Electrostatics Question 482
Question: Two parallel plates separated by a distance of $ 5mm $ are kept at a potential difference of $ 50V. $ A particle of mass $ {{10}^{-15}}kg $ and charge $ {{10}^{-11}}C $ enters in it with a velocity $ 10^{7}m/s. $ The acceleration of the particle will be [MP PMT 1997]
Options:
A) $ 10^{8}m/s^{2} $
B) $ 5\times 10^{5}m/s^{2} $
C) $ 10^{5}m/s^{2} $
D) $ 2\times 10^{3}m/s^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ a=\frac{qE}{m}=\frac{q}{m}( \frac{V}{d} ) $
$ =\frac{{{10}^{-11}}}{{{10}^{-15}}}\times \frac{50}{5\times {{10}^{-3}}}=10^{8}m/sec^{2} $
