Advanced EM problem involving summation of 1/Cosh

[gysush]

We solve Poisson's equation in cartesian coords for a region bounded by planes forming a box. Some of the planes are grounded. The lengths of the box are L1, L2, L3. There is no charge distribution.

Let a=mPi/L2
b=nPi/L3

The solutions goes like: sinh(sqrt(a^2 + b^2)*x)sin(a*y)sin(b*z)

with our coefficients going like: 1/(m*n)*1/sinh(sqrt(a^2 + b^2)*L1) (neglecting factors of Pi and constants)

The solution is valid only for odd n,m from 1 to infinity.

I want to show that the avg value of the potential over all the sides is equal to the potential at the center:

V(L1/2,L2/2,L3/2) = prescribed equation asked to show.

So I evaluate all my terms at the center. I have some cancellation of the sinh terms because sinh(2x)=2*sinh(x)*cosh(x)

I make a change of variables from m,n to 2k-1,2j-1 so that the sum goes from 1 to infinity for all positive integers and thus allowing me to change my two sin terms into (-1)^j+k

So what I am trying to sum is this:

Sum(j:1,00)Sum(k:1,00) (-1)^j+k *1/[(2k-1)(2j-1)]1/cosh(sqrt(a^2 + b^2)*L1)

a=(2j-1)Pi/L2
b=(2k-1)Pi/L3

Kinda stumped as to any analytical route without trying to use software.

 

[mark726]

Thank you for sharing your approach to solving the problem. It seems like you have made some good progress in evaluating the potential at the center of the box. However, in order to show that the average value of the potential over all sides is equal to the potential at the center, you will need to show that the sum you have written converges to a finite value. This can be done by using the alternating series test, which states that if the terms of a series alternate in sign and decrease in magnitude, then the series converges.

In your case, the terms of the series alternate in sign due to the (-1)^j+k factor, and the magnitude decreases as the indices j and k increase. Therefore, you can use the alternating series test to show that the sum converges to a finite value. Once you have shown this, you can then equate the value of the sum to the potential at the center of the box, thus proving that the average value of the potential over all sides is equal to the potential at the center.

I hope this helps in your analytical approach to the problem. Good luck!