# Using Laplace Equation in one dimension to solve for a charge-free slab

1. Oct 25, 2012

### hansbahia

1. The problem statement, all variables and given/known data
An empty (charge-free) slab shaped region with walls parallel to the yz-plane extends from x=a to x=b; the (constant) potential on the two walls is given as Va and Vb , respectively. Starting with LaPlace's equation in one dimension, derive a formula for the potential at any point in the region

2. Relevant equations

d2V/dx2=0

3. The attempt at a solution

I get the whole concept how they get the average potential since if we were to integrate twice laplace equation we would get two integration constant

d2V/dx2=0

dV/dx= a
V = ax + b

and if we were to average from V(x+a) and V(x-a) we would get

V (x) =1/2[V (x + a) + V (x-a)] and you can use for any point within

how did the answer to this problem came out to be

V(x)=[Vb-Va)x+Vab-Vba]/b-a

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

2. Relevant equations

3. The attempt at a solution

2. Oct 25, 2012

### Staff: Mentor

You are using the symbols a and b to represent two different things. If you represent the potential by

V = Cx + D

and use the two boundary conditions to solve for C and D, you get

V = Va + (Vb - Va) (x -a)/(b -a)

which is the same as the answer you were trying to match.

3. Oct 26, 2012

### hansbahia

So if at we have Q+ at a and Q- at b, from a to b Vb=0 and at a the potential will be Va or we have a distance of d from a to b, Va=Qd/Aε0.

So V(x)=C(x)+D
at V(0)=0+D
D=Vb

but you got D as Va?

How did start solving for C? What boundary conditions?

4. Oct 26, 2012

### hansbahia

Sorry, I totally forgot that is a charge free slab. Therefore I'm still stuck on the boundary conditions

How come your results is equal to the answer? Va is not part of it

5. Oct 26, 2012

### aralbrec

Va is the voltage at coordinate x=a.

V = Cx + D
Va = Ca + D

6. Oct 26, 2012

### Staff: Mentor

This is no longer a physics problem. It has been boiled down to strictly a math problem. The question is, are you capable of solving the following second order ordinary differential equation:

d2V/ dx2 = 0

subject to the following split boundary conditions:

V = Va at x = a

V = Vb at x = b

Well,....can you?

7. Oct 30, 2012

### hansbahia

Yes I can

Since we have the boundary conditions

V = Va at x = a

V = Vb at x = b

Va=Ca+D

Vb=Cb+D

Set D=Va-Ca

Solve for C on Vb

Then solve for D. Then just plug in Vx

Thanks a lot!!