Consider an isolated uniformly charged spherical shell of radius R. A.) Explain
ID: 2077753 • Letter: C
Question
Consider an isolated uniformly charged spherical shell of radius R. A.) Explain why the potential at the center of this shell is V = kQ/R. B.) As we have seen, the potential outside this shell is V = kQ/r, where r is measured from the center of the shell. This implies that the potential at the surface of the shell is V = kQ/R, the same result as at the center of the shell. What does this suggest about the potential at other points inside the shell? Explain how this is consistent with Gauss's Law.Explanation / Answer
We take reference potential = 0 at infinity
a) We know that potential is additive and scalar quantity
and V = kq/r
At center of sphere
=> V(due to each point on sphere) = k * dq/R where dq is the charge contained in the point considered
=> V = k Q/R (since sum of dq = Q)
b) We see that potential is constant inside and on the surface of sphere
So V (center of sphere) = V (inside sphere) = V (at r = R)
By gauss law
Q/epsilon = phi = integral(E.dA)
For a charged sphere
Q/epsilon = E*(4*pi*r^2)
=> E = Q/(4*epsilon*pi*r^2)
V = -integral (E.dr)
= Q/(4*epsilon*pi*r)
At r = R
V = Q/(4*epsilon*pi*R)
Inside sphere E = 0
So V is constant
=> V (center of sphere) = V (inside sphere) = V (at r = R)
V (center of sphere) = Q/(4*epsilon*pi*R)
We see that both results are same and consistent with each other