Question

A hollow metal ball carries an electric charge. At which of the following points is the electric field zero?
A) outside the sphere
B) on its surface
C) inside the sphere
D) at a distance more than twice

Answer 1

Hint: Free charges only lie on the surface of a conductor. The hollow ball is made up of metal and therefore can act as a conductor.

Formulas used:
The electric field experienced at a distance $r$ from the point at which a charge $q$ is placed is given by, $E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}\hat r$ and it is along the direction of the position vector $r$

Complete step by step answer:
Given a hollow metal ball that carries an electric charge. Since the metal ball is a conductor any free charges present inside the ball will move onto reside on the surface of the conductor i.e., there are no free charges present inside the ball.
The electric field experienced at a distance $r$ from the point at which a charge $q$ is placed is given by, $E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}\hat r$ and it is along the direction of the position vector $r$.
Let $R$ be the radius of the hollow metal ball that carries a charge $q$. Let $r$ be the distance at which an electric field is experienced. The charge resides on the surface for a conductor.
Then for $r > R$ i.e., outside the ball, we have $E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}$ and is directed along the position vector.
Now for $r < R$ i.e., inside the ball, we have $E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}} = 0$ since inside the sphere no charge is present ( $q = 0$ ).

Therefore, the electric field is zero inside the surface of the hollow metal ball that carries an electric charge.

Additional information:
Gauss’ law states that the electric flux through a closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the charge enclosed by the surface. The expression $E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}\hat r$ also represents the electric field present at a distance $r$ from the centre of the sphere and is an application of Gauss’ law.

Note:
This can also be explained based on spherical symmetry. We assume that the hollow metal ball has a uniform charge distribution on its surface. Now we consider a point inside the sphere. All the points on the surface have the same nature and charge distribution and thus all the points on the surface will create the same electric field and will be directed opposite to each other. Thus inside the ball, the electric field will be zero.