Electrostatics

JEE Main level, single correct, medium

58 of 311

Question 58 of 311

A simple pendulum has a bob with mass mm and charge qq. The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗\vec{E} is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ____\_\_\_\_. (g: acceleration due to gravity)

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Answer: (A) m2g2+q2E2\sqrt{m^{2} g^{2}+q^{2} E^{2}}

The charged bob experiences two forces at equilibrium:
Gravitational force: Fg=mgF_g = mg (downward)
Electric force: FE=qEF_E = qE (horizontal, in direction of field)
These forces are perpendicular to each other. The string tension must balance their resultant.
At equilibrium, the string makes angle θ\theta with vertical such that:
Vertical: Tcos⁡θ=mgT\cos\theta = mg
Horizontal: Tsin⁡θ=qET\sin\theta = qE
The net force from gravity and electric field is (mg)2+(qE)2\sqrt{(mg)^2 + (qE)^2}
Therefore, the tension in the string is: T=m2g2+q2E2T = \sqrt{m^2g^2 + q^2E^2}

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