Let's solve the problem step by step. --- ### **Step 1: Analyze the motion when only gravity acts** - **Given:** Both spheres have identical mass (\(m\)), radius (\(r\)), and charge (\(-q\)), but one is conducting and the other is non-conducting. - **If only gravity acts:** - Both spheres are dropped from the same height (\(h\)). - Time to reach ground (ignoring air resistance and other forces) is independent of mass, charge, or whether they are conducting or non-conducting. **Explanation:** Both spheres will fall for the same time (option a) and will have the same velocity just before they land (option b), as per the basic laws of free fall. --- ### **Step 2: Effect of ground with uniform charge density (\(+\sigma\))** - **Ground with charge density \(+\sigma\)** creates an electric field (\(E = \frac{\sigma}{2\epsilon_}\)) above the ground (using Gauss's Law for a large plate). - **Force on each sphere:** - Both have charge \(-q\), so electric force on each: \(F_E = -qE\). - **For a conducting sphere:** The charge may redistribute due to induction, possibly increasing the force. - Conducting spheres can experience an additional attractive force due to induced charges, making them accelerate faster and reach the ground earlier. **Explanation:** So, (c) is correct: the conducting sphere will reach earlier if the ground is charged. And (d) is also correct: the conducting sphere will have a greater velocity when it lands in this scenario. --- ## **Final Answer** **Correct options:** **(a), (b), (c), and (d)** - (a) and (b): True if only gravity acts. - (c) and (d): True if the ground has a uniform charge density \(+\sigma\); the conducting sphere reaches earlier and with greater velocity.

## Problem Analysis Two spheres, each with radius \(r\) and mass \(m\), are released from a height \(h\). Both spheres have the same charge (\(-q\), with one being conducting and the other non-conducting. We will analyze the motion of these spheres under gravity and the influence of a charged ground. --- ### **Step 1: Motion Under Gravity Alone** - **Scenario:** Both spheres are dropped from height \(h\). - **Key Points:** - The time taken for both spheres to reach the ground is given by the formula for free fall: \[ t = \sqrt{\frac{2h}{g}} \] - This time is independent of mass and charge. **Explanation:** Since both spheres are affected equally by gravity alone, they will fall for the same amount of time (option a) and will have the same velocity just before hitting the ground (option b). --- ### **Step 2: Effect of Ground with Uniform Charge Density (+σ)** - **Scenario:** The ground has a uniform charge density denoted as \(+\sigma\). - **Electric Field Calculation:** - The electric field (\(E\)) above an infinite sheet of charge is given by: \[ E = \frac{\sigma}{2\epsilon_} \] - This field exerts a force on the charged spheres: \[ F_E = -qE \] - **Conducting Sphere Behavior:** - The conducting sphere can redistribute its charge due to the electric field, leading to a stronger effective force acting on it. **Explanation:** The conducting sphere will experience an additional attractive force due to induced charges, causing it to accelerate faster compared to the non-conducting sphere. Therefore, the conducting sphere will reach the ground earlier (option c) and will have a greater velocity upon landing (option d). --- ## **Final Answer** **Correct options:** - **(a)** Both will fall for the same time. - **(b)** The velocity of both spheres will be the same just before they land on the ground. - **(c)** If the ground has a uniform charge density \(+\sigma\), then the conducting sphere will reach the ground earlier. - **(d)** The conducting sphere will have greater velocity upon landing. Thus, all options **(a), (b), (c), and (d)** are correct.