Question

A simple pendulum has a bob with mass \(m\) and charge \(q\). The pendulum string has negligible mass. When a uniform and horizontal electric field \( \vec{E} \) is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is _______.
(\( g \): acceleration due to gravity)

• A. \( \sqrt{m^2 g^2 - q^2 E^2} \)
• B. \( \sqrt{m^2 g^2 + q^2 E^2} \)
• C. \( mg + qE \)
• D. \( mg - qE \)

Hint:

When forces act perpendicular to each other, the resultant magnitude is obtained using Pythagoras theorem.

Answer 1

The Correct Option is B

To determine the final tension in the string of a pendulum with a charged bob in an electric field, we need to consider all forces acting on the bob:

1. Gravitational Force: This force acts vertically downward and has a magnitude of \(mg\) where \(m\) is the mass of the bob and \(g\) is the acceleration due to gravity.
2. Electric Force: Due to the charge \(q\) on the bob and the horizontal electric field \(\vec{E}\), there is a force \(qE\) acting horizontally in the direction of the field.
3. Tension in the String: Let \(T\) be the tension in the string when the system is in equilibrium.

In equilibrium, the resultant of these forces will be balanced by the tension in the string. The tension can be resolved into vertical and horizontal components:

• The vertical component of tension balances the gravitational force: \(T_y = mg\)
• The horizontal component of tension balances the electric force: \(T_x = qE\)

The tension \(T\) in the string is the resultant of these components:

\(T = \sqrt{T_x^2 + T_y^2} = \sqrt{(qE)^2 + (mg)^2}\)

Substituting the forces, we get:

\(T = \sqrt{m^2 g^2 + q^2 E^2}\)

This gives us the final tension in the string when the pendulum attains an equilibrium position.

Conclusion: The correct answer is \(\sqrt{m^2 g^2 + q^2 E^2}\).