# Electric Field and Potential in Spherical Shells

1. Jul 10, 2010

### rizardon

1. The problem statement, all variables and given/known data
Consider two thin conducting spherical shells. The inner shell has a radius r1=15cm and a charge of 10.0 nC. The outer shell has a radius r2=30cm and a charge of -15.0 nC.
Find
a)the electric field E and
b)the electric potential V in A, B and C, with V=0 t r=infinity
A: r<r1
B: r1<r<r2
C: r>r2

2. Relevant equations
E=kq/r2
V=kq/r
3. The attempt at a solution
Using the above formula, I get

For the region A
E=0 and V=0 because the charge within a conducting shell is 0

For the region B
E= k(10 nC)/r2, r1<r<r2
V= k(10 nC/r

For the region C
E=k (10 nC + -15 nc)/r2=k(-5 nC)/r2
V=k(-5 nC)/r

Am I doing it the right way?

2. Jul 10, 2010

### hikaru1221

E looks okay, though I'm not sure if you think in the right way or not. But V is not Remember that only V(infinity) = 0. You should find the potential difference instead.
Have you learned the Gauss law yet?

3. Jul 10, 2010

### rizardon

Yes, I have, but using the law is quite confusing. How will finding the potential difference make any difference, since my Va would be 0 and what's left is Vb which is kq/r?

4. Jul 10, 2010

### hikaru1221

The formula E = kq/r^2 only applies to charge points, not a distribution of charges. In this case, where symmetry is present, you should apply the Gauss law. Consider a spherical surface concentric with the shells.

Again, the formula V=kq/r only applies to charge points. To find V, you should apply the original formula: dV = -Edr. Va=0 only if a=infinity, so you must do the integration from r to infinity.

EDIT: Forgot to mention something. After using the Gauss law, you should find that the formula of E looks similar to E of a charge point. That's why you got right answers for E

Last edited: Jul 10, 2010
5. Jul 10, 2010

### graphene

The equations you have mentioned are not appropriate here.

You'll need to use Guass Law to find the field in the three regions you've mentioned.
Do you know how to use Gauss law? Look into your book if you don't.

Do you know how potential is related to the field?

6. Jul 10, 2010

### rizardon

The potential difference is V=k(integral)dq/r

V=-(integral)Eds=-(integral)Edr=-(integral)kQ/r2dr=-kQ(integral from a to b)dr/r2=kQ(1/b - 1/a)

Is this right

7. Jul 10, 2010

### hikaru1221

Only when E=kQ/r^2 from r=b to r=a
What if E=kq1/r^2 from r=b to r=c, and E=kq2/r^2 from r=c to r=a?