Question

The bob of a pendulum of mass 8$\mu g$ carries an electric charge of $39.2 \times 10^{- 10}$ coulomb in an electric field of $20 \times 10^{3}volt/meter$and it is at rest. The angle made by the pendulum with the vertical will be

• A. 27⁰
• B. 45⁰
• C. 87⁰
• D. 127⁰

Answer 1

Answer: (B)

45⁰

Answer 2

$T\sin\theta = qE$, $T\cos\theta = mg$

$\therefore$ $\tan\theta = \frac{qE}{mg}$

$\tan\theta = \frac{39.2 \times 10^{- 10} \times 20 \times 10^{3}}{8 \times 10^{- 6} \times 9.8} = 1$

⇒ $\theta = 45^{o}$