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# 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
3 years, 1 month 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 $$\displaystyle \theta \approx 0$$, $$\triangle Q \approx \frac{2 \epsilon_{0} L^2 \zeta }{d} \bigg(1 - \frac{\theta}{2}\bigg)$$ · 3 years, 1 month ago