NCERT Books Free Download for Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

Some frequently asked 1-mark questions for the board exam include:

• Defining equipotential surfaces and stating one key property.
• What is the S.I. unit of capacitance? (Answer: Farad).
• Why is the electrostatic potential constant throughout the volume of a conductor?
• Drawing the equipotential surfaces for a single point charge or an electric dipole.
• What is the net work done in moving a test charge over a closed equipotential surface? (Answer: Zero).

For a 3-mark question, you should prepare the following key derivations:

• Deriving the expression for the electric potential at a point due to an electric dipole (both at an axial and equatorial point).
• Deriving the expression for the potential energy of a system of two point charges in the absence of an external field.
• Establishing the relationship between the electric field and electric potential (E = -dV/dr).
• Deriving the formula for the equivalent capacitance when capacitors are connected in series or parallel.

A classic 5-mark question involves the parallel plate capacitor. It is often asked in parts, such as:

• Part (a): Define capacitance and derive the expression for the capacitance of a parallel plate capacitor with air as the medium between the plates.
• Part (b): Explain how the capacitance changes when a dielectric slab of dielectric constant 'K' is introduced between the plates. Derive the new expression for capacitance.
• Part (c): A short numerical problem based on the derived formulas.

Mastering this entire topic is crucial for scoring full marks in the long answer section.

The electric field must be perpendicular to an equipotential surface because if it were not, there would be a component of the field along the surface. This component would exert a force on a test charge moving along the surface, meaning work would have to be done (W = F.d). However, by definition, an equipotential surface is one where the potential is constant, so the potential difference between any two points on it is zero. Since work done (W) is equal to charge (q) times potential difference (ΔV), the work done is zero. For the work done to be zero despite moving a distance, the angle between the force (and thus the electric field) and the displacement must be 90°. Therefore, the electric field is always normal to the equipotential surface.

Both increase the capacitance, but through different mechanisms:

• Dielectric Slab: When a dielectric is inserted, it gets polarized. An internal electric field is induced within the dielectric, which opposes the external field between the plates. This reduces the net electric field and consequently the potential difference (V = Ed) between the plates. Since capacitance C = Q/V, a decrease in V leads to an increase in capacitance by a factor 'K' (the dielectric constant).
• Conducting Slab: A conductor has free charge carriers. When placed in the electric field, these charges redistribute to make the electric field inside the conductor zero. This effectively reduces the distance between the plates where the electric field exists. As capacitance is inversely proportional to the distance (C ∝ 1/d), reducing the effective distance significantly increases the capacitance. For a conductor, the dielectric constant 'K' is considered infinite.

No, this is a common misconception. The electric field inside a charged hollow spherical conductor is zero. However, the electrostatic potential is not zero; it is constant and equal to the potential on the surface. Since E = -dV/dr, an electric field of zero (E=0) implies that the rate of change of potential with distance (dV/dr) is zero. This means the potential (V) must be constant. Therefore, the potential is the same everywhere inside the sphere and on its surface, and its value is given by V = kQ/R, where R is the radius of the sphere.

To excel in numericals from this chapter, you must be thorough with the following formulas:

• Electric Potential: V = kQ/r (for a point charge).
• Potential due to a Dipole: Formulas for axial and equatorial points.
• Capacitance: C = Q/V.
• Parallel Plate Capacitor: C = ε₀A/d (air-filled) and C = Kε₀A/d (with dielectric).
• Combination of Capacitors: 1/C_s = 1/C₁ + 1/C₂ + ... (Series) and C_p = C₁ + C₂ + ... (Parallel).
• Energy Stored in a Capacitor: U = ½CV² = ½Q²/C = ½QV.

The potential energy of a dipole in an external field is a measure of the work done in orienting it against the field's torque. The field naturally exerts a torque (τ = pE sinθ) that tries to align the dipole moment (p) with the electric field (E). A system is in its most stable, lowest energy state when it is aligned with the forces acting on it. When the dipole aligns with the field (θ = 0°), the torque on it becomes zero, and it reaches a state of stable equilibrium. To move it away from this alignment would require external work, which would be stored as potential energy. Therefore, the state of minimum potential energy occurs when the dipole is fully aligned with the field.

The Van de Graaff generator is an important long-answer question because it tests several core concepts of electrostatics.
Principle: It works on two main principles: (1) The action of sharp points (corona discharge), where charge density is very high, and (2) The property that charge given to a hollow conductor resides on its outer surface and the potential inside is constant.
Working: A conveyor belt made of insulating material carries charge sprayed onto it from a sharp comb (spray comb) at the bottom. The belt moves up, carrying the charge to the inside of a large hollow metallic sphere. Another comb (collecting comb) inside the sphere removes the charge from the belt and transfers it to the outer surface of the sphere. This process is repeated, accumulating a very large amount of charge on the sphere, thereby generating a very high potential of several million volts.