Question:
The bob of a simple pendulum has mass 2g and a charge of \(5.0\ \mu\mathrm{C}\). It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is : (take \(g = 10\ \mathrm{m/s^2}\))
The bob of a simple pendulum has mass 2g and a charge of \(5.0\ \mu\mathrm{C}\). It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is : (take \(g = 10\ \mathrm{m/s^2}\))
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At equilibrium \(T\cos\theta = mg\) and \(T\sin\theta = qE\), so \(\tan\theta = \frac{qE}{mg}\).
Updated On: Nov 5, 2026
- \(\tan^{-1}(5.0)\)
- \(\tan^{-1}(2.0)\)
- \(\tan^{-1}(0.5)\)
- \(\tan^{-1}(0.2)\)
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The Correct Option is C
Solution and Explanation
Electric force \(qE = 5.0\times 10^{-6}\times 2000 = 1.0\times 10^{-2}\ \mathrm{N}\).
Weight \(mg = 2\times 10^{-3}\times 10 = 2.0\times 10^{-2}\ \mathrm{N}\).
Balance of forces on the bob: \(T\cos\theta = mg\), \(T\sin\theta = qE\).
\(\tan\theta = \frac{qE}{mg} = \frac{1.0\times 10^{-2}}{2.0\times 10^{-2}} = 0.5\), so \(\theta = \tan^{-1}(0.5)\).
Answer: (3) \(\tan^{-1}(0.5)\)
Weight \(mg = 2\times 10^{-3}\times 10 = 2.0\times 10^{-2}\ \mathrm{N}\).
Balance of forces on the bob: \(T\cos\theta = mg\), \(T\sin\theta = qE\).
\(\tan\theta = \frac{qE}{mg} = \frac{1.0\times 10^{-2}}{2.0\times 10^{-2}} = 0.5\), so \(\theta = \tan^{-1}(0.5)\).
Answer: (3) \(\tan^{-1}(0.5)\)
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