# Electric potential inside a pipe

1. Oct 21, 2012

### aftershock

1. The problem statement, all variables and given/known data

2. Relevant equations

2V = 0

3. The attempt at a solution

I'm guessing I need to use Laplace's equation in 2 dimensions since the potential depends on x and y. I have no idea how I would do this.

2. Oct 21, 2012

### vela

Staff Emeritus
This is a pretty standard type of problem. Consult your textbook for similar examples, and then post your work.

3. Oct 21, 2012

### aftershock

The reason I'm having trouble is my textbook doesn't really have any examples for this. I figured I could at least find a similar example online but I'm not even sure what to search for.

4. Oct 21, 2012

### vela

Staff Emeritus
That's surprising. You could check another E&M text. It's also covered in math methods texts, like Arfken, as well. Look up "separation of variables" and "Laplace's equation".

5. Oct 21, 2012

### aftershock

There actually is at least one example in the separation of variable section, I didn't think I needed to go into that for this problem, thanks.

6. Oct 23, 2012

### aftershock

Ok so I tried working out a solution but I went wrong somewhere.

From separation of variables I got:

V(x,y) = (Aekx+Be-kx)(Csin(ky)+Dcos(ky))

Boundary conditions:

V=0 when x=0
V=0 when y=0
V=0 when y=b
V=v0(y) when x=a

D=0 by the second condition, and I can then absorb C into A and B:

(Aekx+Be-kx)sin(ky)

-A = B by the first condition:

A[ekx-e-kx]sin(ky) = Asinh(kx)sin(ky)

k = n∏/b by the third condition:

Asinh(n∏x/b)sin(n∏y/b)

Now I'm pretty sure I'm supposed to do some sort of linear combination here to solve for A.

I did this instead. I'm pretty sure it's wrong but I'm not sure why?

By the 4th boundary condition:

Vo(y) = Asinh(n∏a/b)sin(n∏y/b)

solved for A, then plugged that into Asinh(n∏x/b)sin(n∏y/b)

I guess I can't do that since Vo(y) isn't constant? Still though... why?

7. Oct 24, 2012

### aralbrec

Sure you can do that as long is A comes out to be a constant. It must be a constant because you made the assumption that it was a constant.

The variation with y of Vo(y) is taken care of by something other than A in your solution.

8. Oct 24, 2012

### vela

Staff Emeritus
Each value of n corresponds to a solution Vn, so you should write
$$V_n(x,y) = A_n \sinh (k_nx) \sin (k_ny).$$ The general solution is a linear combination of the individual solutions
$$V(x,y) = \sum_{n=1}^\infty A_n \sinh (k_nx) \sin (k_ny)$$ where $k_n = n\pi/b$.