Q171 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Physics
A parallel-plate air capacitor has a capacity '$C$', distance with a separation '$x$', a dielectric material with a dielectric constant '$k$' between plates, and a potential difference with a '$V$' volt applied between plates. The force of attraction between the plates is
- Correct answerA$\dfrac{kCV^2}{2x}$
- B$\dfrac{kCV^2}{2}$
- C$\dfrac{kCV}{2x}$
- D$\dfrac{kCV^2}{x}$
Solution
The capacitance of the air capacitor is given by $C = \dfrac{\epsilon_0 A}{x}$, where $A$ is the area of the plates.
When a dielectric material of dielectric constant $k$ is introduced between the plates, the new capacitance becomes $C' = kC$.
The charge on the capacitor plates when a potential difference $V$ is applied is $Q = C'V = kCV$.
The magnitude of the force of attraction between the plates of the capacitor is given by $F = \dfrac{Q^2}{2k\epsilon_0 A}$.
Substituting $Q = kCV$ and $\epsilon_0 A = Cx$, we get:
$F = \dfrac{(kCV)^2}{2k(Cx)}$
$F = \dfrac{k^2C^2V^2}{2kCx}$
$F = \dfrac{kCV^2}{2x}$
Answer: $\dfrac{kCV^2}{2x}$
When a dielectric material of dielectric constant $k$ is introduced between the plates, the new capacitance becomes $C' = kC$.
The charge on the capacitor plates when a potential difference $V$ is applied is $Q = C'V = kCV$.
The magnitude of the force of attraction between the plates of the capacitor is given by $F = \dfrac{Q^2}{2k\epsilon_0 A}$.
Substituting $Q = kCV$ and $\epsilon_0 A = Cx$, we get:
$F = \dfrac{(kCV)^2}{2k(Cx)}$
$F = \dfrac{k^2C^2V^2}{2kCx}$
$F = \dfrac{kCV^2}{2x}$
Answer: $\dfrac{kCV^2}{2x}$
Hence, the correct answer is option: 'A'.