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Electric Potential from a uniformly charged sphere

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A nonconducting sphere has radius R = 2.70 cm and uniformly distributed charge q = +7.00 fC. Take the electric potential at the sphere's center to be V0 = 0.

    (a) What is V at radial distance r = 1.45 cm?

    (b) What is V at radial distance r = R?

    2. Relevant equations
    E = Vdv
    V = k (q / r)

    3. The attempt at a solution
    I was about to just integrate E from zero to r1 and then r2, but then I realized that as r increases, so does q so I can't just have a simple single integration. And then I didn't know what to do. Help? Thanks.
     
  2. jcsd
  3. Sep 21, 2009 #2

    rl.bhat

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    Charge density in the sphere = ρ = Q/[4/3*π*R^3]
    Charge enclosed in the sphere of radius r = ρ* volume of the sphere of radius r
    Q' = { Q/[4/3*π*R^3]}*4/3*π*r^3
    = Q*r^3/R^3
    Using Gauss theorem, if the electric field E at a distance r is E, then
    4πr^2E = Q/εο*r^3/R^3
    Or E = 1/4πεο*Qr/R^3 = - dV/dr. Now find the integration.
     
  4. Sep 21, 2009 #3
    Aaaaaahhhhhhhhhhhhh. Thank you so much! I do particularly thank you because you helped me recognize that I need to do much more variable rewriting than I've been doing.
     
  5. Apr 29, 2011 #4
    i had a problem in this question,
    i got E = 1/4πεο*Qr/R^3 (using gauss law)
    when i applied - dV/dr. , i could not got the answer,I used the basic defination of electric potential that said bring charge from infinity to that pt , i integrated it (-E.dr) from infinity to r,as evident i gt an infinite term in numerator ,plz help ???
     
  6. Apr 29, 2011 #5

    rl.bhat

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    I used the basic defination of electric potential that said bring charge from infinity to that pt , i integrated it (-E.dr) from infinity to r,as evident i gt an infinite term in numerator ,plz help ???

    To find the potential at r, you have to consider the electric field outside and inside the sphere separately.
    So V(r) = -int[1/4πεο*Q/r^2*dr] from infinity to R - int[1/4πεο*Qr/R^3*dr] from R to r.
     
  7. Apr 30, 2011 #6
    thanks :) but can you explain in detail that why we follow this approach and whats wrong with d other one?
     
  8. Apr 30, 2011 #7

    rl.bhat

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    Which one is the other approach?
    You have tried to find the potential at r using the same expression for E from infinity to r. But it is wrong. Out side the sphere E = 1/4πεο*Q/r^2 and inside the sphere E = 1/4πεο*Qr/R^3. Using these expression find the integration to find the potential at r.
     
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