Solving Laplace's Eqn: Setting Zero of Potential

by bcjochim07
Tags: laplace, potential, setting, solving
bcjochim07 is offline
Nov20-10, 01:56 PM
P: 374
1. The problem statement, all variables and given/known data
I have two quick related questions that I think will help clear something up for me.

(1) An uncharged metal sphere of radius R is placed in an otherwise uniform
electric field E = E0 zhat. The field induces charge. Find the potential in the region outside the sphere.

(2) Find the potential outside an infitely long metal pipe of radius R placed at right angle to an otherwise uniform electric field E0.

2. Relevant equations

3. The attempt at a solution

Okay, so I am mostly comfortable with the solutions to these problems, with the exception of one key concept.

Problem 1: If E=E0 zhat (for r>>R, then V for r>>R is -E0z + C)
The one hang-up I have is that for problem (1), you set V=0 on the xy-plane "by symmetry." Thus we can say C=0. I'm not sure I understand the reasoning behind setting V=0 for all z=0.

Similarly, from problem (2), if we suppose that the electric field is pointing along z direction, so again, the form of the potential is -E0z + C. Again, one should set V=0 on the xy-plane, so C=0. I don't have a clear picture in my head of why you set V=0 on the xy-plane.

Any explanations of this concept would be greatly appreciated
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fzero is offline
Nov20-10, 02:15 PM
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Since the electric field is the gradient of the potential, [tex]\vec{E} = - \nabla V[/tex] (for electrostatics), a constant shift in the potential [tex]V' = V + c[/tex] doesn't change the electric field. Therefore we are usually free to define the location of [tex]V=0[/tex] by adding a constant term to the potential. In your examples, choosing [tex]V(z=0) = 0[/tex] is just the most convenient choice, but it was not the only one.

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