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Finding the potential outside a charge sphere

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the potential outside a charged metal sphere(charge Q , radius R) placed in an otherwise uniform electric Field E0. Explain clearly where you are setting the zero of potential .


    2. Relevant equations



    3. The attempt at a solution

    At the boundaries of the sphere V=0. Since a sphere is a 3 dimensional object,

    1/r^2*d/dr(r^2*dV/dr)+(1/r^2*sin(theta))*d/d(theta)*(sin(theta)*dV/d(theta))+1/r^2*1/(sin(theta)^2)*d^2V/dphi^2=0
    V(r,theta)=R(r)*[tex]\Theta[/tex]([tex]\theta[/tex]R

    1/R*d/dr*(r^2*dR/dr)+(1/([tex]\Theta[/tex]*sin[tex]\theta[/tex])*d/d[tex]\theta[/tex]*(sin([tex]\theta[/tex]*d[tex]\Theta[/tex]/d[tex]\theta[/tex])=0

    1/R*d/dr*(r^2*dR/dr)=l(l+1)
    (1/([tex]\Theta[/tex]*sin[tex]\theta[/tex])*d/d[tex]\theta[/tex]*(sin([tex]\theta[/tex]*d[tex]\Theta[/tex]/d[tex]\theta[/tex]=-l(l+1)

    V(r,[tex]\theta[/tex])=(Arlrl+Bl/(rl+1))*Pl*cos([tex]\theta[/tex])

    wouldn't I be setting the potential to zero because the sphere might be an equipotential?

    Wait a minute , in the exterior region, wouldn't V(r,theta)=kR^3*cos(theta)/3epilison0*r^2
     
    Last edited: Oct 26, 2008
  2. jcsd
  3. Oct 26, 2008 #2

    gabbagabbahey

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    The potential on the surface of the sphere certainly will be an epuipotential; and you could set it to zero if you choose to; however in this case the problem is much easier if you don't.

    Have you seen the solution for an uncharged metal sphere in a uniform electric field? What is the solution for that problem and where did they set the potential to zero?

    If you choose a clever spot to set the potential to zero in this problem, you can use the superposition principle with the solution for the uncharged sphere and a uniformly charged spherical shell of charge Q.
     
  4. Oct 26, 2008 #3
    Couldn't I set the potential to be zero at the boundary condition? Maybe at the edge of the sphere. Since the shape is the surface of a sphere, it wouldn't matter where I would set my boundary condition at the edge of the spheres?

    Outside a sphere V(r,theta)= (R^3/r^2)*cos(theta)

    Law of superposition principle :

    V=V(outside)+V(inside) ?

    I don't know the solution for a uniformly uncharged sphere. Why is the solution of a uncharged sphere relevant since I am asked to find the solution for a uniformily charged sphere?
     
  5. Oct 26, 2008 #4

    gabbagabbahey

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    Start with the solution for the uncharged sphere in an uniform electric field... What is that solution? What are the boundary conditions for that problem?
     
  6. Oct 19, 2010 #5
    help anyone?

    I was wondering if we should simply add kQ/r to the potential given for the uncharged sphere or if there was a way to obtain this result more formally.
     
  7. Oct 20, 2010 #6

    gabbagabbahey

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    Does your total potential satsify Laplace's equation then? Does it satsify the correct Boundary conditions? What does the uniqueness theorem tell you about that solution?
     
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