Q182 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
A parallel plate capacitor filled with oil of dielectric constant 3 between the plates, has a capacitance 'C'. If the oil is completely removed, then the capacitance of the capacitor will be
- A$\sqrt{3}\,\text{C}$
- Correct answerB$\dfrac{\text{C}}{3}$
- C$\dfrac{\text{C}}{\sqrt{3}}$
- D$3\,\text{C}$
Solution
The capacitance of a parallel plate capacitor filled with a dielectric of constant $K$ is given by $C = \dfrac{K \epsilon_0 A}{d}$.
Given $K = 3$, the initial capacitance is $C = \dfrac{3 \epsilon_0 A}{d}$.
When the oil is completely removed, the medium between the plates becomes air or vacuum, for which the dielectric constant is $1$.
The new capacitance becomes $C' = \dfrac{\epsilon_0 A}{d}$.
From the initial equation, we have $\dfrac{\epsilon_0 A}{d} = \dfrac{C}{3}$.
Therefore, the new capacitance is $C' = \dfrac{C}{3}$.
Answer: $\dfrac{\text{C}}{3}$
Given $K = 3$, the initial capacitance is $C = \dfrac{3 \epsilon_0 A}{d}$.
When the oil is completely removed, the medium between the plates becomes air or vacuum, for which the dielectric constant is $1$.
The new capacitance becomes $C' = \dfrac{\epsilon_0 A}{d}$.
From the initial equation, we have $\dfrac{\epsilon_0 A}{d} = \dfrac{C}{3}$.
Therefore, the new capacitance is $C' = \dfrac{C}{3}$.
Answer: $\dfrac{\text{C}}{3}$
Hence, the correct answer is option: 'B'.