Question

A simple pendulum has mass 2 gram and charge \(2 \mu \mathrm{C}\). In a uniform horizontal electric field of intensity \(1000 \mathrm{~V} / \mathrm{m}\), pendulum is at rest. At equilibrium, the angle made by the pendulum with the vertical is ( \(\mathrm{g}=\) acceleration due to gravity \(=10 \mathrm{~m} / \mathrm{s}^2\) )

• A. \(\mathrm{tan}^{-1}(0.2)\)
• B. \(\mathrm{tan}^{-1}(0.1)\)
• C. \(\mathrm{tan}^{-1}(0.4)\)
• D. \(\mathrm{tan}^{-1}(0.5)\)

Hint:

Balance the horizontal electric force and the weight, then use tan of the angle.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At equilibrium, \(T\sin\theta=qE\) and \(T\cos\theta=mg\).

Step 2: Ratio
\(\tan\theta=\dfrac{qE}{mg}\).

Step 3: Substitute
\(q=2\times10^{-6}\ \text{C}\), \(E=1000\ \text{V/m}\), \(m=2\times10^{-3}\ \text{kg}\), \(g=10\).

Step 4: Calculation
\(\tan\theta=\dfrac{2\times10^{-6}\times1000}{2\times10^{-3}\times10}=\dfrac{2\times10^{-3}}{2\times10^{-2}}=0.1\).

Final Answer:
The angle with the vertical is \(\tan^{-1}(0.1)\). \[ \boxed{\text{(B) }\tan^{-1}(0.1)} \]