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 are in contact, their charges are shared equally. The total charge on both spheres is: \[ Q_{\text{total}} = 4 \times 10^{-6} \, \text{C}. \] Thus, the charge on each sphere after they are in contact will be: \[ Q = \frac{4 \times 10^{-6}}{2} = 2 \times 10^{-6} \, \text{C}. \] Using Coulomb's law for the force of repulsion between the two spheres: \[ F = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{r^2}. \] Substitute the known values for the force and charge and solve for the distance \( r \): \[ 9 \times 10^{-3} = \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{r^2}. \] Solving for \( r \), we find \( r = 4 \, \text{cm} \). 
Final Answer: \( 4 \, \text{cm} \).

Answer 2

Step 1: Understand the problem setup.
We are given the following information:
- One of the spheres is initially uncharged, and the other has a charge of \( 4 \times 10^{-6} \, \text{C} \),
- The spheres are placed in contact, and then removed to a distance such that the force of repulsion between them is \( 9 \times 10^{-3} \, \text{N} \),
- The constant \( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \).
We are asked to find the distance between the two spheres when the force of repulsion is \( 9 \times 10^{-3} \, \text{N} \).

Step 2: Concept of charge distribution after contact.
When two conducting spheres come into contact, they share the total charge equally. Since one sphere starts with a charge of \( 4 \times 10^{-6} \, \text{C} \), and the other is uncharged, after contact, each sphere will have half of the total charge.
Thus, after contact, each sphere will have a charge of:
\[ Q = \frac{4 \times 10^{-6}}{2} = 2 \times 10^{-6} \, \text{C}. \]

Step 3: Apply Coulomb's law for the force of repulsion.
Coulomb's law gives the electrostatic force between two point charges as:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{Q_1 Q_2}{r^2}, \] where \( Q_1 \) and \( Q_2 \) are the charges on the spheres, and \( r \) is the distance between them.
Substitute the known values:
- \( Q_1 = Q_2 = 2 \times 10^{-6} \, \text{C} \),
- \( F = 9 \times 10^{-3} \, \text{N} \),
- \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \).
We have the equation:
\[ 9 \times 10^{-3} = \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{r^2}. \] Simplifying:
\[ 9 \times 10^{-3} = \frac{9 \times 10^9 \times 4 \times 10^{-12}}{r^2}. \] \[ 9 \times 10^{-3} = \frac{36 \times 10^{-3}}{r^2}. \] Solve for \( r^2 \):
\[ r^2 = \frac{36 \times 10^{-3}}{9 \times 10^{-3}} = 4. \] Taking the square root of both sides:
\[ r = 2 \, \text{m}. \] Thus, the distance between the spheres is \( 2 \, \text{m} \), or \( 4 \, \text{cm} \).

Final answer:
The distance between the spheres is \( 4 \, \text{cm} \).
\[ \boxed{4 \, \text{cm}}.\]