Q109 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area A/3 and separation between the plates is d, 2d and 3d respectively. The equivalent capacity of the combination is ($\epsilon_0$-permittivity of free space)
- A$\dfrac{9\epsilon_0 A}{17d}$
- B$\dfrac{11\epsilon_0 A}{17d}$
- Correct answerC$\dfrac{11\epsilon_0 A}{18d}$
- D$\dfrac{9\epsilon_0 A}{14d}$
Solution
The capacitance of a parallel plate capacitor is given by $C = \dfrac{\epsilon_0 A'}{d'}$
For the three capacitors, the capacitances are:
$C_1 = \dfrac{\epsilon_0 (A/3)}{d} = \dfrac{\epsilon_0 A}{3d}$
$C_2 = \dfrac{\epsilon_0 (A/3)}{2d} = \dfrac{\epsilon_0 A}{6d}$
$C_3 = \dfrac{\epsilon_0 (A/3)}{3d} = \dfrac{\epsilon_0 A}{9d}$
Since the capacitors are connected in parallel, the equivalent capacitance is the sum of individual capacitances:
$C_{eq} = C_1 + C_2 + C_3$
$C_{eq} = \dfrac{\epsilon_0 A}{3d} + \dfrac{\epsilon_0 A}{6d} + \dfrac{\epsilon_0 A}{9d}$
$C_{eq} = \dfrac{\epsilon_0 A}{d} \left( \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{9} \right)$
$C_{eq} = \dfrac{\epsilon_0 A}{d} \left( \dfrac{6 + 3 + 2}{18} \right) = \dfrac{11\epsilon_0 A}{18d}$
Answer: $\dfrac{11\epsilon_0 A}{18d}$
For the three capacitors, the capacitances are:
$C_1 = \dfrac{\epsilon_0 (A/3)}{d} = \dfrac{\epsilon_0 A}{3d}$
$C_2 = \dfrac{\epsilon_0 (A/3)}{2d} = \dfrac{\epsilon_0 A}{6d}$
$C_3 = \dfrac{\epsilon_0 (A/3)}{3d} = \dfrac{\epsilon_0 A}{9d}$
Since the capacitors are connected in parallel, the equivalent capacitance is the sum of individual capacitances:
$C_{eq} = C_1 + C_2 + C_3$
$C_{eq} = \dfrac{\epsilon_0 A}{3d} + \dfrac{\epsilon_0 A}{6d} + \dfrac{\epsilon_0 A}{9d}$
$C_{eq} = \dfrac{\epsilon_0 A}{d} \left( \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{9} \right)$
$C_{eq} = \dfrac{\epsilon_0 A}{d} \left( \dfrac{6 + 3 + 2}{18} \right) = \dfrac{11\epsilon_0 A}{18d}$
Answer: $\dfrac{11\epsilon_0 A}{18d}$
Hence, the correct answer is option: 'C'.