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ExamsJEE AdvancedPhysics

A simple pendulum of length L hangs between the plates of a parallel-plate capacitor that produces a uniform horizontal electric field E. The bob has mass m and charge q. What is the time period of small oscillations of this pendulum?

  1. 2*pi*sqrt(L / sqrt(g² + (qE/m)²))
  2. 2*pi*sqrt(L / (g + qE/m))
  3. 2*pi*sqrt(L / (g - qE/m))
  4. 2*pi*sqrt(L / sqrt(g² - q²*E²/m²))

Correct answer: 2*pi*sqrt(L / sqrt(g² + (qE/m)²))

Solution

The bob experiences gravity (downward, acceleration g) and an electric force (horizontal, acceleration qE/m). These are perpendicular, so the effective acceleration is g_eff = sqrt(g² + (qE/m)²). The pendulum oscillates about the new equilibrium (tilted) line with period T = 2*pi*sqrt(L/g_eff) = 2*pi*sqrt(L / sqrt(g² + (qE/m)²)).

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