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MHT CET 2024 (11 May, Shift 1) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from MHT CET 2024 (11 May, Shift 1)

Q9MHT CET 2024 · 11 MAY · SHIFT 1Medium

Motion of Charges in an Electric Field

A small particle carrying a negative charge of 1.6×10−19 C1.6 \times 10^{-19}\text{ C} is suspended in equilibrium between two horizontal metal plates 8 cm apart having a potential difference of 980 V across them. The mass of the particle is [g=9.8 m/s2]\left[g = 9.8\text{ m/s}^2\right]

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Answer: (a)

  1. Balance: qE=mgqE = mg with E=VdE = \frac{V}{d}, so m=qVgdm = \frac{qV}{gd}.
  2. m=1.6×10−19×9809.8×0.08=1.568×10−160.784m = \frac{1.6 \times 10^{-19} \times 980}{9.8 \times 0.08} = \frac{1.568 \times 10^{-16}}{0.784}
  3. m=2×10−16 kgm = 2 \times 10^{-16}\text{ kg}
  4. So the answer is (a) 2×10−16 kg2 \times 10^{-16}\text{ kg}.
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