Question

A small uncharged conducting sphere is placed in contact with an identical sphere but having \( 4 \times 10^{-6} \, \text{C} \) charge and then removed to a distance such that the force of repulsion between them is \( 9 \times 10^{-3} \, \text{N} \). The distance between them is (Take \( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \) in SI units):

• A. \( 2 \, \text{cm} \)
• B. \( 4 \, \text{cm} \)
• C. \( 1 \, \text{cm} \)
• D. \( 3 \, \text{cm} \)

Hint:

When two identical conductors come into contact, the charge is evenly distributed. Use Coulomb's law to find the distance between them based on the force of repulsion.

Answer 1

The Correct Option is B

When two identical conducting spheres touch, their charges distribute equally. The combined charge of both spheres is: \[ Q_{\text{total}} = 4 \times 10^{-6} \, \text{C}. \] Consequently, the charge on each sphere post-contact is: \[ Q = \frac{4 \times 10^{-6}}{2} = 2 \times 10^{-6} \, \text{C}. \] Applying Coulomb's law for the repulsive force between the spheres: \[ F = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{r^2}. \] Inputting the given force and charge values to determine the distance \( r \): \[ 9 \times 10^{-3} = \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{r^2}. \] Solving for \( r \) yields \( r = 4 \, \text{cm} \). Final Answer: \( 4 \, \text{cm} \).