# Discussion: Rotated Capacitor Plate

Consider a charged capacitor made with two square plates of side length $L$, uniformly charged, and separated by a very small distance $d$. The EMF across the capacitor is $\xi$. One of the plates is now rotated by a very small angle $\theta$ to the original axis of the capacitor. Find an expression for the difference in charge between the two plates of the capacitor, in terms of (if necessary) $d$, $\theta$, $\xi$, and $L$.

Also, approximate your expression by transforming it to algebraic form: i.e. without any non-algebraic functions. For example, logarithms and trigonometric functions are considered non-algebraic. Assume $d << L$ and $\theta \approx 0$.

Hint: You may assume that $\frac {\theta L}{d}$ is also very small.

Note: This problem was originally proposed by Trung Phan for the IPhOO.

Note by Ahaan Rungta
5 years, 2 months ago

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Draw two rotated squares and find area common to both. Find the new capacitance, and apply $Q = C \zeta$, to get $\triangle Q = Q - (-Q ) = 2C \zeta$

It is easy to find the overlapping area as $\displaystyle A = L^2 \bigg(1 - \frac{2 \tan \theta}{(1+ \tan \theta + \sec \theta)^2}\bigg)$

Hence, $\displaystyle C = \frac{\epsilon_{0} A }{d} = \frac{\epsilon L^2}{d} \bigg(1 - \frac{2 \tan \theta}{(1+ \tan \theta + \sec \theta)^2}\bigg)$

Thus, $\displaystyle \triangle Q = \frac{2 \epsilon_{0} L^2 \zeta }{d} \bigg(1 - \frac{2 \tan \theta}{(1+ \tan \theta + \sec \theta)^2}\bigg)$

When $\displaystyle \theta \approx 0$, $\triangle Q \approx \frac{2 \epsilon_{0} L^2 \zeta }{d} \bigg(1 - \frac{\theta}{2}\bigg)$

- 5 years, 2 months ago