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Question

A simple pendulum of length $l$ has a bob of mass $m$, with a charge $q$. On it a vertical sheet of charge, with surface charge density '$\sigma$' passes through the point of suspension. At equilibrium, if the string makes an angle $\theta$ with the vertical, then

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$\tan\theta = \frac{\sigma q}{2 \varepsilon_0 m g}$

Forces on Charged Pendulum Bob

At equilibrium, the pendulum bob experiences three forces:

  • Gravitational Force ($F_g$): acting vertically downwards, $F_g = mg$.
  • Electrostatic Force ($F_e$): acting horizontally due to the electric field of the charged sheet.
  • Tension ($T$): acting along the string towards the point of suspension.

Electric Field and Electrostatic Force

The electric field ($E$) produced by a vertical sheet of charge with surface charge density $\sigma$ at a distance from the sheet is given by:

$E = \frac{\sigma}{2 \varepsilon_0}$

This electric field exerts a horizontal electrostatic force ($F_e$) on the charged bob ($q$):

$F_e = qE = q \left( \frac{\sigma}{2 \varepsilon_0} \right) = \frac{\sigma q}{2 \varepsilon_0}$

Equilibrium Condition

When the string makes an angle $\theta$ with the vertical, the forces are balanced. Resolving the tension force into horizontal and vertical components:

  • Vertical component: $T \cos\theta = mg$
  • Horizontal component: $T \sin\theta = F_e = \frac{\sigma q}{2 \varepsilon_0}$

Dividing the horizontal component equation by the vertical component equation:

$\frac{T \sin\theta}{T \cos\theta} = \frac{\frac{\sigma q}{2 \varepsilon_0}}{mg}$

$\tan\theta = \frac{\sigma q}{2 \varepsilon_0 m g}$

Result

The equilibrium condition is met when $\tan\theta = \frac{\sigma q}{2 \varepsilon_0 m g}$.

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