) + q_2 V(r_2) + \frac{q_1 q_2}{4\pi\epsilon_0 r_{12}}$$

9. Dipole in a Uniform Electric Field

When a dipole is placed in a uniform field $E$:
Net Force = 0.
Torque $\tau = pE \sin\theta$.
If we rotate the dipole against this torque from angle $\theta_0$ to $\theta_1$, we do work.
$$W = \int_{\theta_0}^{\theta_1} pE \sin\theta d\theta = pE [-\cos\theta]_{\theta_0}^{\theta_1}$$

By convention, we take potential energy to be zero at $\theta = 90^\circ$ ($\pi/2$).
The Potential Energy at angle $\theta$ is:
$$U(\theta) = – pE \cos\theta = – \vec{p} \cdot \vec{E}$$

Figure-4: A dipole placed in a uniform electric field. The torque acts to align the dipole with the field.

Stability:

• Stable Equilibrium: $\theta = 0^\circ$ (p parallel to E), $U = -pE$ (Minimum Energy).
• Unstable Equilibrium: $\theta = 180^\circ$ (p antiparallel), $U = +pE$ (Maximum Energy).

10. Electrostatics of Conductors

This section is theoretical but extremely important for conceptual questions. Here are the 6 key properties of a conductor in electrostatic equilibrium:

1. Inside a conductor, Electric Field is ZERO.
   Why? Conductors have free electrons. If there were a field inside, electrons would experience a force and drift. They move until they accumulate on surfaces in such a way that they cancel the internal field.
2. At the surface, Electric Field is Normal (Perpendicular).
   Why? If it weren’t normal, there would be a tangential component. This would push surface charges to move, creating a current. In static conditions, currents are zero, so the tangential component must be zero.
3. No Excess Charge in the Interior.
   Any excess charge given to a conductor resides only on the outer surface. This is a consequence of Gauss’s Law ($E=0$ inside $\implies$ Flux=0 $\implies$ $q_{enclosed}=0$).
4. Potential is Constant throughout the Conductor.
   Since $E=0$ inside, no work is done moving a charge inside. Therefore, potential is constant inside and equal to the surface potential. The conductor is an equipotential volume.
5. Electric Field at Surface:
   $$E = \frac{\sigma}{\epsilon_0} \hat{n}$$
   Where $\sigma$ is surface charge density.
6. Electrostatic Shielding:
   If a conductor has a cavity inside it, the field inside that cavity is always zero, regardless of the field outside. This is used to protect sensitive instruments.

11. Dielectrics and Polarization

Dielectrics are insulators (non-conductors). They don’t have free electrons.

• Non-polar molecules ($H_2, O_2$): Centers of positive and negative charge coincide.
• Polar molecules ($H_2O, HCl$): Have a permanent dipole moment.

When placed in an external Electric Field ($E_0$):

• Conductors: Cancel the field completely ($E_{in} = 0$).
• Dielectrics: The field stretches the molecules (induces dipoles) or aligns existing dipoles. This creates an internal field that opposes but does not fully cancel the external field.

Result: The net field inside is reduced.
$$E_{net} = \frac{E_0}{K}$$
Where $K$ is the Dielectric Constant (also called Relative Permittivity $\epsilon_r$).
Polarization (P): Dipole moment per unit volume.

12. Capacitors and Capacitance

A Capacitor is a device used to store electric charge and energy. It usually consists of two conductors separated by an insulator.

Capacitance (C):
When we put charge $Q$ on a capacitor, the potential difference $V$ increases proportionally.
$$Q \propto V \Rightarrow Q = CV$$
$$C = \frac{Q}{V}$$

Unit: Farad (F). (1 F is a very large unit, we usually use $\mu F$ or $pF$) .
Dependence: $C$ depends ONLY on geometry (shape, size, separation) and the medium (dielectric), NOT on $Q$ or $V$.

13. The Parallel Plate Capacitor

Structure: Two plates of Area $A$, separated by distance $d$.

Figure-5: Parallel Plate Capacitor showing plates with area A, separation d, and the uniform electric field E between them.

Derivation of Capacitance:
Assume surface charge density $\sigma = Q/A$.
Electric field between plates (vacuum): $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$.
Potential difference $V = E \times d$ (since field is uniform).
$$V = \left( \frac{Q}{\epsilon_0 A} \right) d$$

Calculate C:
$$C = \frac{Q}{V} = \frac{Q}{ \frac{Qd}{\epsilon_0 A} }$$
$$C = \frac{\epsilon_0 A}{d}$$

Effect of Dielectric:
If we fill the space with a dielectric of constant $K$, the field reduces ($E = E_0/K$).
Potential reduces ($V = V_0/K$).
Since $C = Q/V$, if $V$ becomes smaller, C becomes larger.
$$C_{new} = K C_{vacuum}$$
$$C = \frac{K \epsilon_0 A}{d}$$

14. Combination of Capacitors

1. Series Combination

Arrangement: Chain-like.

• Charge (Q) is the same on all capacitors.
• Potential divides: $V = V_1 + V_2$.

Formula:
$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$$
Note: The equivalent capacitance is smaller than the smallest individual capacitor.

Figure-6: (Top) Capacitors connected in Series showing charge conservation. (Bottom) Capacitors connected in Parallel showing voltage equality.

2. Parallel Combination

Arrangement: Connected across same points.

• Potential (V) is the same for all.
• Charge adds up: $Q = Q_1 + Q_2$.

Formula:
$$C_{eq} = C_1 + C_2 + \dots$$
Note: Capacitance simply adds up.

15. Energy Stored in a Capacitor

Charging a capacitor requires work (moving charges against the repulsive force of charges already on the plate). This work is stored as potential energy.

Formula:
$$U = \frac{1}{2} CV^2 = \frac{Q^2}{2C} = \frac{1}{2} QV$$

Energy Density (u):
Energy per unit volume in the electric field.
$$u = \frac{1}{2} \epsilon_0 E^2$$

Solved Numericals (Teacher’s Walkthrough)

Numerical 1 (Potential):

Question: A regular hexagon of side 10 cm has a charge $5 \mu C$ at each of its vertices. Calculate the potential at the center.

Solution:
Distance $r$ from center to any vertex in a hexagon = side length = $10 \text{ cm} = 0.1 \text{ m}$.
Total potential $V = 6 \times (\text{Potential due to one charge})$.
$V = 6 \times \frac{kq}{r}$.
$V = 6 \times \frac{(9 \times 10^9) \times (5 \times 10^{-6})}{0.1}$.
$V = \frac{270 \times 10^3}{0.1} = 2.7 \times 10^6 \text{ Volts}$.

Numerical 2 (Energy):

Question: Two charges $7 \mu C$ and $-2 \mu C$ are placed at $(-9 \text{ cm}, 0, 0)$ and $(9 \text{ cm}, 0, 0)$. External field is zero. Calculate Potential Energy.

Solution:
Distance $r = 9 – (-9) = 18 \text{ cm} = 0.18 \text{ m}$.
Formula: $U = \frac{k q_1 q_2}{r}$.
$U = \frac{(9 \times 10^9) \times (7 \times 10^{-6}) \times (-2 \times 10^{-6})}{0.18}$.
$U = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \text{ Joules}$.
Significance: Negative energy means the system is bound (attractive).

Numerical 3 (Capacitor Dielectric):

Question: A parallel plate capacitor has capacitance 8 pF. Distance is reduced by half ($d’ = d/2$) and filled with dielectric $K=6$. Find new capacitance.

Solution:
Initial $C_0 = \frac{\epsilon_0 A}{d} = 8 \text{ pF}$.
New $C’ = \frac{K \epsilon_0 A}{d’}$.
Substitute $d’ = d/2$: $C’ = \frac{K \epsilon_0 A}{d/2} = 2K \left( \frac{\epsilon_0 A}{d} \right)$.
$C’ = 2K \times C_0$.
$C’ = 2 \times 6 \times 8 = 96 \text{ pF}$.

Real-Life Examples (CBSE Context)

• Lightning Rods: Conductors that shield buildings. They provide a path for charge to flow, utilizing the principle that potential is constant on a conductor and charges accumulate at sharp points (Corona discharge).
• Touch Screens: Capacitive touch screens work because your finger acts as a conductor. When you touch the screen, you alter the capacitance at that point, which the processor detects.
• Spark Plugs (Cars): Use high potential difference to create a strong electric field that ionizes air (dielectric breakdown) to ignite fuel.
• Faraday Cages: Your car protects you during lightning not because of rubber tires, but because the metal body acts as a conducting shell (Electrostatic Shielding), keeping the field inside zero.

Important Board Exam Derivations Summary

Students, please practice these three specifically for the 3 or 5-mark questions:

• Potential due to a Dipole: Remember the approximation $1/r_1 – 1/r_2 \approx 2a \cos\theta / r^2$.
• Potential Energy of a Dipole in Uniform Field: The integration of torque $pE \sin\theta$.
• Capacitance of Parallel Plate Capacitor: The logic chain: $\sigma \rightarrow E \rightarrow V \rightarrow C$.

Multiple Choice Questions (CBSE Pattern)

1. Q1. The work done in moving a charge of 5 C across two points having a potential difference of 2 V is:
   A) 2.5 J
   B) 10 J
   C) 0.4 J
   D) 0 J
2. Q2. Which of the following is true for an equipotential surface?
   A) Work done moving charge on it is maximum.
   B) Electric field is parallel to the surface.
   C) Electric field is perpendicular to the surface.
   D) It is always spherical.
3. Q3. The electric potential inside a hollow conducting sphere of radius R carrying charge Q is:
   A) Zero
   B) $kQ/r$ (where r < R)
   C) $kQ/R$
   D) Infinite
4. Q4. A dipole is placed in a uniform electric field. The potential energy is minimum when the angle between dipole moment and field is:
   A) $0^\circ$
   B) $90^\circ$
   C) $180^\circ$
   D) $45^\circ$
5. Q5. If the distance between plates of a parallel plate capacitor is halved, its capacitance:
   A) Halves
   B) Doubles
   C) Remains same
   D) Becomes four times
6. Q6. The unit of Electric Potential is:
   A) Newton/Coulomb
   B) Joule/Coulomb
   C) Coulomb/Joule
   D) Joule-Coulomb
7. Q7. Dielectric constant K for a metal is:
   A) 0
   B) 1
   C) Infinite
   D) Between 0 and 1
8. Q8. A capacitor stores energy in the form of:
   A) Magnetic Field
   B) Electric Field
   C) Heat
   D) Chemical Energy
9. Q9. Three capacitors of $2\mu F$ each are connected in series. The equivalent capacitance is:
   A) $6 \mu F$
   B) $2/3 \mu F$
   C) $3/2 \mu F$
   D) $1.5 \mu F$
10. Q10. The potential due to a point charge varies with distance r as:
    A) $r$
    B) $1/r$
    C) $1/r^2$
    D) $1/r^3$
11. Q11. Electrostatic shielding is possible because:
    A) Electric field inside a cavity of a conductor is zero.
    B) Potential is zero inside a conductor.
    C) Surface charge density is uniform.
    D) Conductors are insulators.
12. Q12. Work done in rotating a dipole from stable equilibrium to unstable equilibrium in a uniform electric field E is:
    A) $pE$
    B) $2pE$
    C) $-pE$
    D) Zero
13. Q13. If a dielectric slab is inserted in a charged capacitor (disconnected from battery), the potential difference across plates:
    A) Increases
    B) Decreases
    C) Remains same
    D) Becomes zero
14. Q14. The value of potential at a point on the equatorial line of a dipole is:
    A) $kp/r^2$
    B) $2kp/r^2$
    C) Zero
    D) $kp/r$
15. Q15. Which physical quantity has the unit Farad?
    A) Electric Flux
    B) Capacitance
    C) Charge
    D) Potential

MCQ Answers:

• (B) $W = qV = 5 \times 2 = 10$.
• (C) Field is always normal to equipotential surface.
• (C) Inside potential = Surface potential = $kQ/R$.
• (A) Stable equilibrium is at $0^\circ$.
• (B) $C \propto 1/d$. If d is halved, C doubles.
• (B) Volt = Joule/Coulomb.
• (C) Infinite (Metals conduct, so field inside is zero, implying infinite permittivity effect relative to vacuum).
• (B) Stored in the electric field between plates.
• (B) $1/C = 1/2 + 1/2 + 1/2 = 3/2 \implies C = 2/3$.
• (B) $V \propto 1/r$.
• (A) Definition of shielding.
• (B) $U_{initial} = -pE$, $U_{final} = +pE$. Change = $2pE$.
• (B) $V = V_0/K$. Since $K>1$, V decreases.
• (C) Equidistant from +q and -q.
• (B) Capacitance.