Potential outside charged metal sphere.

[-Dragoon-]

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

My only problem here is that I don't know where to set the zero of potential to zero. The surface doesn't work, neither does r = ∞. Any help or hints? Thanks.

 

[vanhees71]

You can use the asymptotic condition
$$U(\vec{r}) \simeq -\vec{E}_0 \cdot \vec{r} \quad \text{for} \quad |\vec{r}|\rightarrow \infty$$
to define the arbitrary additive constant of the potential. This determines the part of the potential that originates from the sphere (charge Q + influence surface-charge distribution) to tend to 0 at infinity.

Hint: This problem is solved very elegantly using the electrostatic multipole expansion. You'll see that together with the appropriate boundary conditions at the sphere's boundary you'll get a closed solution with this ansatz pretty easily.

 

[scoobmx]

You should be able to use the superposition principle and add terms from both the external field and the charged sphere into a guess solution: $V (\vec{r}) = AE_{0}rsin(\theta) + B \frac{Q}{r}$ then use some boundary conditions including the asymptotic condition mentioned above which determines the constant A.

 

[-Dragoon-]

You should be able to use the superposition principle and add terms from both the external field and the charged sphere into a guess solution: $V (\vec{r}) = AE_{0}rsin(\theta) + B \frac{Q}{r}$ then use some boundary conditions including the asymptotic condition mentioned above which determines the constant A.

Wait, isn't the correct approach to use the general solution to Laplace's equation in spherical coordinates (assuming azimuthal symmetry)? There's an example like this in the book (Griffith's) except the sphere that is placed in the external field is uncharged.

The boundary conditions for that problem were:
$V = 0$ when r = R
$V \rightarrow -E_{0}r\cos{\theta}$ for r >> R

So, for this problem I assume the second boundary condition would become:
$V \rightarrow -E_{0}r\cos{\theta} + \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r}$ for r >> R.

The only problem I seem to be having is where to set the zero of potential, and that doesn't seem at all be clear from the previous answer and from everything else that I've searched for on the web.

 

[scoobmx]

Wait, isn't the correct approach to use the general solution to Laplace's equation in spherical coordinates (assuming azimuthal symmetry)? There's an example like this in the book (Griffith's) except the sphere that is placed in the external field is uncharged.

The boundary conditions for that problem were:
$V = 0$ when r = R
$V \rightarrow -E_{0}r\cos{\theta}$ for r >> R

So, for this problem I assume the second boundary condition would become:
$V \rightarrow -E_{0}r\cos{\theta} + \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r}$ for r >> R.

The only problem I seem to be having is where to set the zero of potential, and that doesn't seem at all be clear from the previous answer and from everything else that I've searched for on the web.

Can you quote this 'general solution'? I suggested superposition because the external electric field doesn't affect the charge distribution of the sphere (at least I assumed it was all bound charges) so in the electric field point of view it's just the sum of the 2 parts. You do not need to find any 0 of potential because the value of potential is meaningless - only the gradient matters.

 

[Jilang]

It's a metal sphere so the potential will be the same over its surface. I would set the surface of the sphere to be V=0. (The charge distribution over the sphere will be arranged so this is the case).

 

[BruceW]

for this question, I think they probably just want to see that you are aware of where you have set the zero of the potential. as scoobmx says, it's not important what place you actually choose.

Also, I'd guess that you are meant to assume the charges are not bound, since the question doesn't mention anything like that. So it will be a bit more difficult to find the answer than the simple superposition method.

 

[scoobmx]

Ah, my mistake, I just read the problem again to find it's a metal sphere. You're going to want to add a dipole-like term into the superposition solution. This is because a metal sphere screening a uniform external field will move its surface charge to produce dipole-like fields. The following superposition solution is in spherical coordinates.

$V( \vec{r} ) = A rcos( \theta ) + B \frac{1}{r} + C \frac{cos( \theta )}{r^2}$

The general solution would have an extra constant in it which I left out because it can be anything. This is the reason you cannot just arbitrarily set 0 potential somewhere. You now need 3 boundary conditions.

The first you should use is the asymptotic condition that at $r = \infty$ the electric field is $E_0 \hat{z}$ which determines the constant A.

Next because the sphere is metal the potential at the surface is constant. If you included the extra constant term in the solution you could have set the surface potential to 0 at this step, but instead you must now set it to be an arbitrary constant. If the metal sphere had NO net charge you could set its potential to 0 because of symmetry. However this problem is not antisymmetric on the z axis. This should let you relate the A and C terms by an equality because both the A and C terms depend on $\theta$ and so are not constant; therefore at r = R they must cancel each other out so that V(R) does not depend on $\theta$. This step is equivalent to requiring the electric field parallel to the surface to be 0.

The final boundary condition is Gauss's law in the integral form using a spherical surface of radius R to require that the charge enclosed is Q.

 

[-Dragoon-]

The final boundary condition is Gauss's law in the integral form using a spherical surface of radius R to require that the charge enclosed is Q.

This was exactly the boundary condition that I was missing from my solution. Thank you and all the others in this thread for all the help.