# Homework Help: Griffiths Problem 3.15

1. Mar 10, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
This question refers to Griffiths E and M book.

2. Relevant equations

3. The attempt at a solution
Obviously this is a 3D problem and we should use Example 3.5 as a model. However, since the region of interest is bounded in all 3 directions doesn't that mean that all of C_1, C_2, and C_3 need to be negative? But then C_1+C_2+C_3=0 is impossible! The constants C_i that I am referring to are on page 135.

2. Mar 11, 2008

### Mindscrape

Laplacian

$$\nabla^2 V = 0$$

After separation of variables:

$$\frac{1}{F}\frac{d^2 F}{dx^2} + \frac{1}{G}\frac{d^2 G}{dy^2} + \frac{1}{H}\frac{d^2 H}{dz^2} = 0$$

Argument: How can a function of x and y and z added together be zero always?

3. Mar 11, 2008

### ehrenfest

That is my point! If
$$\frac{1}{F}\frac{d^2 F}{dx^2}=C_1$$
$$\frac{1}{G}\frac{d^2 G}{dy^2}=C_2$$
$$\frac{1}{H}\frac{d^2 H}{dz^2} =C_3$$

Then C_1 +C_2+C_3 must be 0 which means that not all the C_i can be negative, which is absurd because we need trig functions not exponentials since the region is bounded in all three directions!?!?

4. Mar 11, 2008

### Mindscrape

Absolutely, positive constants will give the trivial solution.

Why not say -C1-C2-C3=0? Does it really matter whether the constants we make are negative or positive, aren't they allowing the functions to *sum* to zero in either case?

5. Mar 11, 2008

### nicksauce

Why not exponentials? It seems like the potential will start at Vo at z = a, then decay (exponentially) down the z axis.

6. Mar 11, 2008

### Mindscrape

Try to apply the boundary conditions to exponentials. It may be easier if you write them in terms of sinh and cosh in this case.