Electrostatics Question 382

Question: Two metal pieces having a potential difference of $ 800V $ are $ 0.02m $ apart horizontally. A particle of mass $ 1.96\times {{10}^{-15}}kg $ is suspended in equilibrium between the plates. If $ e $ is the elementary charge, then charge on the particle is [MP PET 1999]

Options:

A) $ e $

B) $ 3e $

C) $ 6e $

D) $ 8e $

Show Answer

Answer:

Correct Answer: B

Solution:

For equilibrium mg = qE $ 1.96\times {{10}^{-15}}\times 9.8=q\times ( \frac{800}{0.02} ) $

therefore $ q=\frac{1.96\times {{10}^{-15}}\times 9.8\times 0.02}{800} $

therefore $ n\times 1.6\times {{10}^{-19}}=\frac{1.96\times {{10}^{-15}}\times 9.8\times 0.02}{800} $

therefore n = 3.



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