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The potential field outside of a charged hemispherical shell

Physics Asked on April 9, 2021

I have a hemispherical shell of radius a held at potential Vo. Using Laplace’s equation, with appropriate boundary conditions, I’ve been able to find the potential at any point within the hemisphere as:

U(r,$Theta$) = $sum_{n=0}^infty B_n (frac ra)^n P_n[cosTheta]$

where r is radial distance from the centre of the hemisphere
$Theta$ is polar angle
$B_n$ is a coefficient determinable from initial conditions
$P_n$ is the legendre polynomial wrt cos$Theta$

How would one go about calculating the potential outside of this hemisphere at any point? Could the general result above be useful?

Thanks in advance.

One Answer

The outside must have it's own boundary conditions. If no external fields are present, then the usual boundary conditions for such problems are:

Boundedness or 0 at infinity: $$U(r=infty)=0$$

Continuity (of the potential) at the boundary : $$U(r=a^+)=U(r=a^-)$$

The calculation steps should be about the same as inside the hemisphere.

Correct answer by Papa Jonathan on April 9, 2021

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