Question

A small spherical ball having charge \( q \) and mass \( m \), is tied to a thin massless non-conducting string of length \( l \). The other end of the string is fixed to an infinitely extended thin non-conducting sheet with uniform surface charge density \( \sigma \). Under equilibrium, the string makes an angle of 45° with the sheet as shown in the figure. Then \( \sigma \) is given by \[ g \text{ is the acceleration due to gravity and } \epsilon_0 \text{ is the permittivity of free space.} \] 

• A. \( \frac{mg\epsilon_0}{q} \)
• B. \( \sqrt{2} \frac{mg\epsilon_0}{q} \)
• C. \( 2 \frac{mg\epsilon_0}{q} \)
• D. \( \frac{mg\epsilon_0}{q\sqrt{2}} \)

Hint:

When dealing with forces in equilibrium, resolve the forces into components and use trigonometric identities to find the desired quantities.

Answer 1

The Correct Option is C

Step 1: Understanding the forces involved.
The spherical ball with charge \( q \) experiences two main forces: - The gravitational force \( F_g = mg \), acting downward. - The electrostatic force due to the surface charge on the sheet, which creates an electric field \( E \). The electric field due to a uniformly charged infinite sheet is given by: \[ E = \frac{\sigma}{2 \epsilon_0} \] where \( \sigma \) is the surface charge density of the sheet and \( \epsilon_0 \) is the permittivity of free space.
Step 2: Forces on the spherical ball.
At equilibrium, the forces on the spherical ball are balanced. The ball experiences the following: - The electrostatic force \( F_e = qE = \frac{q \sigma}{2 \epsilon_0} \), acting to the right (towards the sheet). - The gravitational force \( F_g = mg \), acting downward. - The tension \( T \) in the string, which has two components: one vertical (balancing \( mg \)) and one horizontal (balancing \( F_e \)). Since the string makes a 45° angle with the sheet, we can resolve the tension into horizontal and vertical components: \[ T \cos(45^\circ) = F_e \quad \text{(horizontal component)} \] \[ T \sin(45^\circ) = mg \quad \text{(vertical component)} \]
Step 3: Solving for \( \sigma \).
Using the fact that \( \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we can equate the horizontal and vertical components of tension: From the vertical component: \[ T \sin(45^\circ) = mg \quad \Rightarrow \quad T = \frac{mg}{\sin(45^\circ)} = \frac{mg}{\frac{1}{\sqrt{2}}} = \sqrt{2} mg \] From the horizontal component: \[ T \cos(45^\circ) = F_e \quad \Rightarrow \quad \sqrt{2} mg \cdot \frac{1}{\sqrt{2}} = \frac{q \sigma}{2 \epsilon_0} \] Simplifying: \[ mg = \frac{q \sigma}{2 \epsilon_0} \] \[ \sigma = \frac{2 mg \epsilon_0}{q} \] Thus, the correct expression for \( \sigma \) is: \[ \sigma = \frac{mg \epsilon_0}{q \sqrt{2}} \]
Step 4: Conclusion.
The correct answer is (D) \( \frac{mg \epsilon_0}{q \sqrt{2}} \).