Finding the potential outside a charge sphere

In summary: Yes, the potential satisfies Laplace's equation. The boundary conditions for that problem are that the potential is zero at the edge of the sphere. The uniqueness theorem tells us that the solution is unique and it results in a charge on the surface of the sphere that is equal to Q.
  • #1
Benzoate
422
0

Homework Statement



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 .

Homework Equations


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
 
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  • #2
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.
 
  • #3
gabbagabbahey said:
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.

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?
 
  • #4
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?
 
  • #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.
 
  • #6
erogard said:
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.

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?
 

1. What is the potential outside a charge sphere?

The potential outside a charge sphere is the amount of energy needed to move a unit charge from infinity to a point outside the sphere, relative to a reference point. It is also known as the electric potential.

2. How is the potential outside a charge sphere calculated?

The potential outside a charge sphere can be calculated using the equation V = kQ/r, where V is the potential, k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), Q is the charge of the sphere, and r is the distance from the center of the sphere.

3. Does the potential outside a charge sphere depend on the distance from the sphere?

Yes, the potential outside a charge sphere is inversely proportional to the distance from the center of the sphere. This means that as the distance increases, the potential decreases.

4. How does the potential outside a charge sphere change with an increase in the charge of the sphere?

The potential outside a charge sphere is directly proportional to the charge of the sphere. This means that as the charge increases, the potential also increases.

5. Can the potential outside a charge sphere be negative?

Yes, the potential outside a charge sphere can be negative if the reference point is chosen to be inside the sphere. This means that the energy needed to move a unit charge from infinity to that point is negative, indicating a decrease in potential energy.

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