Question on a conducting ball in the middle of a constant electric field.

[yungman]

I am referring to David Griffiths "Introduction to Electrodynamics" 3rd edition, page 141, example 3.8.

The example is regarding to a conductive ball radius = R and contain no charge being put in the middle of a constant electric field $\vec {E_0} = \hat z E_0$.

My problem with the example is the equation the book use:

$$V(r,\theta) = \sum _{l=0}^{\infty} [A_l \; r^l \;+\; B_l \; r^{-(l+1)}]\; P_l (cos \theta )$$

This is the general solution of $\nabla^2V=0$.

This is all looking nice and fine...EXCEPT we know the potential inside the ball is 0 or some constant voltage same as the potential on the surface of the ball. For simplicity, let the voltage be zero.

With that, I set $A_l = 0$ and the equation become:

$$V(r,\theta) = \sum _{l=0}^{\infty} \; B_l \; r^{-(l+1)}\; P_l (cos \theta )$$


If you look at the book how they derive the potential, they keep the $A_l$. This I cannot see with the boundary condition. Below is what the book has in case you don't have the book:

i) $V=0 \;\;\; r=R$
ii) $V=-E_0\;r cos \theta \;\;\; r>>R$

At r=R, potential is zero:

$$A_l \;+\;\frac{B_l}{R^{l+1}} = 0 \Rightarrow B_l \;=\; -A_l \; R^{2l+1}$$

$$V(r,\theta) = \sum _{l=0}^{\infty} \;[A_l (r^l \;-\; \frac{R^{2l+1}}{r^{l+1}})\; P_l (cos \theta )$$

Am I wrong to assume $A_l = 0$?

Thanks

Alan

 

[yungman]

I think I can answer my question, I was wrong to assume $A_l$ is only for inside the sphere where r<R. In this case, because we have an external field, field don't go to zero as r>>R. We need to keep $A_l$ for this. It is not a set rule that $A_l$ is for r<R and $B_l$ for r>R.


I have more question.

1) Why can we assume $V(r,\theta) = 0 \;$ at the surface. I know for fact if I use a HV supply and put 10KV on one large plate and ground the second plate. Put the plate in parallel and distance Z apart, the $\vec E = -\hat z \frac{V}{Z}$. If you put the ball in the middle, it is not going to be 0 volt! I know from working with high voltage for years that floating metal pieces arc to ground and we had to be very careful to make sure no floating conductor ( metal ) inside any high voltage enviroment.

2) Another part of the question is the problem 3.20 on page 145. It asked if the ball has initial charge Q on the ball and put into the field, what is the potential $V(r,\theta)$? They just simply add $\frac{Q}{4\pi \epsilon_0 r}$ into the original answer of the uncharged ball in the example. So we can just sum them together? Why? So the ball is no long and zero volt?

 

[Born2bwire]

Ground reference is whatever you desire it to be. Setting the surface of the sphere to be zero is a convenient reference. It is easy to see that a constant offset to the potential does not affect the resulting electric fields or Laplace's equation. It's the voltage differences that we are concerned about.

They add because when the charge is zero, we are solving for the homogeneous solution to the Poisson equation (Laplace's Equation). When we have a non-zero charge, the effects of the non-zero charge are taken care of in the particular solution. Obviously since the result when the Poisson operator acts on the homogeneous solution is zero, we can always add the homogeneous solution to the particular solution to achieve the full solution.

 

[sweet springs]

Hi, yungman

i) $V=0 \;\;\; r=R$
ii) $V=-E_0\;r cos \theta \;\;\; r>>R$

ii) requires that Al except A1 are zero.
Regards.