# Potential at Center of Insulating Spherical Shell

1. Jul 7, 2012

### waters

1. The problem statement, all variables and given/known data
The inner radius of a spherical insulating shell is c=14.6 cm, and the outer radius is d=15.7 cm. The shell carries a charge of q=1451 E−8 C, distributed uniformly through its volume. The goal of this problem is to determine the potential at the center of the shell (r=0), assuming the potential is zero infinitely far from the sphere. In order to do this, we will first determine the electric field everywhere in space due to this charge distribution, and then use the relationship between the electric field and the change in potential to find the potential at the center.

Finally, apply the relationship between the change in electric potential and the electric field and the value of the electric potential at r=c, to find a numerical value for the electric potential at r=0.

2. Relevant equations
dQ = ρdV
∫E●dA = q/ε
ε = 8.85 E-12 F/m
-∫E●dl = V
r(unit vector)●dl = dr
V = kq/r

3. The attempt at a solution
The charge density, ρ
ρ = q/((4/3)pi(d^3-c^3)) = 3q/(4pi(d^3-c^3)

The charge differential, dq
dV = 4pi(r^2)dr
dq = ρ(dV) = (3q(r^2)dr)/(d^3 - c^3)
q = ∫dq

The electric field is either:
E = q/(εA)

or

E = kq/(r^2)

However, if I evaluate ∫E*dr for both equations, I get the wrong answer for the potential at c. I don't even know if I should use dq, or q. The potential at c is the same potential at c (I don't understand why or how). I just know that

-∫E●dl from infinity to c = (-∫E●dl from infinity to d) + (-∫E●dl from d to c)

should give me the potential at c (and the center), but it doesn't. The potential at d is (8.30*10^2 V), which I got right.
E = q/εA
-∫E●dl = V = ∫(qdr)/(4*pi*ε*(r^2)) from infinity to d

The answer to the potential at both the center and when r = c is 8.60 * 10^2 V.

Please help me out. I have a test on Monday, and I need to know how to do this problem.

2. Jul 7, 2012

### TSny

You'll need to use Gauss' law to find the electric field in the three different regions:
(1) r>d (2) c<r<d (3) r<c

You should find that Gauss' law gives E = kq/r^2 for r>d, which is why you got the potential at r = d correct.

The toughest part of the problem is using Gauss' law to find E for c<r<d. But you already have the charge density for this region. That will be useful when using Gauss' law.

For r<c, Gauss' law will give you a simple result for E.

You have the right idea for finding V from E.

Last edited: Jul 7, 2012