A capacitor consists of two stationary plates shaped as a semi-circle of radius R and a movable plate made of dielectric with permittivity ‘ $ε$ ’ and capable of rotating about an a axis O between the stationary plate. The thickness of movable plate is equal to d which is practically the separation between the stationary plates. A potential difference V is applied to the capacitor. Find magnitude of the moment of forces relative to the axis O acting on the movable plate in the position shown

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$- \frac{(ε - 1) ε_{0} R^{2} V^{2}}{4 d}$

$- \frac{(ε - 1) ε_{0} R^{2} V^{2}}{2 d}$

$- \frac{(ε - 1) ε_{0} R^{2} V^{2}}{d}$

$- \frac{(ε - 1) ε_{0} R^{2} V^{2}}{6 d}$

answer is A.

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Detailed Solution

Moment of force $= \frac{dU}{d ϕ} = \frac{1}{2} V^{2} \frac{dC}{d ϕ}$

$C = \frac{ε_{0} R^{2} ϕ}{2 d} + \frac{ε_{0} ε (π - ϕ) R^{2}}{2 d}$

$\frac{dC}{dϕ} = \frac{ε_{0} R^{2}}{2 d} - \frac{{εε}_{0} R^{2}}{2 d} = - (ε - 1) \frac{ε_{0} R^{2}}{2 d}$

$\frac{dU}{d ϕ} = - \frac{V^{2} (ε - 1) ε_{0} R^{2}}{4 d}$

-ve sign indicates moment of force is clockwise (ie., trying to suck in dielectric)

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