# Homework Help: A non-conducting sphere, e-field and potential.

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1. Jan 31, 2015

### timnswede

1. The problem statement, all variables and given/known data
A non-conducting sphere of radius R has volume charge density ρ = B/r. for r<R and ρ = - for r>R. B is a constant.
a) Calculate E-field for r>R.
b) Calculate E-field for r<R.
c) Calculate potential for r>R.
d) Calculate potential for r=R.
e) Calculate potential for r<R

2. Relevant equations
V=-∫Edr
∫Eda=Qin/ε

3. The attempt at a solution
Well first of all I was a bit confused on the whole ρ = 0 when r>R, but apparently that just means that all of the charge is contained within the sphere.
So since it is a nonuniform charge, I first found the charge of the sphere. ∫(0 to R)ρ4πr^2dr and got 2BπR^2.
I'm pretty confident about all my answers except for part e) For part a) I did E4πr^2=2BπR^2/ε and got E=BR^2/2r^2ε. Same thing for part b, but this time the radius cancels out so B/2ε
So for part c I took the negative integral of the e-field outside of R. V=-∫(BR^2/2r^2ε)dr and ended up with BR^2/2rε. Same thing for part d, but the radius cancels out.
Now part e I did the same thing -∫B/2εdr and got -Br/2ε. I feel like it might be wrong since it's negative, and I don't really understand why the potential inside the sphere would be negative

2. Jan 31, 2015

### ehild

Replace r=R. The radius does not cancel.

You get the potential with an unknown constant, C. So U = C - Br/2ε. The potential is continuous function, and you know it at R. From that, you can find C.

3. Feb 1, 2015

### timnswede

Woops you are right, so for part d it is actually BR/2ε
Why would I integrate from R to r? What my thinking was is that I am integrating the e-field to find the potential everywhere inside the sphere, from the center, zero to the radius of the sphere R.

4. Feb 1, 2015

### ehild

You do not know the potential at the centre of the sphere, but you know it at r=R.

The electric field is the negative gradient of the potential. dU/dr=-E.
If you integrate it between a and b you get $\int _a^b(dU/dr )dr = -\int_a^b Edr \rightarrow U_b-U_a= -\int_a^b Edr$
The potential was assumed zero at infinity. You do not know that it is zero at r=0 again. But it is known at r=R. Choose one boundary at R.

5. Feb 1, 2015

### ehild

You do not know the potential at the centre of the sphere, but you know it at r=R.

The electric field is the negative gradient of the potential. dU/dr=-E.
If you integrate it between a and b you get $\int _a^b(dU/dr )dr = -\int_a^b Edr \rightarrow U_b-U_a= -\int_a^b Edr$
The potential was assumed zero at infinity. You do not know that it is zero at r=0 again. But it is known at r=R. Choose one boundary at R.

6. Feb 1, 2015

### timnswede

Oh OK, that makes sense. So would it work if I did VR-Vr=-∫(from r to R)B/2ε dr. So after some integration I would get Vr=BR/ε -Br/2ε?

7. Feb 1, 2015

Correct!

8. Feb 1, 2015

### timnswede

Great, thanks! Your explanation helped a lot.

9. Feb 1, 2015

### ehild

You are welcome:)