The charge always lies on the outer surface of a conductor. If there is a cavity inside a conductor, we can also prove that the charge always lies on the outer surface of the conductor. A conductor with a cavity is shown in Figure 1. The conductor initially has charge \(+q\) and in electrostatic situation, that is charges not in motion, the electric field inside the conductor should be zero.

Figure 1 The charge \(+q\) inside the cavity insulated form the cavity surface induces negative charge surrounding it and therefore the charge \(+q_i\) is added to the charge on conductor's surface.

Now we put charge \(+q\) inside the cavity insulated form the conductor. The charge \(+q\) induces charge \(-q_i\) around the cavity surface and again the electric field inside the conductor is zero. The additional \(+q_i\) charge is added to the charge on the conductor's surface. Now the conductor has charge \(q+q_i\). Note that the subscript \(i\) represents the induced charge. Since in electrostatic situation, there should not be any electric field inside the conductor and hence no charge (using Gauss law) and the additional charge is added to the conductor's original charge. We test this with Gauss's law. Make a Gaussian surface like the one shown in Figure 2.

If the electric field inside the Gaussian surface is zero, the net charge enclosed by the Gaussian surface is also zero. If the excess charge tries to go towards the interior of the conductor there will be electrostatic interaction between the excess charge and the free electrons within the conductor which results in the motion of the excess charge but the excess charge ultimately remains at rest and for this reason the outer surface is the best place to go (electrostatic situation).

What's the electric field at the surface?

Now let's determine the electric field at the surface of a conductor. The electric field is radially outward at every point on the conductor's surface; it means the electric field is perpendicular to the conductor's surface. If the electric field has any component parallel to the surface, then we need to explain why it has that component. Also that component affects the electrostatic equilibrium. Make a Gaussian surface as shown in Figure 3 which is a cylinder and let the surface charge density of the conductor be \(\sigma \).

Figure 2 The electric field of a charged conductor is perpendicular to the surface of the conductor.

The electric field is radially outward so it is parallel to the curved Gaussian surface i.e. the curved part of the cylinder and perpendicular to one end of the Gaussian cylinder outside the conductor. The electric flux through the curved part of the cylinder is zero. The electric field within the conductor is zero, so there is no electric flux through the end within the conductor. Now the total charge enclosed by the Gaussian surface is the product of the surface charge density and the area of the conductor's surface enclosed by the Gaussian surface which is \(A\) (equal to the area of one end of the cylinder). So the total net charge enclosed is \(q = \sigma A\). Also the total electric flux through the Gaussian surface is \(EA\). Now applying Gauss's Law,

\[\begin{align*} EA &= \frac{{\sigma A}}{{{\epsilon_0}}}\\ \therefore E &= \frac{\sigma }{{{\epsilon_0}}} \tag{1} \end{align*}\]

This is the same expression as that of between two oppositely charged parallel plates.