Question

A metallic rod is placed in a uniform electric field. Select the CORRECT option.

• A. Inside the rod there will be an induced electric field from B to A which cancels external uniform field in the rod.
• B. Free electrons will accumulate at the end B of the rod.
• C. The potential of the end A will be more than that at B.
• D. The electric field outside the rod will not change due to the induced charges in the rod.

Answer 1

Answer: (A)

Inside the rod there will be an induced electric field from B to A which cancels external uniform field in the rod.

Answer 2

When a metallic rod (a conductor) is placed in a uniform external electric field, the free electrons within the conductor redistribute themselves until the net electric field inside the conductor becomes zero.

1. Charge Redistribution: Let the uniform external electric field ($\vec{E}_{ext}$) point from end A to end B of the rod. Free electrons, being negatively charged, experience a force opposite to the direction of the electric field. Therefore, they will move towards end A (against the direction of $\vec{E}_{ext}$). This causes an accumulation of negative charges at end A and leaves a net positive charge at end B (due to the deficiency of electrons).

2. Induced Electric Field: These accumulated charges (negative at A and positive at B) create an induced electric field ($\vec{E}_{ind}$) inside the rod. This induced field points from the positive end (B) to the negative end (A).

3. Electrostatic Equilibrium: The charge redistribution continues until the induced electric field ($\vec{E}_{ind}$) inside the conductor becomes equal in magnitude and opposite in direction to the external electric field ($\vec{E}_{ext}$). At this point, the net electric field inside the conductor is zero ($\vec{E}_{net} = \vec{E}_{ext} + \vec{E}_{ind} = 0$).

The induced electric field is from B to A and cancels the external field.