- #1

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The charge q(t) on the plates of a capacitor at time t is shown as a saw-tooth wave with period [itex]2\pi[/itex] and its peak is at [itex]t = \pi[/itex], where q(t) = Q.

At first, I tried to use the expressions for open boundary conditions (as q(t) is 0 at t = 0 and t = 2 pi). I got stuck, so I started again, with

[tex]q(t) = \frac{a_{0}}{\sqrt{L}} + \sum_{n=1}^\infty a_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} + \sum_{n=1}^\infty b_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L}[/tex]

where

[tex]a_{0} = \int_{0}^{L} \left(\frac{1}{L}\right)^\frac{1}{2} q(t) dt[/tex]

[tex]a_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} q(t) dt[/tex]

[tex]b_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L} q(t) dt[/tex]

Now, since the saw-tooth is symmetric, there was no need to calculate [itex]b_{n}[/itex], since they'd have been 0 anyway.

I represented q(t) using top hat functions:

[tex]q(t) = \left[\theta(t) - \theta(t - \frac{L}{2})\right]\frac{2tQ}{L} + \left[\theta(t - \frac{L}{2}) - \theta(t - L)\right]\frac{2(L - t)Q}{L}[/tex]

Then I used that to calculate [itex]a_{0}[/tex] and [itex]a_{n}[/itex], using [itex]L = 2\pi[/itex]

After pages of working out, I got:

[tex]a_{0} = Q\pi \left(\frac{1}{2\pi}\right)^\frac{1}{2}[/tex]

[tex]a_{n} = \left(\frac{1}{\pi}\right)^\frac{1}{2} \frac{Q}{\pi} \frac{2}{n^2} (\cos n\pi - 1)[/tex]

So,

[tex]q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt \cos (n\pi - 1)}{n^2}\right][/tex]

The book's answer is [tex]q(t) = Q\left[\frac{1}{2} - \frac{4}{\pi^2} \sum_{n=1}^\infty \frac{\cos (2n -1)t}{(2n - 1)^2}\right][/tex]

I don't know where I've gone wrong. I can't post all my working, as it's a lot and would take forever to LaTeX.