Question

A simple pendulum of length \( l \) has a bob of mass \( m \), with a charge \( q \). On it a vertical sheet of charge with surface charge density \( \sigma \) passes through the point of suspension. At equilibrium, if the string makes an angle \( \theta \) with the vertical, then

• A. \( \tan\theta = \frac{\sigma q}{2\varepsilon_0 mg} \)
• B. \( \tan\theta = \frac{\sigma q}{\varepsilon_0 mg} \)
• C. \( \cot\theta = \frac{\sigma q}{2\varepsilon_0 mg} \)
• D. \( \cot\theta = \frac{\sigma q}{\varepsilon_0 mg} \)

Hint:

For any particle under gravity and a constant horizontal force \( F_h \), the angle \( \theta \) of suspension with the vertical at equilibrium always satisfies \( \tan\theta = \frac{F_h}{mg} \). Simply substituting the electric force as \( F_h \) gives the result immediately.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The bob of the pendulum is in equilibrium, meaning the net force acting on it is zero.
Three distinct forces act on the bob:
1. The force of gravity (\(mg\)) acting vertically downwards.
2. The tension (\(T\)) in the string acting along the length of the string towards the point of suspension.
3. The electrostatic force (\(F_e\)) due to the electric field of the vertical charged sheet.
A vertical infinite sheet of charge produces a uniform horizontal electric field.
The equilibrium position is reached when the string tilts by an angle \(\theta\) such that the horizontal and vertical components of the tension balance the other forces.
Step 2: Key Formula or Approach:
Electric field of an infinite non-conducting sheet: \(E = \frac{\sigma}{2\epsilon_0}\).
Electrostatic force on charge \(q\): \(F_e = qE\).
Equilibrium conditions: \(\sum F_x = 0\) and \(\sum F_y = 0\).
Step 3: Detailed Explanation:
First, let's identify the direction and magnitude of the electric field. The sheet is vertical, so the field it produces is horizontal and perpendicular to the sheet.
The magnitude of the field is:
\[ E = \frac{\sigma}{2\epsilon_0} \]
The electrostatic force acting horizontally on the bob is:
\[ F_e = qE = \frac{q\sigma}{2\epsilon_0} \]
At equilibrium, the string makes an angle \(\theta\) with the vertical. Resolving the tension \(T\) into components:
Vertical component = \(T \cos\theta\)
Horizontal component = \(T \sin\theta\)
Applying the equilibrium condition in the vertical direction:
\[ T \cos\theta = mg \quad \text{--- (Eq. 1)} \]
Applying the equilibrium condition in the horizontal direction:
\[ T \sin\theta = F_e = \frac{q\sigma}{2\epsilon_0} \quad \text{--- (Eq. 2)} \]
To find \(\theta\), we divide Eq. 2 by Eq. 1:
\[ \frac{T \sin\theta}{T \cos\theta} = \frac{(q\sigma) / (2\epsilon_0)}{mg} \]
The tension \(T\) cancels out, and \(\sin\theta / \cos\theta = \tan\theta\):
\[ \tan\theta = \frac{q\sigma}{2\epsilon_0 mg} \]
This matches option (A).
Step 4: Final Answer:
The angle of deflection is determined by the ratio of the electrostatic force to the weight. Substituting the field for a single non-conducting sheet yields \(\tan\theta = \frac{\sigma q}{2\epsilon_0 mg}\).