# Thick spherical shell Question.

1. Sep 15, 2014

### twinpair

1. The problem statement, all variables and given/known data
A thick spherical shell with inner radius R and outer radius S has a uniform charge density d.(A) What is the total charge on the shell? Express your answer in terms of R, S, d, and π. (B) Express the electric field as a function of distance from the center of the sphere r, R, S, d, and the permitivity of free space p for each of the following regions: 0<r<R , R<r<S, S<r

2. Relevant equations
E = kQ/r2

V=(4/3)πr3

3. The attempt at a solution
(For part A)
Since its a thick shell the volume would be V = (4/3)π(S^3 - R^3)

and dV = Qenc => d=Q/V

so

d = 3Q/(4π(S^3 - R^3))

(For part B)
0<r<R
the electric field will be 0 because the electric field can't close on itself

R< r< S
E=kQenc/r2 = (4πkρ(r^3 - R^3))/3r^2

S<r
E = kQ/r2 because the sphere can be treated as a point charge and the electric field is symmetric on a sphere

Is this correct?

2. Sep 15, 2014

### nrqed

Everything looks good. But note that for part A they want Q. And for the other problems, you must give the answer in terms of $d , R, S$

3. Sep 15, 2014

### twinpair

So for part A would it be Q = dv --> Q = d * (4/3)π(S^3 - R^3)

And for other problems what does it mean to give the answer in terms of d,R,S

4. Sep 15, 2014

### nrqed

That's correct.

What it means is that your answer must contain only those parameters (your answer cannot contain Q).

5. Sep 15, 2014

### twinpair

what does it mean to give the answer in terms of d,R,S?

6. Sep 15, 2014

### collinsmark

Hello twinpair,

Welcome to Physics Forums!

That's not technically wrong, but it doesn't answer the question. You are already given the density, d. The problem statement is asking you to solve for the total charge, Q.

Correct.

That's the first introduction of ρ, the Greek letter "rho." Is that supposed to be d?

You're supposed to answer in terms of r, R, S, d and p.

Whatever the case, rather than fiddle with substitutions, I suggest starting over. Use Gauss' law.

This part of the problem is very straightforward if you start with Gauss' law; that way it doesn't require any fancy-schmancy substitutions.

That's also correct, but is using the wrong set of variables. You could make the substitution of $k = \frac{1}{4 \pi p}$, but as it turns out, the solution is actually easier to come by if you apply Gauss' law from start to finish, solving for E as the final step*.

*[Edit: although you can leverage the total charge found in part A.]

Last edited: Sep 15, 2014