# Electrical Potential of a uniformly charged spherical shell

• icesalmon
C = -R*e^2/hThis is confusing because it seems like the author is saying that the potential inside the sphere is the same as the potential at the surface, but then in the next sentence they say that the potential inside the sphere is sensitive to what's going on outside the sphere. What's going on here?f

## Homework Statement

From Griffiths Third Edition: "Introduction to Electrodynamics" p.p. 81 ex. 2.6

"Find the potential inside and outside a spherical shell of radius R, which carries a uniform surface charge. Set the reference point at infinity.

V(r) = -∫E⋅dl

## The Attempt at a Solution

The electric field inside the spherical shell is 0, so I'm looking at
V(r) = 0∫dl = 0 which attacks this problem by working from the center of the sphere outward. The issue is that, and this is what is confusing to me, something the author mentions: "It is tempting to suppose that you could figure out the potential inside the sphere alone but this is false: The potential inside the sphere is sensitive to what's going on outside the sphere as well" p.p 82 because what is going on outside the sphere in this situation? absolutely nothing except that the field generated by the charge distribution on the surface of the sphere radiates uniformly away from the sphere not towards it...so what is having an effect on the inside of the sphere?

For one thing, the field inside the sphere may not be zero; it doesn't say it is a conductor.

The field and potential due to the uniform surface charge would be zero inside, but f there are other charges nearby, they would produce a field which would be superimposed on the field due to that surface charge.

The field and potential due to the uniform surface charge would be zero inside...if what were true? If it were, in fact, a conductor? But there are no charged objects nearby so we're talking about the effects of the one object in question, which as you've pointed out may be either a conductor or an insulator. In the solution the author does say the potential inside the shell is NOT zero while the electric field IS zero.

OK, there are not other charges nearby. Still you have the charge on the surface, creating a potential.

Your integral ##\int \vec{E} \cdot dl## only gives the difference between the potential at the two endpoints of the integration. What this shows is that the potential is the same at the center as it is at the surface; it doesn't show that it is zero. You could integrate from infinity (where the potential is assumed to be zero) up to the surface.

The potential inside the sphere is sensitive to what's going on outside the sphere as well"
That's not a helpful way to look at it.
Potentials are relative to some arbitrarily defined zero. If it is defined as zero at infinity then the potential at any other point should be found by integrating from infinity to that point.
As Gene notes, the potential inside the sphere will be the same as at its surface.

V(r) = 0∫dl = 0
The result of an indefinite integral has a constant.

## \int 0 dt = 0*t + C##

The value of this constant potential is the same one as the surface's.