Charge density on inner surface of conducting shell

Question

The following exercise has brought to my mind something that confuses me:

Here, I don't see why there isn't positive charge residing on the inner surface of the conductor at $r=a$.

The way I see it, at first the conductors were neutrally charged. They were then given additional positive charge. As the outer shell is positively charged, it will exert a force on the positively charged inner shell, such that this added positive charge will be pushed to the inner surface of the conductor and the negative charge (which existed when the inner conductor was electrically neutral and hasn't gone anywhere despite being outnumbered by the positive charge) moves to the outer surface of the conductor, attracted to the outer shell.

Why is this not happening here?

Answer 1

Take the coordinate system to be the center of your arrangement. Your arrangement has spherical symmetry so any field $\vec E$ must be spherically symmetric, i.e. $\vec E= E(r)\hat r$, where $\hat r$ is the unit vector in the radial direction.

Use a Gaussian spherical surface with center coinciding with your arrangement and compute $$ \oint \vec E\cdot d\vec S= E(r_0) 4\pi r_0^2=\frac{q_{\rm encl}}{\epsilon}\, , \tag{1} $$
where $a<r_0<b$ is the radius of the Gaussian sphere, which is between the inner and outer shells. Note that, by construction, $E(r_0)$ is the constant magnitude of $\vec E$ on that Gaussian sphere.

Now, by Gauss's law, we know that the right hand side of (1) is proportional to the net charge enclosed by the Gaussian sphere. We also know that $E(r_0)=0$ since there can be no static electric field in a conductor at equilibrium. Thus, it must follow that $q_{\rm encl}=0$ since $r_0\ne 0$, i.e. the net charge enclosed is $0$. Since there is no charge inside the cavity, the only net charge would have to be on the inner surface of the conductor (since there is no charge in the bulk of a conductor at static equilibrium), so this next charge on the inner surface must therefore be $0$.

The entire charge $+Q$ must therefore reside on the exterior surface of the smaller conductor, and be symmetrically distributed. Then, any Gaussian surface with $r_0<b$ will enclose no charge, indicating by symmetry there would be no field inside that conductor, which is correct.

Answer 2

The Shell Theorem tells us that the electric field due to a uniformly charged spherical shell is zero inside of the shell, so the outer shell does not apply a net electric field to the inner shell. The reason that the charge on the inner shell is on the exterior is that each charge carrier repels the other charge carriers (of the same sign), thus pushing the charges farther away. Since the exterior is where the charge is most spread out, that's where it goes.

Answer 3

The outer charge only gives rise to a field outside the spheres, the inner charge only outside the inner sphere. Thus the charges have no effect in ide the inner sphere.