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

[hansbahia]

Homework Statement

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

Homework Equations

d²V/dx²=0

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)=[V_b-V_a)x+V_ab-V_ba]/b-a

??

 

[Chestermiller]

Homework Statement

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

Homework Equations

d²V/dx²=0

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)=[V_b-V_a)x+V_ab-V_ba]/b-a

??

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 = V_a + (V_b - V_a) (x -a)/(b -a)

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

 

[hansbahia]

V = Cx + D

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

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

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

but you got D as V_a?

How did start solving for C? What boundary conditions?

 

[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

 

[aralbrec]

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

Va is the voltage at coordinate x=a.

V = Cx + D
V_a = Ca + D

 

[Chestermiller]

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

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 = V_a at x = a

V = V_b at x = b

Well,...can you?

 

[hansbahia]

d²V/ dx² = 0

subject to the following split boundary conditions:

V = V_a at x = a

V = V_b at x = b

Well,...can you?

Yes I can

Since we have the boundary conditions

V = V_a at x = a

V = V_b at x = b

V_a=Ca+D

V_b=Cb+D

Set D=V_a-Ca

Solve for C on V_b

Then solve for D. Then just plug in V_x

Thanks a lot!