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.

#Physics #IPhOO #PhysicsProblem

Easy Math Editor

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Comments

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

jatin yadav - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$