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

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance provides concise notes, important formulas, step-by-step derivations, and textbook solutions covering electric potential, equipotential surfaces, dielectrics, capacitance, and energy storage.

Overview of Topics & Subtopics (Click to Jump)

1. Electrostatic Potential & Potential Energy

Short Note:

• Electrostatic Potential (V): The amount of work done by an external force in bringing a unit positive charge from infinity to a specific point without acceleration: V = W / q₀. SI unit is Volt (V) or Joules per Coulomb (J/C).

• Potential Difference (ΔV): Difference in potential between two points B and A: V_B - V_A = W_AB / q₀.

• Electrostatic Potential Energy (U): Work done to assemble a system of charges. For two point charges q₁ and q₂ separated by distance r: U = (1 / 4πε₀) × (q₁ q₂ / r).

• Relation between Field and Potential: Electric field is the negative gradient of potential: E = -dV / dr.

2. Potential due to a Point Charge & Dipole

Short Note:

• Potential due to Point Charge: V = (1 / 4πε₀) × (q / r). Note that V ∝ 1/r (whereas E ∝ 1/r²).

• Potential due to an Electric Dipole: At a distance r making angle θ with dipole moment p: V = (1 / 4πε₀) × (p cos θ / r²).

  - On Axial Line (θ = 0°): V_axial = (1 / 4πε₀) × (p / r²)

  - On Equatorial Line (θ = 90°): V_eq = 0

• Potential due to System of Charges: Superposition principle applies directly as algebraic sum: V = V₁ + V₂ + V₃ + ...

3. Equipotential Surfaces & Electrostatics of Conductors

Short Note:

• Equipotential Surface: Any surface that has a constant electrostatic potential at every point on it.

  - No work is done in moving a test charge over an equipotential surface (W = q ΔV = 0).

  - Electric field lines are always perpendicular to equipotential surfaces.

  - Equipotential surfaces are closer in regions of strong electric field and farther apart in weak fields.

• Electrostatics of Conductors:

  - Electric field inside a conductor is zero (E = 0).

  - Electrostatic potential is constant throughout the volume and equal to its value on the surface.

  - Excess charges reside entirely on the outer surface of a conductor.

  - Electrostatic Shielding: The field inside a cavity of a conductor is always zero, shielding the interior from external electric fields.

4. Capacitors, Capacitance & Dielectrics

Short Note:

• Capacitance (C): Measure of ability to store electric charge: C = Q / V. SI unit is Farad (F).

• Parallel Plate Capacitor: Consists of two conducting plates of area A separated by distance d: C₀ = ε₀ A / d.

• Dielectric Constant (K): Ratio of original field to reduced field: K = E₀ / E = C / C₀.

• Capacitance with Dielectric Slab: When a slab of dielectric constant K fully fills space between plates: C = K × C₀ = (K ε₀ A) / d.

• Dielectric Polarization (P): Induced dipole moment per unit volume inside a dielectric under an external electric field.

5. Energy Stored in a Capacitor & Combinations

Short Note:

• Combinations of Capacitors:

  - Series Combination: Charge Q is same on each capacitor, 1 / C_eq = 1 / C₁ + 1 / C₂ + ...

  - Parallel Combination: Potential V is same across each capacitor, C_eq = C₁ + C₂ + ...

• Energy Stored in Capacitor (U): Work done in charging a capacitor is stored as electrostatic potential energy:

  - U = (1 / 2) Q V = (1 / 2) C V² = Q² / (2 C)

• Energy Density (u): Energy stored per unit volume between capacitor plates: u = (1 / 2) ε₀ E².

Intext & Conceptual Questions

Q1: Why are electric field lines always perpendicular to equipotential surfaces?
Answer:
The work done in moving a charge q by distance dr along an equipotential surface is zero because potential remains constant (dW = q dV = 0). Since dW = F · dr = q (E · dr) = q E dr cos θ = 0, and neither q, E, nor dr is zero, cos θ must be 0, which means θ = 90°. Thus, the electric field lines are always perpendicular to the equipotential surface.
Q2: What happens to the capacitance, charge, potential difference, and stored energy of a parallel plate capacitor when a dielectric slab is inserted while (i) battery remains connected, and (ii) battery is disconnected?
Answer:
• (i) When Battery Remains Connected:
  - Potential Difference (V): Remains constant (V = V₀).
  - Capacitance (C): Increases by factor K (C = K C₀).
  - Charge (Q): Increases by factor K (Q = K Q₀).
  - Stored Energy (U): Increases by factor K (U = K U₀).

• (ii) When Battery is Disconnected:
  - Charge (Q): Remains constant (Q = Q₀).
  - Capacitance (C): Increases by factor K (C = K C₀).
  - Potential Difference (V): Decreases by factor K (V = V₀ / K).
  - Stored Energy (U): Decreases by factor K (U = U₀ / K).

Textbook Chapter End Exercise Questions

Q1: Two charges 5 × 10⁻⁸ C and -3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Answer:
Let q₁ = 5 × 10⁻⁸ C and q₂ = -3 × 10⁻⁸ C placed distance d = 16 cm = 0.16 m apart.

Case 1: Point P lies between the charges at distance x from q₁
V_total = V₁ + V₂ = 0
(1 / 4πε₀) × (q₁ / x) + (1 / 4πε₀) × (q₂ / (d - x)) = 0
q₁ / x = -q₂ / (0.16 - x)
(5 × 10⁻⁸) / x = (3 × 10⁻⁸) / (0.16 - x)
5(0.16 - x) = 3x => 0.80 - 5x = 3x => 8x = 0.80 => x = 0.10 m = 10 cm

Case 2: Point P lies outside the line segment beyond q₂ at distance x from q₁
q₁ / x = -q₂ / (x - 0.16)
(5 × 10⁻⁸) / x = (3 × 10⁻⁸) / (x - 0.16)
5(x - 0.16) = 3x => 5x - 0.80 = 3x => 2x = 0.80 => x = 0.40 m = 40 cm

Therefore, the electric potential is zero at a distance of 10 cm (between the charges) and 40 cm (outside towards the negative charge) from q₁.
Q2: Distinguish between Series and Parallel Combination of Capacitors.
Answer:
Feature Series Combination Parallel Combination
Charge (Q) Same charge Q flows through each capacitor. Total charge splits across individual capacitors (Q = Q₁ + Q₂).
Voltage (V) Total potential divides (V = V₁ + V₂). Same potential difference V across each capacitor.
Equivalent Capacitance Reciprocal sum: 1/C_eq = 1/C₁ + 1/C₂ + ... Direct sum: C_eq = C₁ + C₂ + ...
Net Capacitance Value Decreases (C_eq is smaller than smallest individual C). Increases (C_eq is larger than largest individual C).
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Class 12 › Physics › Electrostatic Potential and Capacitance
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Electrostatic Potential & Potential EnergyPotential due to a Point Charge & DipoleEquipotential Surfaces & Electrostatics of ConductorsCapacitors, Capacitance & DielectricsEnergy Stored in a Capacitor & Combinations