Two charges inside and outside a hollow conductor

[Selab]

Homework Statement

Two charge q1 and q2 are located respectively inside and outside a hollow conductor. Charge q2 experience a force due to q1 but not vice versa. Explain this appardnt violation of Newton's third law. There is no charge on the conductor.

Relevant Equations

F=Ke×q1×q2/r^2

Please help

 

[Delta2]

Hmmm, not sure how to resolve the apparent violation. I think we should think about the forces between each charge and the hollow conductor. Though the hollow conductor has no net charge, i believe the aforementioned forces will not all be zero. I think if we do the math correctly we can prove that the force from the hollow conductor to the inner charge q1 will be equal to $-K\frac{q_1q_2}{r^2}$.

(in view of @etotheipi post #3 i cancel out the last statement, it is not true).

EDIT 2: Now that i think of it again, i just did a mistake in the sign at first place.

 

[PhDeezNutz]

This is my take on it. I'm far from the most knowledgeable poster on these forums but I've been recently been gaining a lot of confidence. (hopefully it's not misplaced)

We have a conductor with a net charge of $0$ and then we put (somehow) $+q_1$ inside of it.

$+q_1$ inside the conductor induces total charge $-q_1$ on the inner surface of the conductor in order to cancel out the field in the "meat" of the conductor. Because the conductor is overall neutral there must be $+q_1$ on the outer surface of the conductor. This $+q_1$ on the outer shell exerts a force on $q_2$ and visa-versa.

So the cavity charge doesn't directly exert a force on $q_2$ but does so through charge redistribution on the outer surface.

But why doesn't $q_2$ exert a force on $q_{1,cavity}$? (The crux of the issue) Because $-q_{1}$ on the inner wall of the conductor kills off the field of the cavity charge (The field has to be zero inside a conductor) $q_{1,cavity}$ for exterior points.

This is easily reconcilable if you think of a spherically symmetric conducting shell with $q_1$ smack dab in the center of the cavity...but the truth is that your problem statement/conclusion applies to more than just spherically symmetric configurations.

I will try to think of a more general proof using relevant uniqueness theorems while I'm waiting on my computer to finish running a really long program lol.

Anyway, even if my reasoning isn't complete/general I think it gives you a starting point.

TLDR; The cavity charge doesn't directly exert a force on the external charge. Instead it exerts a force by redistributing charge on the conductor (both on the inner surface and outer surface).

 

[haruspex]

The apparent paradox arises through assumptions one makes using expressions like "due to".
There are three charged bodies here. (Ok, the shell has no net charge, but its charges create a field. )We can consider the force on one as due to the sum of forces from the other two, or as due to the net field the other two create.
So we can say q1 experiences no force, or we can say it experiences equal and opposite forces from q2 and the shell.

 

[Delta2]

I think this apparent paradox, is partly due to how we think that the shell (hollow conductor) does the shielding: It does not "kill" the external electric field in an immediate and explicit way, it rather produces (though it has net charge 0, the charge density is not everywhere zero on it) an equal and opposite to the external field, field, so that the total field in its interior is zero.