Question

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

• A. \( T=\sqrt{(qE)^2+(mg)^2} \)
• B. \( T=mg+qE\tan\theta \)
• C. \( T=\sqrt{(qE)^2-(mg)^2} \)
• D. \( T=mg-qE\tan\theta \)

Hint:

For equilibrium under perpendicular forces:
Resultant force magnitude is found using vector addition
Tension balances the resultant of all external forces

Answer 1

The Correct Option is A

Concept:
In equilibrium, the bob is acted upon by three forces:
Weight \(mg\) acting vertically downward
Electric force \(qE\) acting horizontally
Tension \(T\) in the string along the string direction The tension balances the vector sum of the weight and the electric force.
Step 1: Resolve forces along vertical and horizontal directions. Vertical equilibrium: \[ T\cos\theta = mg \] Horizontal equilibrium: \[ T\sin\theta = qE \]
Step 2: Square and add the two equations. \[ (T\cos\theta)^2 + (T\sin\theta)^2 = (mg)^2 + (qE)^2 \] \[ T^2(\cos^2\theta+\sin^2\theta) = (mg)^2 + (qE)^2 \] \[ T^2 = (mg)^2 + (qE)^2 \]
Step 3: Solve for tension. \[ T=\sqrt{(mg)^2+(qE)^2} \] \[ \boxed{T=\sqrt{(qE)^2+(mg)^2}} \]