Potential Difference of Inverted Hemispherical Shell

[Jonathan K]

The problems states: An inverted hemispherical bowl of radius R carries a uniform surface charge density σ. Find the potential difference between the north pole and the center.

I was able to do the problem and got the correct answer the book gives, which is (Rσ/2ε₀)(√2 - 1). My professor, however asks that we do an assessment, or check, of our answer for every problem. This is where I am having troubles. I figured I could use Poisson's Equation to show that the Laplacian of the answer is -σ/ε₀, but I can't seem to get this. Any other ideas?

 

[rude man]

The problems states: An inverted hemispherical bowl of radius R carries a uniform surface charge density σ. Find the potential difference between the north pole and the center.

I was able to do the problem and got the correct answer the book gives, which is (Rσ/2ε₀)(√2 - 1). My professor, however asks that we do an assessment, or check, of our answer for every problem. This is where I am having troubles. I figured I could use Poisson's Equation to show that the Laplacian of the answer is -σ/ε₀, but I can't seem to get this. Any other ideas?

Well, sice no one else has responded in a week ..for what it's worth, probably not much ...

Poisson's equation is ∇²V = -ρ/ε with ρ = volume, not surface, density, so your units would be incorrect.

Also, the entire volume inside the hemisphere has ρ = 0. But I don't see that solving for ∇²V = 0 would be pleasant, since V varies not only along the z axis (assuming the hemisphere axis is the z axis and the bottom rests symmetrically on the x-y plane), but also with x and y (or similar problem if you went cylindrical or spherical.) The only alternative that comes to mind is solving for the E field directly and integrating from the pole to the bottom center.