# Charge kept inside a conducting sphere not at its center

Assume a conducting sphere has a charge which is placed at its center like in the figure given below. In this setup charge is only induced in the walls of the sphere, right? Why cannot the charge be induced inside the actual solid part of the sphere, I have two questions regarding this.

1. The electric field inside a conducting sphere should be zero, right, but if we take a gaussian surface at any point from the center which is less than the radius and is not inside the walls, we have a charge inside that. Does that mean that the electric field in this part is non-zero?

2. Also why and how can I prove that the outer charges will not have any effect on the inside placed charge?

## 2 Answers

This is a conductor and in static equilibrium all charges are on the surfaces of the conductor, and the $$\vec E$$ inside the conductor is $$0$$.

This follows because, if there’s an $$\vec E$$-field inside, the (negative) charges would move, thus violating the hypothesis that we have static equilibrium. Since there’s no field inside, an arbitrary Gaussian surface anywhere inside the conductor would always have no flux and thus enclose no net charge, so there can be no net charge inside the conductor. Thus any induced charge must be on the physical surface as it cannot be inside the conductor.

If you can a Gaussian sphere inside the hollow sphere, but which excludes the charge, then it will contain no charge but field will not be zero: this is because, from the charge distribution, it is clear that the field is not constant on the surface of your Gaussian sphere so that $$\oint \vec E\cdot d\vec S \ne \vert \vec E\vert \times S\, .$$

To prove your second point, note that the surface of a conductor must be an equipotential so you have the outer sphere to be an equipotential, irrespective of the location of the charge inside. Moreover, the net charge on the outer portion must be exactly $$+q$$ since the induced charge on this inner surface is $$-q$$ so that a Gaussian sphere, enclosing the inner surface and concentric with the hollow sphere, satisfies $$\oint \vec E\cdot d\vec S =0 = q_{net}/\epsilon_0$$ since there is no field inside the conductor, i.e. the Gaussian sphere must enclose no net charge (hence $$-q$$ on the inner surface). In other words, irrespective of the location of the charge inside, thee is always $$+q$$ distributed uniformly on the outer surface.

You can then solve Poisson’s equation in spherical coordinates outside the spherical conductor to get the potential, with boundary condition $$V=V_0$$ on the surface, so the potential outside the spherical conductor is $$V(r)$$ irrespective of the location of the charge inside. Once you have $$V(r)$$ (it will be the same as for a spherical distribution), you recover $$\vec E=-{\hat r} dV(r)/dr$$ everywhere outside, and since $$V(r)$$ does not depend on the location of the inside charge, $$\vec E(r)$$ will not depend on this either once you are outside the arrangement.

for your first question you are right, the electric field inside a conductor is $$0$$, but not its convex hull. To make things clear, I’ll write $$r_i$$ and $$r_e$$ the interior and exterior radii of the conducting shell. When the gaussian surface is within the ball of radius less than $$r_i$$, the there is a non trivial flux if it encloses the point charge. In fact, this shows that the induced charged on the interior face of the shell is equal and opposite to the point charge.

For your second question, in the case of of spherical symmetry, it is simply a consequence of Newton’s theorem (https://en.m.wikipedia.org/wiki/Shell_theorem), applicable also for gravitation. There is a geometrical proof, but you can also see it using Gauss’ law.