What Is the Electric Potential Inside a Pipe?

[aftershock]

Homework Statement

Homework Equations

∇²V = 0

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.

 

[vela]

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

 

[aftershock]

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

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.

 

[vela]

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".

 

[aftershock]

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".

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.

 

[aftershock]

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

From separation of variables I got:

V(x,y) = (Ae^kx+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:

(Ae^kx+Be^-kx)sin(ky)

-A = B by the first condition:

A[e^kx-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:

V_o(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 V_o(y) isn't constant? Still though... why?

 

[aralbrec]

By the 4th boundary condition:

V_o(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 V_o(y) isn't constant? Still though... why?

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.

 

[vela]

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

From separation of variables I got:

V(x,y) = (Ae^kx+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:

(Ae^kx+Be^-kx)sin(ky)

-A = B by the first condition:

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

k = n∏/b by the third condition:

Asinh(n∏x/b)sin(n∏y/b)

Each value of n corresponds to a solution V_n, 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 $k_n = n\pi/b$.

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:

V_o(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 V_o(y) isn't constant? Still though... why?