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Calculating the potential of a uniformily charged spherical solid

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the potential inside and outside uniformily charged spherical solid whose radius R and whose total charge is q.use infinity as your reference point

    2. Relevant equations
    [tex]V=-\int E* dl[/tex]

    gauss law = [tex] \int E *da=q/\epsilon_ 0[/tex]
    3. The attempt at a solution

    This should be easy. Inside a solid sphere, E=0 so the potential inside sphere is zero. The electric field of a sphere is : [tex] E_sphere=(1/(4*\pi*\epsilon_ 0))*q/R^2 [/tex] => [tex] V=-(1/(4*\pi*\epsilon_ 0))*q/R[/tex]. Hmm... my solution is too easy; I know this solution was worked out in one of the examples found in my textbooks. Should I apply gauss law I take into account that [tex] dq=\rho*d\tau=\sigma*da[/tex] where [tex] d\tau=(4/3)*\pi R^3 [/tex]and[tex] da=4*\pi*r^2[/tex] ?
     
  2. jcsd
  3. Sep 19, 2009 #2
    The electric field is only zero inside a conducting sphere. This is because all the charge migrates to the surface and it acts likes a hollow charged shell. In the question it does not say the sphere is conducting, it says it is uniformly charged. Therefore it must be acting as an insulating sphere, because a conducting sphere is not uniformly charged. Work it out assuming there is a field inside and you will get the right answer.

    Also the potential inside a conducting sphere won't be zero, it will assume the value of the potential at the surface of the sphere.
     
  4. Sep 19, 2009 #3
    I think I got it; for r>R, E(4*pi*r^2)=rho*(4/3*pi*R^3)/epilison_0 => and for R<r E(4*pi*r^2)==rho*(4/3*pi*R^3)/epilison_0
     
    Last edited: Sep 19, 2009
  5. Sep 20, 2009 #4
    Yep that looks right.
     
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