Discussion: Rotated Capacitor Plate

Consider a charged capacitor made with two square plates of side length L L , uniformly charged, and separated by a very small distance d d . The EMF across the capacitor is ξ \xi . One of the plates is now rotated by a very small angle θ \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 d , θ \theta , ξ \xi , and L 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
5 years, 11 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 - 5 years, 11 months ago

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