(adsbygoogle = window.adsbygoogle || []).push({}); 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 E_{0}. 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])=(Ar^{l}r^{l}+B_{l}/(r^{l+1}))*P_{l}*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

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# Homework Help: Finding the potential outside a charge sphere

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