How can I plot equipotentials from this series?....

[Daniel Sellers]

Homework Statement

The problem is given in the attached picture, but I already have a solution to part a) which I am confident in (I have checked it carefully, compared to other students and confirmed it with my graduate-TA).

Part b) asks us to plot the equipotentials but I cannot figure out how. I cannot get this infinite series into a form which I can use to plot. I have tried several online calculators, I have tried separating the series into two different series and re-writing it a number of different ways but I cannot get it into a usable form.

Homework Equations

V(r,Θ) = Σ (2V₀/a)[(1-(-1)^n)/cosh(n*pi)]cosh(n*pi*y/a)sin(n*pi*x/a)

The Attempt at a Solution

See above. I think I can answer part c) and I have an answer to part d) I just need a form of this solution which can be used to plot equipotentials in the x-y plane.

 

[tnich]

V(r,Θ) = Σ (2V₀/a)[(1-(-1)^n)/cosh(n*pi)]cosh(n*pi*y/a)sin(n*pi*x/a)

As near as I can tell, you mean to say that your result is
$V(x,y) = \sum_{n=1}^{\infty} \frac {2V_0 [1-(-1)^n] \cosh(n\pi \frac y a) \sin(n\pi \frac x a)} {a \cosh n\pi}$

I don't see how this could give you $V = V_0$ at $y = a$. At $y= a$ your solution would look something like
$V(x,a) = \frac {4V_0} a ( \sin(\pi \frac x a)+ \sin(3\pi \frac x a)+ \sin(5 \pi \frac x a) + \dots)$

but what you need is a square wave. You are almost there, but it is not quite right.

 

[vela]

Homework Statement

The problem is given in the attached picture, but I already have a solution to part a) which I am confident in (I have checked it carefully, compared to other students and confirmed it with my graduate-TA).

Your answer to (a) looks wrong just from looking at the units. Your expression has units of volts/meter, but it should just be volts.

As far as plotting the equipotentials go, you might simply be expected to use software that can do that for you.

 

[Daniel Sellers]

Thanks to both of you who replied. You are both correct. There is no 'a' in the denominator, my mistake. I plotted 1000 terms of the (correct) series in Mathematica in order to read off the equipotentials.

The series does converge to a constant at y = a.