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JEE Mains · Physics · STD 12 - 1. Electric charges and fields

A vertical electric field of magnitude \(4.9 \times 10^{5} N / C\) just prevents a water droplet of a mass \(0.1\, g\) from falling. The value of charge on the droplet will be  ........ \(\times 10^{-9} \;C\) \(\left(\right.\) Given \(\left.g =9.8 m / s ^{2}\right)\)

  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 \(0.5 \times 10^{-9} C\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2.0 \times 10^{-9} C\)

Step-by-step Solution

Detailed explanation

\(Mg = qE\) \(\left(0.1 \times 10^{-3}\right)(9.8)=4.9 \times 10^{5} q\) \(\frac{2 \times 10^{-4}}{10^{5}}= q\) \(q=2 \times 10^{-9} C\)
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