Question

A hollow conducting sphere is placed in an electric field produced by a point charge as shown. Let $V_A,V_B,V_C$ be the electric potentials at points A,B,C respectively and $V_0$ is the potential at centre O due to induced charge on shell.

• A. $V_0=0$
• B. $V_A>V_B>V_C$
• C. $V_A=V_B=V_C$
• D. $V_A<V_B<V_C$

Answer 1

Answer: (A), (C)

The correct options are
A $V_0=0$
C $V_A=V_B=V_C$

$\text{Step 1: Potential at points A, B \& C}$

$\bullet$ As a property of conductor, In static situation, potential at all the points of the conductor is same i.e. whole conductor is equipotential.

$\bullet$ Since the sphere is a conductor, and all three points A, B and C lie on the its surface, which is equipotential. Therefore, $V_A=V_B=V_C$

$\text{Step 2: Total charge on the sphere[Refer Figure]}$

On placing a charge $q$ at point P.

Due to coulomb's Electric Force, charge redistribution will happen on the sphere, negative and positive charges will induce unevenly on the surface of the sphere, as shown in the figure.

Therefore, total charge on the sphere will remain zero, as it is isolated.

$\text{Step 3: Potential at Point O due to Induced charges}$

As the distance of all the charges on the sphere from the centre is R.

So, Potential at Centre of sphere, $V=\frac{K{Q}_{total}}{R}$

Since, Total charge on the sphere is zero as proved in Step 2 i.e. $Q_{total}=0$

Therefore, Potential at point O due to induced charges is Zero.

Hence Correct answers are Option A and C