Non-convergent series for surface charge density

[musemonkey]

1. This is a 2D Laplace eqn problem. A semi-infinite strip of width a has a conductor held at potential V(0,a) = V_0 at one end and grounded conductors at y=0 and y=a. Find the induced surface charge \sigma (y) on the conductor at x=0.



2. Homework Equations .

The potential is

V(x,y) = \frac{4V_0}{\pi}\sum_{n=0}^{\infty}\frac{e^{-(2n+1)\pi x / a}}{2n+1} \sin(\frac{(2n+1)\pi y}{a} ).


3. I tried

\sigma = \epsilon_0 E_{norm} = -\epsilon_0 \left . \frac{\partial V}{\partial x} \right |_{x=0}

which yielded the bounded but non-convergent series

\sigma(y) = \frac{4V_0\epsilon_0}{a} \sum_{n=0}^{\infty} \sin ( \frac{(2n+1)\pi y}{a} ).

Differentiation with respect to x killed the 1/(2n+1) factor and then evaluating at x=0 killed the exponential, leaving nothing to cause the terms to decay with higher n. It makes me doubt that using the normal derivative of the potential is the right way to get the field, but everything I've read on it states with demonstration that the field at the surface of a conductor is - \epsilon_0 \partial V / \partial n. So what to do?

Thanks for reading!

 

[weejee]

Evaluate for x>0 and then take the x->0 limit. It'll converge.

 

[musemonkey]

Thank you Weejee. A further question:

Before taking the limit, the series is

-\frac{\partial V}{\partial x} = \frac{4V_0}{a}\sum_{n=0}^{\infty}e^{-(2n+1)\pi x / a} \sin(\frac{(2n+1)\pi y}{a} ).

I don't know how to get a closed form for the sum of this series, and even if I could, there are no doubt cases in which it can't be done; so are you saying that in general one writes the solution as

\sigma(y) = -\epsilon_0 E = \lim_{x->0} \epsilon_0 \frac{4V_0}{a}\sum_{n=0}^{\infty}e^{-(2n+1)\pi x / a} \sin(\frac{(2n+1)\pi y}{a} ) ?

I can see how this is not equivalent to just plugging in x=0 inside the series, and I can see that by the ratio test this series converges, but is there really no closed form for \sigma?

 

[weejee]

One can write the summand as the imaginary part of something. You'll find that it is just a geometric series. Obtaining the sum is quite trivial, but taking its imaginary part needs some algebraic manipulation.

 

[musemonkey]

oh I see, Euler's formula -- nice. I'll give this a try. Thanks!