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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2026 · 22 January · Morning Shift · Physics · Electrostatics

A simple pendulum has a bob with mass m and charge q . The pendulum...

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)

Options

  1. Am2g2+q2E2\sqrt{m^{2} g^{2}+q^{2} E^{2}}
  2. Bm g+qEm \mathrm{~g}+q E
  3. Cmg−qEm g-q E
  4. Dm2g2−q2E2\sqrt{m^{2} g^{2}-q^{2} E^{2}}

Correct answer

A. m2g2+q2E2\sqrt{m^{2} g^{2}+q^{2} E^{2}}

Step-by-step solution

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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