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Electrostatic Potential inside sphere of uniform charge.

  1. Nov 9, 2009 #1
    1. Given a sphere of uniform charge Q, radius R: Find an expression for the electrostatic potential V as a function of r for r ≤ R.
    Prior proof: E= Q(r)/(4πϵ0)r^2 ,for r≤R where Q(r) is the excess charge in the spherical volume of radius r

    2. Relevant equation: V = Q/(4πϵ0)r

    3. I tried integrating for spheres of radius ri over 0 - R where ∆Qi is each small bit of charge and ∆Qi = (Q/V)∆Vi giving (3Qr2/R3)∆r.
    My eventual answer resulted in (Q/4πϵ0)(R3 - r3)

    I know there's some form of integral involved, but i'm not sure what/where.
    The answer (shown on the answers) is shown to be V (r) = [Q/(8πϵ0R3)](3R2 - r2) but no working is provided, and I can't for the life of me figure out what they did. Thanks.
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 9, 2009 #2


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  4. Nov 9, 2009 #3
    I tried this method:
    [tex] V(r) - V(R) = \int_{R}^r E.dr [/tex] where [tex] E = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2} [/tex]

    But I only end up with factors of [tex] \frac{1}{R} [/tex] and [tex] \frac{1}{r} [/tex]

    Am I approaching the problem wrong? Should I be integrating over a different domain?

    Considering the answer: [tex] V(r) = \frac{Q}{8\pi\epsilon_{0}R^3}(3R^2 - r^2) [/tex]
    I don't get which integral could give the difference in scalar in front of r and R, nor where the bottom factor of [tex] R^3 [/tex] comes from?

    I also tried:
    [tex] \eta = \frac{Q}{V} = \frac{3Q}{4\pi R^3} \Rightarrow \Delta Q_{i} = \frac{3Q}{4\pi R^3}\Delta V_{i} = \frac{3Q}{4\pi R^3}4\pi r^2\Delta r [/tex]

    [tex] \Rightarrow V(r) = \frac{3Q}{4\pi\epsilon_{0}R^3}\int_{R}^r r^2 dr [/tex]

    Which is closer, but still evades the factor of 3 infront of R, and the factor of 8 on the bottom, and the powers of r and R are incorrect. =S
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