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 d << L and θ0 \theta \approx 0 .

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

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

Note by Ahaan Rungta
6 years, 3 months ago

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

It is easy to find the overlapping area as A=L2(12tanθ(1+tanθ+secθ)2)\displaystyle A = L^2 \bigg(1 - \frac{2 \tan \theta}{(1+ \tan \theta + \sec \theta)^2}\bigg)

Hence, C=ϵ0Ad=ϵL2d(12tanθ(1+tanθ+secθ)2)\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, Q=2ϵ0L2ζd(12tanθ(1+tanθ+secθ)2)\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 θ0\displaystyle \theta \approx 0, Q2ϵ0L2ζd(1θ2)\triangle Q \approx \frac{2 \epsilon_{0} L^2 \zeta }{d} \bigg(1 - \frac{\theta}{2}\bigg)

jatin yadav - 6 years, 3 months ago

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