A non-conducting sphere, e-field and potential.

  • #1
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Homework Statement


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

Homework Equations


V=-∫Edr
∫Eda=Qin/ε

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
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement


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

Homework Equations


V=-∫Edr
∫Eda=Qin/ε

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.
Replace r=R. The radius does not cancel.

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
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
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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
ehild
Homework Helper
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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.
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
ehild
Homework Helper
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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.
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.
 
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  • #6
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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.
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
ehild
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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?
Correct!
 
  • #8
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Great, thanks! Your explanation helped a lot.
 
  • #9
ehild
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You are welcome:)
 

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