(adsbygoogle = window.adsbygoogle || []).push({}); 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

d^{2}V/dx^{2}=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

d^{2}V/dx^{2}=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_{a}b-V_{b}a]/b-a

??

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

2. Relevant equations

3. The attempt at a solution

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

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