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Electrostatics question

2022 · 24 Jun · Shift 1 · Q51
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  5. /2022 · 24 Jun · Shift 1 · Q51

Electrostatics question

2022 · 24 Jun · Shift 1 · Q51

JEE MainPhysicsElectrostaticsMCQ+4 / −1
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)
  1. A
    1.6 ×\times× 10 −-− 9 C
  2. B
    2.0 ×\times× 10 −-− 9 C
  3. C
    3.2 ×\times× 10 −-− 9 C
  4. D
    0.5 ×\times× 10 −-− 9 C
View written solutionFree

Correct answer: B

  1. Condition for the droplet to just remain suspended

Since the electric field just prevents the droplet from falling, the upward electric force balances the downward gravitational force.

So, qE=mgqE = mgqE=mg

Hence, q=mgEq = \frac{mg}{E}q=Emg​

  1. Given data
  • Mass of droplet: 0.1 g=0.1×10−3 kg=10−4 kg0.1\,\text{g} = 0.1 \times 10^{-3}\,\text{kg} = 10^{-4}\,\text{kg}0.1g=0.1×10−3kg=10−4kg
  • Electric field: E=4.9×105 N/CE = 4.9 \times 10^5\,\text{N/C}E=4.9×105N/C
  • Acceleration due to gravity: g=9.8 m/s2g = 9.8\,\text{m/s}^2g=9.8m/s2
  1. Substitute the values

q=(10−4)(9.8)4.9×105q = \frac{(10^{-4})(9.8)}{4.9 \times 10^5}q=4.9×105(10−4)(9.8)​

q=9.8×10−44.9×105q = \frac{9.8 \times 10^{-4}}{4.9 \times 10^5}q=4.9×1059.8×10−4​

q=2×10−9 Cq = 2 \times 10^{-9}\,\text{C}q=2×10−9C

  1. Match with the options

This corresponds to:

Option B: 2.0×10−9 C2.0 \times 10^{-9}\,\text{C}2.0×10−9C

  1. Verification with stored answer

Stored correct answer is B, which matches the derived answer.

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