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

  1. Jan 31, 2015 #1
    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. jcsd
  3. Jan 31, 2015 #2

    ehild

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    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.
     
  4. Feb 1, 2015 #3
    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.
     
  5. Feb 1, 2015 #4

    ehild

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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.
     
  6. Feb 1, 2015 #4

    ehild

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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.
     
  7. Feb 1, 2015 #5
    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?
     
  8. Feb 1, 2015 #6

    ehild

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    Correct!
     
  9. Feb 1, 2015 #7
    Great, thanks! Your explanation helped a lot.
     
  10. Feb 1, 2015 #8

    ehild

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    You are welcome:)
     
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