Question

A simple pendulum with bob (mass $m$ and charge $q$) is in equilibrium in the presence of a horizontal electric field $E$. Then the tension in the thread is:

• A. $mg + qE$
• B. $\sqrt{m^2 g^2 + q^2 E^2}$
• C. $\sqrt{mg + qE}$
• D. $mg + qE \tan\theta$

Hint:

Whenever multiple forces act perpendicular to each other in equilibrium, the tension balances their \textbf{vector resultant}, not their arithmetic sum.

Answer 1

The Correct Option is B

Concept: When a charged pendulum bob is placed in a horizontal electric field, it experiences two external forces:
Gravitational force $mg$ acting vertically downward
Electric force $qE$ acting horizontally In equilibrium, the tension $T$ in the string balances the resultant of these two mutually perpendicular forces. Key principles used:
Resolution of forces in perpendicular directions
Resultant of perpendicular vectors: $R = \sqrt{F_1^2 + F_2^2}$
Step 1: Identify all forces acting on the bob.
Weight of bob: $mg$ (downward)
Electric force: $qE$ (horizontal)
Tension $T$ along the string
Step 2: Resolve the tension into vertical and horizontal components. \[ T \cos\theta = mg \quad \text{(vertical equilibrium)} \] \[ T \sin\theta = qE \quad \text{(horizontal equilibrium)} \]
Step 3: Square and add the two equations. \[ (T \cos\theta)^2 + (T \sin\theta)^2 = (mg)^2 + (qE)^2 \]
Step 4: Use the identity $\cos^2\theta + \sin^2\theta = 1$. \[ T^2 = m^2 g^2 + q^2 E^2 \]
Step 5: Take square root to find the tension. \[ T = \sqrt{m^2 g^2 + q^2 E^2} \]