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JEE Main 2022 (24 Jun, Shift 1) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2022 (24 Jun, Shift 1)

Q19JEE Main 2022 · 24 JUN · SHIFT 1Easy

Motion of Charges in an Electric Field

A vertical electric field of magnitude 4.9 ×\times 105 N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be:

(Given: g = 9.8 m/s2)

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Answer: (b)

  1. Balance: qE=mg⇒q=mgEqE = mg \Rightarrow q = \frac{mg}{E}, with m=0.1m = 0.1 g =10−4= 10^{-4} kg.
  2. q=10−4×9.84.9×105=2.0×10−9q = \frac{10^{-4} \times 9.8}{4.9 \times 10^5} = 2.0 \times 10^{-9} C
  3. So the answer is (b) 2.0×10−92.0 \times 10^{-9} C.
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