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 \(\left.g =9.8 m / s ^{2}\right)\)
- A \(1.6 \times 10^{-9} C\)
- B \(2.0 \times 10^{-9} C\)
- C \(3.2 \times 10^{-9} C\)
- D \(0.5 \times 10^{-9} C\)
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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