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Potential inside of chrged non-conductive sphere

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A charge is distributed uniformly throughout a non-conducting spherical volume of radius R. Show, that the potential at distance r from center (r<R) is given by:
    V=q(3*R^2-r^2)/(8*PI*e0*R^3)

    2. Relevant equations
    From Gauss low:
    E(r)=q*r/(4*PI*e0*R^3)

    E(r)=-grad(V(r))

    3. The attempt at a solution
    After integrating I got:
    V(r)=-q*R^2/(8*PI*e0*R^3)
    which is in fact stupid, as it yields V=0 at r=0. So probably I integrated with wrong limits.
    How to obtain correct formula?
     
  2. jcsd
  3. Oct 3, 2008 #2
    I made a small mistake in typing the formula on V that I obtained. It should be:
    V(r)=-q*r^2/(8*PI*e0*R^3)
     
  4. Oct 3, 2008 #3
    And R = radius of gaussian surface
    r = radius of your sphere?
    please be precise.
     
  5. Oct 3, 2008 #4
    Never mind, I've solved it.

    You have two different functions for the electric field.
    If r > R, [tex] E = \frac{kQ}{r^2} [/tex] where [tex] k = \frac{1}{4\pi\epsilon_0} [/tex]
    if r < R, [tex] E = \frac{kQr}{R^3} [/tex]

    so V of r < R will be the negative integral of r > R from R to infinity plus the negative integral of r < R from r to R.

    [tex] V = - \int^{R}_{\infty} \frac{kQ}{r^2}dr - \int^{r}_{R} \frac{kQr}{R^3}dr [/tex]

    [tex] V = \frac{kR}{R} - \frac{kR}{2R^3}(r^2-R^2) = \frac{2kQR^2}{2R^3} + \frac{kQR^2}{2R^3} - \frac{kQr^2}{2R^3} = \frac{kQ(3R^2 - r^2)}{2R^3} [/tex]

    subbing k back in = [tex] \frac{Q(3R^2 - r^2)}{8 \pi \epsilon_0 R^3} [/tex]

    phew.. Hope there are no errors in that haha. I have been up for far too many consecutive hours..
     
    Last edited: Oct 3, 2008
  6. Oct 3, 2008 #5
    Thank you very much for the reply!
    I have just one more question: why should I integrate from infinity to zero, not from zero to infinity?
     
  7. Oct 3, 2008 #6
    Sorry again: from inf to r not from r to inf, I meant.
     
  8. Oct 3, 2008 #7
    One more correction:
    Why should I integrate from infinity to r not from zero to r?
    I hope, now it's ok...
     
  9. Oct 3, 2008 #8
    because if we integrated from zero to r, that would give us V for inside the gaussian surface aswell as inside the sphere.

    Sorry for the late reply I was away for sometime.
     
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