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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply