Consider a spherical shell with radius R and surface charge density σ. By integrating the electric...

Consider a spherical shell with radius R and surface charge density σ. By integrating the electric field, find the potential outside and inside the shell. You should find that the potential is constant inside the shell. Why?

Expert Solution

Electric field inside the shell is zero, as there is no enclosed charge inside and due to spherical symmetry from the Gauss's law we get that E is zero inside. And the electric field is otside r>R is

So, the potential outside the shell is

And the potential inside

(As E is zero inside, so the 2nd term vanishes).

So, the potential is constant inside the shell. This is obvious from the fact that the elctric field is zero everywhere inside the spherical shell. This is also true from the fact that from Laplace's equation, we have the result that the potential function V(r) is extremum at the boundary. i.e, the maximum and minimum both are at the boundary. As the value of the function V(r) is constant at the boundary r = R. So, this is the minimum and maximum also for the solution of the Laplace's equation inside the sphere. So, the solution V(r) has to be a constant and its value inside is the same as its value at the boundary.

genius_generous answered 2 years ago

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