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Capacitance, capacitor combinations, parallel-plate capacitor and stored energy: NEET MCQs with answers

89 questions on this topic of Electrostatics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Capacitors of 10 μF and 6 μF are connected in series. The equivalent capacitance is:

  1. 4 μF
  2. 60 μF
  3. 3.75 μF
  4. 16 μF
Show answer and explanation

Answer: (C) 3.75 μF

In series, 1/C = 1/10 + 1/6, so C = 10×6/16 = 3.75 μF.

medium

2. A parallel-plate capacitor has plate area A and separation d. The space between the plates is filled with two slabs of dielectric constants K₁ = 2 and K₂ = 6, each having thickness d/2 and extending over the entire area A of the plates. If C₀ = ε₀A/d is the capacitance with air, what is the capacitance of this modified system?

  1. 3 C₀
  2. 4 C₀
  3. 1.5 C₀
  4. 8 C₀
Show answer and explanation

Answer: (A) 3 C₀

The arrangement represents two capacitors connected in series, each of plate area A and separation d/2. Their capacitances are C₁ = K₁ε₀A/(d/2) = 2K₁C₀ = 4C₀, and C₂ = K₂ε₀A/(d/2) = 2K₂C₀ = 12C₀. The equivalent capacitance is C_eq = (C₁C₂)/(C₁ + C₂) = (4C₀ × 12C₀)/(4C₀ + 12C₀) = 48/16 C₀ = 3C₀.

hard

3. A parallel-plate capacitor has capacitance C = 10 μF when filled with air. The space between the plates is now filled with two identical dielectric slabs of dielectric constants K₁ = 2 and K₂ = 4, each occupying half the area of the plates (A/2) and spanning the full separation d. The new capacitance is:

  1. 60 μF
  2. 15 μF
  3. 20 μF
  4. 30 μF
Show answer and explanation

Answer: (D) 30 μF

When the space is divided along the area, the two regions behave as two capacitors in parallel: C₁ = K₁ ε₀ (A/2) / d = (K₁/2) C = (2/2) × 10 = 10 μF, and C₂ = K₂ ε₀ (A/2) / d = (K₂/2) C = (4/2) × 10 = 20 μF. The total equivalent capacitance is C_eq = C₁ + C₂ = 10 μF + 20 μF = 30 μF.

medium

4. The space between the plates of a parallel-plate capacitor of area A and separation d is filled with two equal slabs of dielectrics of constants K₁ and K₂ such that each slab has area A/2 and thickness d. The equivalent capacitance of the combination is:

  1. (ε₀A / d) × [K₁K₂ / (K₁ + K₂)]
  2. (2ε₀A / d) × [K₁K₂ / (K₁ + K₂)]
  3. (ε₀A / 2d) × (K₁ + K₂)
  4. (ε₀A / d) × (K₁ + K₂)
Show answer and explanation

Answer: (C) (ε₀A / 2d) × (K₁ + K₂)

Dividing the area into two halves across the full separation d corresponds to two capacitors connected in parallel. C₁ = K₁ε₀(A/2)/d and C₂ = K₂ε₀(A/2)/d. The equivalent capacitance is C_eq = C₁ + C₂ = (ε₀A / 2d)(K₁ + K₂).

easy

5. A 8 μF capacitor is charged to 170 V. The energy stored in it is:

  1. 6.8 × 10⁻⁴ J
  2. 0.231 J
  3. 1.16 × 10⁵ J
  4. 0.116 J
Show answer and explanation

Answer: (D) 0.116 J

Energy stored in a charged capacitor is U = ½CV² = ½ × 8 × 10⁻⁶ × 170² = 0.116 J.

hard

6. Three identical parallel conducting plates, each of area A, are placed parallel to each other at equal separation d. The outer two plates are connected together by a thin conducting wire, while a battery of voltage V is connected between the central plate and the outer plates. The total charge supplied by the battery is:

  1. ε₀AV / d
  2. ε₀AV / (2d)
  3. 4ε₀AV / d
  4. 2ε₀AV / d
Show answer and explanation

Answer: (D) 2ε₀AV / d

The system forms two identical capacitors in parallel, each having capacitance C = ε₀A/d, because the central plate forms a capacitor with both the top and bottom outer plates. The equivalent capacitance is C_eq = 2C = 2ε₀A/d. Therefore, the total charge drawn from the battery is Q = C_eq × V = 2ε₀AV/d.

easy

7. Capacitors of 8 μF and 9 μF are connected in parallel. The equivalent capacitance is:

  1. 4.24 μF
  2. 1 μF
  3. 72 μF
  4. 17 μF
Show answer and explanation

Answer: (D) 17 μF

In parallel, capacitances add: 8 + 9 = 17 μF.

easy

8. Capacitors of 12 μF and 9 μF are connected in series. The equivalent capacitance is:

  1. 108 μF
  2. 5.14 μF
  3. 21 μF
  4. 3 μF
Show answer and explanation

Answer: (B) 5.14 μF

In series, 1/C = 1/12 + 1/9, so C = 12×9/21 = 5.14 μF.

easy

9. A 11 μF capacitor is charged to 110 V. The energy stored in it is:

  1. 66550 J
  2. 0.0665 J
  3. 6.05 × 10⁻⁴ J
  4. 0.133 J
Show answer and explanation

Answer: (B) 0.0665 J

Energy stored in a charged capacitor is U = ½CV² = ½ × 11 × 10⁻⁶ × 110² = 0.0665 J.

easy

10. Capacitors of 5 μF and 12 μF are connected in parallel. The equivalent capacitance is:

  1. 17 μF
  2. 3.53 μF
  3. 7 μF
  4. 60 μF
Show answer and explanation

Answer: (A) 17 μF

In parallel, capacitances add: 5 + 12 = 17 μF.

medium

11. Given below are two statements: Statement I: When two charged conductors at different potentials are connected by a conducting wire, there is always a loss of electrostatic potential energy, provided their initial potentials are not equal. Statement II: The loss of electrostatic energy during redistribution of charge is dissipated mainly in the form of heat and electromagnetic radiation in the connecting wire. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

The loss of energy upon sharing charges is given by ΔU = (1/2) [C₁C₂/(C₁+C₂)](V₁ - V₂)² > 0 whenever V₁ ≠ V₂. This lost electrostatic energy appears as heat produced by the current in the connecting wire and as radiated electromagnetic waves. Therefore, both statements are correct.

easy

12. A 6 μF capacitor is charged to 160 V. The energy stored in it is:

  1. 4.8 × 10⁻⁴ J
  2. 0.0768 J
  3. 76800 J
  4. 0.154 J
Show answer and explanation

Answer: (B) 0.0768 J

Energy stored in a charged capacitor is U = ½CV² = ½ × 6 × 10⁻⁶ × 160² = 0.0768 J.

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