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Question about Gauss's law, finding electric field and potential field

  1. May 18, 2014 #1
    ztbty9.png

    I'm having trouble with the second part b) for this problem.

    I used Gauss's Law and for a I got

    r<R

    [tex]E = \frac{{r}^{2}}{4\varepsilon}[/tex]

    and for r>R

    [tex]E = \frac{{R}^{4}}{4\varepsilon{r}^{2}}[/tex]

    and then

    [tex]V = \int{E.dr}[/tex]

    from r to infinity right? So

    for r>R I got

    [tex]V = \frac{{R}^{4}}{4\varepsilon r}[/tex]

    but for r<R

    I get [tex]V = \left[ \frac{{r}^{3}}{12\varepsilon}\right][/tex] from infinity to r so I would get infinity which doesn't make sense,

    can someone help me please?
     
  2. jcsd
  3. May 18, 2014 #2

    BvU

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    Why from infinity ?
     
  4. May 18, 2014 #3
    because isn't the potential field always the integral of the electric field from r to infinity?

    but is it different when you integrate inside the sphere?
     
  5. May 18, 2014 #4

    TSny

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    Yes, to get V at r inside you want the integral [itex]V = \int{E.dr}[/itex] from r to infinity. But you'll need to break this up into an integral from r to R and then from R to infinity. For R to infinity, you can just use your result for outside.
     
  6. May 24, 2014 #5
    First find the potential at distance 0< r < R inside the sphere. (Hint: think of shells) Then we know for electrostatic case ##E=-\nabla V##.

    For outside the sphere, you were right in using gauss's law.
     
  7. May 25, 2014 #6

    vanhees71

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    Hint: It's much easier to use the local form of the electrostatic Maxwell equations (as is almost always the case). Here you just can make the ansatz
    [tex]V(\vec{x})=V(r) \quad \text{with} \quad r=|\vec{x}|[/tex]
    and solve the (now ordinary) differential equation (written in Heaviside-Lorentz units)
    [tex]\Delta V=-\rho \; \Rightarrow\; \frac{1}{r} \frac{\mathrm{d}}{\mathrm{d} r} \left [r V(r) \right ] = -\rho(r).[/tex]
    This reduces the problem to two simple integrations.
     
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