CBSE Class 12 › Unit I: Electrostatics

Electrostatic Potential & Capacitance

NCERT Chapter 2 Weightage: ~9 Marks JEE Main: ~2 Qs

Jump To Section:

1. Electric Potential & Derivations
2. Equipotential Surfaces
3. Potential Energy & Derivations
4. Electrostatics of Conductors
5. Capacitance & Derivations
6. Combinations & Energy

Topic	Question Type	Marks
Derivation: Potential due to Point Charge	Short Answer	2 Marks
Derivation: Potential due to Dipole	Long Answer	5 Marks
Derivation: Parallel Plate Capacitor	Short Answer	3 Marks
Derivation: Energy Stored	Short Answer	2 Marks

1. Electric Potential

Electrostatic potential $V$ at a point is defined as the work done by an external force in bringing a unit positive charge from infinity to that point (without acceleration).

Definition

V_P - V_R = \frac{W_{RP}}{q}

SI Unit: Volt (V) = Joule/Coulomb

Derivation 1: Potential due to a Point Charge 3 Marks

Diagram showing work done in bringing a unit positive test charge from infinity to point P against the repulsive force of charge Q. — Calculating work done to bring a unit positive charge from infinity to point P.

Step 1: Electrostatic Force Consider a source charge $Q$ at origin. Force on unit positive test charge at distance $r'$ :
$F = \frac{1}{4\pi\epsilon_0} \frac{Q \times 1}{(r')^2}$

Step 2: Work Done Work done by external force against this repulsion to move by $dr'$ :
$dW = -F dr' = -\frac{Q}{4\pi\epsilon_0 (r')^2} dr'$

Step 3: Integration Total work done from $\infty$ to $r$ :
$W = -\int_{\infty}^{r} \frac{Q}{4\pi\epsilon_0 (r')^2} dr' = -\frac{Q}{4\pi\epsilon_0} \left[ -\frac{1}{r'} \right]_{\infty}^{r}$

Final Result $W = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{r} - 0 \right)$
$V(r) = \frac{Q}{4\pi\epsilon_0 r}$

⚠️

Relation between E and V: Electric field is the negative gradient of potential.
$E = - \frac{dV}{dr}$ (Field flows from high to low potential).

Derivation 2: Potential due to an Electric Dipole 5 Marks

Geometry diagram for calculating potential at a point P due to a dipole, showing distances r1, r2 and angle theta. — Geometry involved in calculating potential at P due to charges +q and -q separated by 2a.

Step 1: Superposition Potential at P due to $+q$ (at $r_1$ ) and $-q$ (at $r_2$ ):
$V = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$

Step 2: Geometry (Cosine Rule) For $r \gg a$ :
$r_1^2 \approx r^2 \left( 1 - \frac{2a \cos\theta}{r} \right) \implies \frac{1}{r_1} \approx \frac{1}{r} \left( 1 + \frac{a \cos\theta}{r} \right)$
$r_2^2 \approx r^2 \left( 1 + \frac{2a \cos\theta}{r} \right) \implies \frac{1}{r_2} \approx \frac{1}{r} \left( 1 - \frac{a \cos\theta}{r} \right)$

Step 3: Substitution $V = \frac{q}{4\pi\epsilon_0 r} \left[ \left( 1 + \frac{a \cos\theta}{r} \right) - \left( 1 - \frac{a \cos\theta}{r} \right) \right]$
$V = \frac{q}{4\pi\epsilon_0 r} \left( \frac{2a \cos\theta}{r} \right)$

Final Result Using Dipole Moment $p = 2qa$ :
$V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}$

2. Equipotential Surfaces

An equipotential surface is a surface with a constant value of potential at all points.

Equipotential surfaces for a single point charge (concentric spheres) and an electric dipole. — Equipotential surfaces are concentric spheres for a point charge, but distorted for a dipole.

Key Properties

Work Done: No work is done in moving a test charge over an equipotential surface.
Field Direction: Electric field is always normal to the equipotential surface at every point.
Pattern: For a point charge, they are concentric spheres.

3. Potential Energy

Electrostatic potential energy is the work done in assembling a system of charges by bringing them from infinity to their present locations.

Derivation 3: Potential Energy of a System of Charges 3 Marks

Three point charges in space to illustrate the potential energy of a system. — The potential energy of the system arises from the work done in assembling the charges.

Step 1: First Charge Bring $q_1$ from infinity to $r_1$ . Work done $W_1 = 0$ (no external field).

Step 2: Second Charge Bring $q_2$ from infinity to $r_2$ . Work is done against field of $q_1$ .
$W_2 = q_2 V_1(r_2) = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$

Step 3: Third Charge (Optional) Bring $q_3$ against fields of $q_1$ and $q_2$ .
$W_3 = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right)$

Final Result Total $U = W_1 + W_2 + W_3$ :
$U = \frac{1}{4\pi\epsilon_0} \sum \frac{q_i q_j}{r_{ij}}$

Derivation 4: PE of Dipole in External Field 3 Marks

Diagram of an electric dipole in a uniform electric field experiencing a torque. — An external field exerts a torque on the dipole, storing potential energy.

Step 1: Torque External field $E$ exerts torque on dipole $p$ :
$\tau = pE \sin\theta$

Step 2: Work Done Work done to rotate dipole from $\theta_0$ to $\theta_1$ :
$W = \int_{\theta_0}^{\theta_1} \tau d\theta = \int_{\theta_0}^{\theta_1} pE \sin\theta d\theta$
$W = pE [-\cos\theta]_{\theta_0}^{\theta_1} = pE(\cos\theta_0 - \cos\theta_1)$

Final Result Setting reference $\theta_0 = \pi/2$ (where $U=0$ ):
$U(\theta) = -pE \cos\theta = -\vec{p} \cdot \vec{E}$

4. Electrostatics of Conductors

Important Results

Inside: Electrostatic field is zero everywhere inside the conductor.
Surface: Electric field is normal to the surface.
Potential: Constant throughout volume and surface.
Shielding: Electric field inside a cavity is zero.

5. Capacitance & Dielectrics

Capacitance is the ratio of charge to potential difference: $C = Q/V$ .

Derivation 5: Parallel Plate Capacitor 3 Marks

Cross-section of a parallel plate capacitor showing the uniform electric field between plates. — The electric field is uniform between the plates and zero outside.

Step 1: Electric Field Using Gauss’s Law, field between plates with charge density $\sigma$ :
$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$

Step 2: Potential Difference For uniform field $E$ and separation $d$ :
$V = Ed = \left( \frac{Q}{A\epsilon_0} \right) d$

Final Result Using $C = Q/V$ :
$C_0 = \frac{\epsilon_0 A}{d}$

Derivation 6: Effect of Dielectric on Capacitance 3 Marks

Polarization of a dielectric material inside a capacitor reducing the net electric field. — The polarized dielectric produces an opposing field, reducing the net field and increasing capacitance.

Step 1: Polarization Dielectric polarized creates induced surface charge $\sigma_p$ .
Net field $E = \frac{\sigma - \sigma_p}{\epsilon_0} = \frac{\sigma}{K\epsilon_0}$ (since $K = \frac{E_0}{E}$ ).