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Homework Help: Relating Electric Field and Voltage in an Insulating Sphere

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. From the expression for E=kQ/r^2 for r>R and E=kQr/R^3 for r<R, find the expression for the electric potential V as a function of r both inside and outside the uniformly charged sphere. Assume that V = 0 at infinity.


    2. Relevant equations
    [tex]\int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)[/tex]

    2.5. Apparent Answer


    Apologies ... I'm not very knowledgeable in tex, but that last part would read r-squared over R-squared.

    3. The attempt at a solution

    I have already obtained V = kQ/r for r>R. This was a simple antiderivative.
    I'm getting really caught up on r<R. I've tried integrating from 0 to r, 0 to R, r to R, and backwards versions of the same. Nothing comes to the answer in 2.5 above. I'm really at a loss for how this integral should be done. I can understand most of it except for the 3 in the second expression; where the heck does it come from? I've seen this answer in a couple of different places, but none of them actually explain it.

    Any help would be appreciated. Thanks ahead of time.
  2. jcsd
  3. Sep 16, 2012 #2


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    Hello NeonJay. Welcome to PF !

    Extend this to r = R, to get V(R) = kQ/R.

    Then use [itex]\displaystyle \int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)[/itex] to get V(r) for r < R . To do this, A = R, B = r .
  4. Sep 16, 2012 #3


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    The potential is continuous function, and you know it at r=R. V(R)=kQ/R2 . Integrate from r to R:
    substitute kQ/R for V(R) and solve for V(r).

  5. Sep 19, 2012 #4
    I think I get it. I was essentially neglecting part of my equation.

    Thanks for the help! It really clears a lot of things up.
  6. Sep 19, 2012 #5


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    The equation comes from the definition of the electric potential: a function V(r) of position r so as the negative gradient of V is equal to the electric field intensity.


    The electric field is conservative so the work done on unit charge between two points A and B does not depend on the path taken.

    [tex]\int_A^B{-\nabla V d\vec r}=\int_A^B{\vec Ed\vec r}[/tex]

    V dr=dV, the increment of V.

    [tex]\int_A^B{-dV}=V(A)-V(B)=\int_A^B{\vec Ed \vec r}[/tex]

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