Using Separation of Variables to find potential inside of a sphere

[mdwerner]

Homework Statement

A cubical box (sides of length a) consists of five metal plates, which are welded together and grounded (Fig 3.23). The top is made of a separate sheet of metal, insulated from the others, and held at a constant potential V₀. Find the potential inside the box.

Homework Equations

If $$\frac{1}{X}$$ $$\frac{ d^{2} X}{dx^{2}}$$ = C_x , then the solution is in an exponential form, otherwise is is in the trigonometric form.

The Attempt at a Solution

Because in the X and Y directions the potential function must be zero twice, I decided that they must be of the trigonometric solution form. In the Z direction I chose it to have an exponential solution.
I decided that the boundary conditions were the following :
V(x,0,z) = 0
V(0,y,z) = 0
V(x,y,0) = 0
V(x,a,z) = 0
V(a,y,z) = 0
V(x,y,a) = V₀

Applying these I decided that
X(x) = A sin [(n * π / a) x]
Y(y) = C sin [(m * π / a) y]
and
Z(z) = E $$e^{\sqrt{k^{2} + l^{2}} z}$$ - E $$e^{-\sqrt{k^{2} + l^{2}} z}$$

However, I do not know where to go from here. I have not applied the last boundary condition that I listed, but I do not see how it would simplify the expression...are my boundary conditions correct? I don't see how I can solve any farther to find the coefficients A,C, and E.
Any advice on how to solve this things or criticism of my work will be much appreciated.

 

[Mindscrape]

So you have a 3D PDE that is not too different from the 2D. You tell me, what are the boundary conditions?

 

[Mindscrape]

Ah, okay, I see your edit now. The boundary conditions are fine. Try using cosh and sinh as the solution to Z(z). Exponentials are really only good if you are dealing with a boundary condition of infinity. Also, remember how the coefficients all relate to each other. :)

 

[mdwerner]

What do you mean "remember how all the coefficents relate to each other..." ? I am unaware of any relationship between the coefficients.

Also, I don't see how using the cosh helps -

My problem is that I don't understand how the last boundary condition that I listed can be applied. I already have expressions for the variables k and l (From the equations X and Y), so the only variable left when applying V(x,y,a) = V₀ is the coefficient E. Is this correct? I have been reading other similar problems like this and I do not see any which solve to get EXACT values for the coefficients...EDIT : Actually, nevermind - the cosh does help, as it allows boundary condition 3 to cancel out the cosh term, leaving only the sinh term. Thanks for that bit of advice :)

 

[Mindscrape]

Yeah, using cosh and sinh will let you get rid of the cosh from V(z=0)=0.

So back in the very beginning of the problem you would have had something that looked like this

$$\frac{1}{X}\frac{\partial^2 X}{\partial x^2} + \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2} + \frac{1}{Z}\frac{\partial^2 Z}{\partial z^2}=0$$

and the only way for that equation to work out is if each individual piece is a constant. So you made

$$\frac{1}{X}\frac{\partial^2 X}{\partial x^2}=k_x^2$$
$$\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2}=k_y^2$$
$$\frac{1}{Z}\frac{\partial^2 Z}{\partial z^2}=-k_z^2$$

Which gives the constraint that

$$k_x^2+k_y^2-k_z^2=0$$

So for your Z equation I imagine you are stuck here

$$Z(z)=A sinh(k_z z)$$

You know k_z in terms of the other constants k_x and k_y. You can get A from the boundary conditions.

Putting all that together you just use orthogonality in the end to determine the coefficients of n and m (griffiths calls them c_n and c_m respectively).

 

[leonne]

Hey how do we know which constant is neg or positive , where u have k² does it matter ?
thxs