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Electric Potential of a Uniformly Charged Sphere

  1. Mar 25, 2013 #1
    My book gives the electric potential of a uniformly charged sphere as [itex]\frac{1}{4\pi \epsilon_0}[/itex][itex]\frac{Qr^2}{r_0^3}[/itex]. I can't derive this forumula. What I get is [itex]\frac{1}{8\pi \epsilon_0}[/itex][itex]\frac{Qr^2}{r_0^3}[/itex]. I can guess how the book did it which is by taking the equation for voltage of a single point charge [itex]V=\frac{1}{4\pi \epsilon_0}[/itex][itex]\frac{Q}{r}[/itex] then subbing in for Q enclosed [itex] Q\frac{r^3}{r_o^3} [/itex] which is the charge in terms of r. That would give you the book answer. I tried doing it a different way however which I'm not sure why it won't work. I used Va-Vb=[itex]\int _a^{b}[/itex] E dl. Yet when I get to solving for the electric field I get [itex]\frac{Q}{4\pi r^2 \epsilon_0}[/itex]. If I take Q out then I can solve for the book answer. However, Q is in terms of r which I wrote above so I can't take Q out. I must integrate it. That gives me the 1/8 instead of the 1/4. What am I doing wrong? Thanks!
     
  2. jcsd
  3. Mar 26, 2013 #2

    mfb

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    Staff: Mentor

    Do you have a uniformly charged sphere, or ball? The derivation suggests a ball (charge everywhere in the volume), not a sphere. Q is the total charge, and r_0 is the radius of the ball?
    How do you get the electric field? It should not depend on r like that.
     
  4. Mar 29, 2013 #3
    Are you looking for the potential inside the sphere or outside? In other words is r > r0 or is r < r0?
     
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