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If I have a hollow conductor of arbitrary shape, and have a charge placed inside is cavity, then is there anything I can say for sure about the charge distribution on the outer surface? I know that inside surface will be such that field in the conducting material is zero, but can I say that the outer surface charge is somewhat "shielded" and behaves just like any charge placed outside a conductor, and so will not have any role? Can I further imply that the charge distribution will be uniform on the outer surface, as if there is no charge inside?

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  • $\begingroup$ The charge distribution will be uniform, yes, since the negative charge repels itself and thus tries to cover as much area as possible in order to get as far away from each other as possible. $\endgroup$ – Steeven Jun 7 '16 at 11:57
  • $\begingroup$ But then will it not go against the fact that a conductor is an equipotential surface? Can you please explain with an example? $\endgroup$ – Shodai Jun 7 '16 at 12:08
  • $\begingroup$ If the conductor is not grounded, the outer charge will be equal to the inner charge. The distribution will depend on the geometry of the conductor, but if you look at the electric field far away from the conductor you won't notice much difference with the one produced by the charged particle alone. $\endgroup$ – Phoenix87 Jun 7 '16 at 12:27
  • $\begingroup$ Oh so charge density on the outer surface will not be same throughout? $\endgroup$ – Shodai Jun 7 '16 at 12:29
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Your statement is correct. The charge distribution is such that the hollow cavity of the conductor has a equal amount of negative charge induced on its inner part. This distribution is such that field due the cavity (including the charge inside the cavity) cancels out everywhere outside the cavity.

So looking it the other way around external sources do not influence the field inside the cavity so any charge inside the cavity is 'shielded' from external sources.

As to the distribution on the outer surface you cannot argue that it is uniform. It is only uniform jn cases where there is unique symmetry like a spherical conductor with a cavity.

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If the shell is conductive in this case, one thing for sure is that the net charge on the interior surface has the opposite sign and equal amount of charge as the charge placed inside of the cavity. If there are no net charge on the conductor before the charge inside of the cavity was placed, the charge on the conductor equal to the charge placed inside the cavity will flow to the exterior where the potential is lower and to balance the net charge on the conductor. If the conductor is grounded, the charge on the exterior will flow away while charges on the interior will stay. The reason is the following:

  1. Charge can freely flow around on the shell which has the same potential, since it is conductive (equipotential). This implies that there is no net charge between the exterior and interior. You can see this by choosing any two points between the exterior and interior, P and Q, so that the potential between these two points are zero. That is

    \begin{align} \phi(P)-\phi(Q) &= \int_P^Q \nabla\phi\cdot d\mathbf{l} = \int_P^Q \mathbf{E} \cdot d\mathbf{l}=0 \Rightarrow \mathbf{E}=0\\ \nabla\cdot \mathbf{E} &=\frac{\rho}{\epsilon_0} \end{align} Therefore, $\rho=0$ in the conductor for arbitrary path $P\rightarrow Q$. All charges have to stay on the surface of the conductor.

  2. Based on the same equipotential condition, and $\mathbf{E}=0$ inside the conductor body, one can draw a closed surface, S, between the exterior and interior surfaces of the conductor shell and conclude that \begin{align} \oint_S \mathbf{E}\cdot d\mathbf{S} &= \frac{1}{\epsilon_0}\int_V\rho dV= \frac{Q}{\epsilon_0}=0. \end{align} Since the area of S can be arbitrarily close to the interior, there must be the same amount but opposite sign of charge distributed on the interior as the placed inside charge to make the total charge $Q=\int_V\rho dV=0$.

You cannot say anything about the charge distribution pattern if you don't know the shape of the conductor body and the external field. In any case, charge will distributed so that the total potential is the lowest in all possible distribution patterns. For example, if there are electric fields elsewhere, the distribution of charge will balance the impact of the external field to find the lowest potential condition and the screen effect will take place to make sure the potential inside of the cavity won't change.

In sum, one thing for sure is that the charge on the inner surface of the conductor is $-q$ if the charged placed inside of the hollow cavity is $q$.

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Assuming that the conductor was initially uncharged, then the total charge on the outer surface of the conductor equals the charge inside the cavity.

Yes the charge on the outer surface is shielded from the charge distribution inside, with the result that the charge distribution on the outer surface is the same as when the charge is placed on the outer surface instead of inside the cavity. Unlike the distribution on the inner surface, the charge distribution on the outer surface is independent of the shape and position of the cavity and of the position of the charge inside the cavity.

The charge distribution on the outer surface will only be uniform if that surface is spherical.

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