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Determining Electric Potential

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a nonconducting sphere of radius r2. It has a concentric spherical cavity of r1. The material between r1 and r2 has a charge density p (C/m3). Take V=0 and r=infinity. Determine the electrical potential V as a function of the distance r from the center for (a) r>r2, (b) r1<r<r2, and (c)r<r1

    2. Relevant equations
    ρ=Q/V

    3. The attempt at a solution
    a) I have no idea where to start. Would my volume be just (4/3)πr3??

    b) Total volume enclosed = (4/3)(π)(r2)3 - (4/3)(π)(r1)3

    = (4/3)(π)(r23-r13
    ρ=Q/V
    =3Q/4(π)(r23-r13)

    dQ= 3Qr2/(r23-r13)

    V= (1/(4πε)∫dQ/r

    Having trouble doing the integral...

    c) There should be no electric potential inside
     
  2. jcsd
  3. Oct 26, 2014 #2

    Simon Bridge

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    Welcome to PF;
    "no idea" is not acceptable here - you have just done a section of coursework that should have given you some ideas ... go back and look again.

    The volume of the sphere radius r is going to be ##\frac{4}{3}\pi r^3## all right - but what is the volume you are supposed to find? Your next answer suggests you are confused about this.

    Lets say you had a ball of charge radius R and charge Q with uniform density - you want to find the potential at radius r < R from the center ... the rule (for spherical distributions of charge) is that the potential is proportional to the total charge inside the radius and inversely proportional to the distance from the center.

    So you should not be thinking in terms of volumes at all, but in terms of how much charge there is.
    You'll end up doing as volume integral but only because the maths is easier that way.
     
  4. Oct 26, 2014 #3
    Outside of the sphere would we not just use V= ((1/(4πε))(Q/r) ?
     
  5. Oct 27, 2014 #4

    Simon Bridge

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    That's outside the sphere.
    For r > R, the total charge inside r is Q, so V=kQ/r
    For r< R, what is the total charge inside r?
     
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