# Homework Help: Nonconducting spherical shell with uniform charge

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1. Sep 3, 2016

### ooohffff

1. The problem statement, all variables and given/known data
Suppose the nonconducting sphere of Example 22-4 has a spherical cavity of radius r1 centered at the sphere's center (see the figure). Assuming the charge Q is distributed uniformly in the "shell" (between r = r1 and r = r0), determine the electric field as a function of r for the following conditions.

r1 < r < r0

2. Relevant equations
Gauss's law

3. The attempt at a solution

I know the electric field is 0 in the inner cavity. The electric field outside of the shell is Q/(4pi*E0*r^2). How would I derive the equation for the electric field inside the shell? Charge per unit volume should be p=Q/(4/3*pi(r0-r1)^3) right?

Last edited: Sep 4, 2016
2. Sep 3, 2016

### SammyS

Staff Emeritus
That charge density is incorrect.

A3 - B3 ≠ (A - B)3 .

3. Sep 3, 2016

### ooohffff

Mmm yes, your'e right it should be Q/(4pi/3(r0^3-r1^3)). But where do you go from there?

Last edited: Sep 4, 2016
4. Sep 3, 2016

### SammyS

Staff Emeritus
How much charge there interior to a sphere of radius, r, where r1 < r < r0 .

5. Sep 3, 2016

### ooohffff

Could you rephrase that and be more specific?

6. Sep 3, 2016

### ooohffff

Would it be the same as the electric field outside the sphere?

7. Sep 4, 2016

### SammyS

Staff Emeritus
No, electric charge is not equal to electric field.

You want to find what the electric field is interior to the "shell". That's where r1 < r < r0 .

You mention Gauss's Law.

8. Sep 4, 2016

### ooohffff

Right its

∫E⋅da=qenc0

So I keep getting E4πr^2=qenc0

so that would mean E would equal the same as the electric field as the outside of the sphere, which I know is wrong. What am I doing wrong?

9. Sep 4, 2016

### SammyS

Staff Emeritus
For r between r1 and r0, what do use for your Gaussian surface?

10. Sep 4, 2016

### ooohffff

Would it be r1^2 instead of r^2?

11. Sep 4, 2016

### SammyS

Staff Emeritus
No.

If you want to know the electric field at a distance r from the origin, use s sphere of radius r.

How much electric charge is enclosed within this sphere?

In the case of your problem the only charge inside this sphere resides in the spherical shell with inner radius r1 and outer radius r. Right ?

So, how much charge is that ?

Last edited: Sep 4, 2016
12. Sep 4, 2016

### ooohffff

So if charge density =

Q/ (4π/3 ( r03-r13)) = qenc/(4π/3 (r3-r13)).

Then,

Q(r3-r13) = qenc (r03-r13)

qenc = [Q(r3-r13)] / (r03-r13)

E4πr2 = Q(r3-r13)] / ε0(r03-r13)

E = Q(r3-r13)] / 4πr2ε0(r03-r13)

Last edited: Sep 4, 2016
13. Sep 4, 2016

### ooohffff

Can someone verify this?

14. Sep 4, 2016

### SammyS

Staff Emeritus
Patience.

That's the right idea. To be perfectly correct, the entire denominator should be enclosed in parentheses, else use some other grouping symbol.

15. Sep 4, 2016

### ooohffff

Haha, thank you SammyS!