A capacitor consists of two stationary parallel plates shaped as a semi-circle of radius $R$. A mova

Question

A capacitor consists of two stationary parallel plates shaped as a semi-circle of radius $R$. A movable plate made of a uniform dielectric with relative permittivity $K$ and same semi-circular shape is capable of rotating about an axis passing through the centre of the straight edges. The thickness of the movable dielectric plate is $d$, which is practically the separation between the stationary plates. A constant potential difference $V$ is applied across the stationary plates. When the movable plate has been rotated by an angle $	heta$ (where $0 < 	heta < \pi$) into the space between the plates, what is the magnitude of the electrostatic torque acting on the movable plate?

Professor PC Thomas Classes · Answer

The system can be modeled as two parallel capacitors: one filled with dielectric and the other with air. The area of the plates with the dielectric inserted is $A_d = \frac{1}{2} R^2 	heta$. The remaining area with air is $A_a = \frac{1}{2} R^2 (\pi - 	heta)$. The total capacitance as a function of $	heta$ is $C(	heta) = \frac{K \epsilon_0 (\frac{1}{2} R^2 	heta)}{d} + \frac{\epsilon_0 (\frac{1}{2} R^2 (\pi - 	heta))}{d} = \frac{\epsilon_0 R^2}{2d} [K	heta + (\pi - 	heta)]$. Since the voltage $V$ is maintained constant, the electrostatic torque $	au$ is given by the derivative of electrostatic energy with respect to $	heta$: $	au = \frac{\partial U}{\partial 	heta} = \frac{1}{2} V^2 \frac{dC}{d	heta}$. Differentiating $C(	heta)$ yields $\frac{dC}{d	heta} = \frac{\epsilon_0 R^2}{2d} (K - 1)$. Substituting this into the torque equation gives $	au = \frac{1}{2} V^2 \left( \frac{\epsilon_0 R^2 (K - 1)}{2d} ight) = \frac{\epsilon_0 (K-1) R^2 V^2}{4d}$.