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JEE Main 2019 (8 Apr, Shift 1) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2019 (8 Apr, Shift 1)

Q36JEE Main 2019 · 8 APR · SHIFT 1Easy

Motion of Charges in an Electric Field

The bob of a simple pendulum has mass 2g and a charge of 5.0 μC. 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 m/s2)

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Answer: (c)

  1. At equilibrium tan⁡θ=qEmg\tan\theta = \frac{qE}{mg}.
  2. qE=5×10−6×2000=10−2 NqE = 5 \times 10^{-6} \times 2000 = 10^{-2}\text{ N}; mg=2×10−3×10=2×10−2 Nmg = 2 \times 10^{-3} \times 10 = 2 \times 10^{-2}\text{ N}
  3. tan⁡θ=0.5\tan\theta = 0.5
  4. So the answer is (c) tan⁡−1(0.5)\tan^{-1}(0.5).
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