Question

A positive point charge Q is kept (as shown in the figure) inside a neutral conducting shell whose centre is at C. An external uniform electric field E is applied. Then :

• A. Force on Q due to E is zero
• B. Net force on Q is zero
• C. Net force acting on Q and conducting shell considered as a system is zero
• D. Net force acting on the shell due to E is zero.

Answer 1

Answer: (D)

(D) Net force acting on the shell due to E is zero.

Answer 2

1. Force on Q due to E: The force on a charge $q$ in an electric field $\vec{E}$ is $\vec{F} = q\vec{E}$. Since $Q$ is a positive charge and $\vec{E}$ is a uniform external electric field, the force on $Q$ due to $\vec{E}$ is $\vec{F}_{Q,E} = Q\vec{E}$. This force is generally non-zero. Thus, option (A) is incorrect.

2. Net force on the system (Q + shell): The system consists of charge $Q$ and a neutral conducting shell. The total charge of the system is $Q$. When a system with total charge $Q$ is placed in a uniform external electric field $\vec{E}$, the net force on the system is $\vec{F}_{system} = Q\vec{E}$. Therefore, the net force on the system is not zero. Thus, option (C) is incorrect.

3. Force on the shell due to E: A neutral conductor placed in a uniform external electric field experiences no net force. The external field induces charges on the surface of the conductor, but the forces on these induced charges due to the external field cancel out, resulting in zero net force on the conductor. Thus, the net force acting on the shell due to $\vec{E}$ is zero. Option (D) is correct.

4. Net force on Q: The net force on $Q$ is the sum of the force due to the external electric field $\vec{E}$ and the force due to the induced charges on the conducting shell. Let $\vec{F}_{Q,induced}$ be the force on $Q$ due to the induced charges on the shell. The net force on $Q$ is $\vec{F}_{net,Q} = Q\vec{E} + \vec{F}_{Q,induced}$. We know that the total force on the system is $\vec{F}_{system} = \vec{F}_{net,Q} + \vec{F}_{shell}$. We also know that $\vec{F}_{shell} = \vec{F}_{shell,E} + \vec{F}_{shell,Q}$, where $\vec{F}_{shell,E}$ is the force on the shell due to the external field $\vec{E}$, and $\vec{F}_{shell,Q}$ is the force on the shell due to charge $Q$. From point 3, $\vec{F}_{shell,E} = 0$. By Newton's third law, the force exerted by $Q$ on the induced charges of the shell ($\vec{F}_{shell,Q}$) is equal and opposite to the force exerted by the induced charges of the shell on $Q$ ($\vec{F}_{Q,induced}$). So, $\vec{F}_{shell,Q} = -\vec{F}_{Q,induced}$. Substituting these into the equation for $\vec{F}_{system}$: $Q\vec{E} = (Q\vec{E} + \vec{F}_{Q,induced}) + (0 + \vec{F}_{shell,Q})$ $Q\vec{E} = Q\vec{E} + \vec{F}_{Q,induced} + (-\vec{F}_{Q,induced})$ $Q\vec{E} = Q\vec{E}$. This identity confirms our previous findings.

   Now, consider option (B): Net force on $Q$ is zero ($\vec{F}_{net,Q} = 0$). If $\vec{F}_{net,Q} = 0$, then $Q\vec{E} + \vec{F}_{Q,induced} = 0$, which means $\vec{F}_{Q,induced} = -Q\vec{E}$. This implies that the electric field produced by the induced charges on the shell at the location of $Q$ must be equal and opposite to the external field $\vec{E}$. This is not generally true for an arbitrary position of $Q$ inside the cavity, especially when $Q$ is not at the center. For example, if $Q$ is very close to the inner surface, the induced charge distribution will be non-uniform and will create a field that, when added to $\vec{E}$, does not necessarily result in zero total field at $Q$. Therefore, option (B) is generally false.