Electric field of a spherical conductor with a dipole in the center

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Homework Statement:

Calculate the E field outside concentric spherical shell with point dipole in center

Relevant Equations:

Gauss' law/Field of a dipole
In a recent test we were asked to calculate the electric field outside a concentric spherical metal shell, in which a point dipole of magnitude p was placed in the center.
Given values are the outer radius of the shell, R, The thickness of the shell, ##\Delta R## and the magnitude of the dipole moment p

I am very tempted to say that the electric field outside the conductor is 0 due to the plus and minus side of the inner dipole inducing equal amounts of positive and negative charge on the inside of the shell.
This would leave no residual charge to reside on the outer surface of the shell, which by invoking Gauss' law means that $$Q_{enc}=0 \rightarrow E=0$$
I am not entirely sure of this argumentation and would like some input as I am aware that you would normally compute the electrical field generated far from a dipole by the formula $$E_{dip}=\frac{p}{4 \pi \epsilon_{0}r^{3}} \cdot (2 cos{\theta}\hat{r}+sin{\theta}{\hat{\theta}})$$

So my 2nd thought would be that somehow the field from the dipole moment inside the empty space in the sphere would make the sphere reallocate plus and minus charges in a way that effectively makes the sphere itself seem like a dipole when viewed from far away.
However, my guess would be that this is flawed since this would violate the condition that the conductors boundary be an equipotential, and also violating the condition that ##E=0## inside the conductor (right?)

Can someone provide feedback on my line of thinking and correct me where/if I'm wrong?
Thanks
 

Answers and Replies

  • #2
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Your second answer is right. ##Q_{enc} \Rightarrow \vec{E} = 0## only for distributions with symmetry This problem does not lend itself to Gauss's Law. ##\vec{E} = 0## inside a conductor is always true for the static case you need not worry about substantiating it on a test, it is proved generally (i.e. without assuming symmetry) in one of the uniqueness theorems.

Also, one of the corollaries of the uniqueness theorem(s) is that a conductor shields from outside fields but not inside fields.
 
  • #3
TSny
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Your first supposition is correct: E = 0 everywhere outside the sphere whenever the net charge inside the shell is zero. So, no matter where you place the dipole inside the shell, there will be no electric field outside.

There are different ways to show this. One way is essentially your argument that if there were any induced charge on the outer surface, then the net charge on the outer surface would have to be zero (via Gauss' law). But, as you said, any distribution of both positive and negative charge densities on the outer surface would prevent the outer surface of the conductor from being an equipotential surface.
 
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  • #4
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Your first supposition is correct: E = 0 everywhere outside the sphere whenever the net charge inside the shell is zero. So, no matter where you place the dipole inside the shell, there will be no electric field outside.

There are different ways to show this. One way is essentially your argument that if there were any induced charge on the outer surface, then the net charge on the outer surface would have to be zero (via Gauss' law). But, as you said, any distribution of both positive and negative charge densities on the outer surface would prevent the outer surface of the conductor from being an equipotential surface.
On second thought you are right, I just looked in Griffiths

Second Uniqueness Theorem: In a volume ##\mathcal{V}## "surrounded" by conductors and containing a specified a specified charge density ##\rho## (In this case ##\rho_{outside} = 0##), the electric field is uniquely determined if the total charge on each conductors is given. (The region as a whole can be bounded by another conductor, or else unbounded)

The outside of the conducting shell is the boundary surface and it has 0 total charge.
 

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