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.
