Question

A small bob of mass 100 mg and charge +10 µC is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density \( \sigma \) as shown in figure. If the string subtends an angle of 45° with the sheet at equilibrium, the charge density of sheet will be :

• A. 0.885 nC/cm\(^2\)
• B. 17.7 nC/cm\(^2\)
• C. 885 nC/cm\(^2\)
• D. 1.77 nC/cm\(^2\)

Hint:

When calculating the charge density from the angle, consider the forces acting on the object and how the electric field interacts with the charge. Use the equilibrium condition to set up the necessary equations.

Answer 1

The Correct Option is D

From the diagram in the solution, we have the force acting on the charge due to the electric field of the sheet: 
The force is given by: \[ F_e = qE = mg \] where \( q \) is the charge and \( E \) is the electric field due to the sheet. The electric field is related to the charge density \( \sigma \) as: \[ E = \frac{\sigma}{2 \epsilon_0} \] Thus, the equation becomes: \[ q \left( \frac{\sigma}{2 \epsilon_0} \right) = mg \] Rearranging to solve for \( \sigma \): \[ \sigma = \frac{2g m}{q} \] Substitute the known values: \[ \sigma = \frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}} \] \[ \sigma = 17.7 \times 10^{-10} \text{ C/m}^2 \] \[ \sigma = 1.77 \text{ nC/cm}^2 \] Thus, the charge density of the sheet is \( 1.77 \) nC/cm\(^2\).