GATE 2016 EC (EC2) – Question 65
A small positive charge q starts at rest at x=0 between infinite yz metal plates at x=−d and x=+d. Voltages +V and −V are applied respectively at t=0. The charge does not perturb the field. Its time to reach the right plate is proportional to
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Correct answer: (C) $d/\sqrt V$
Explanation
The voltage difference is 2V across distance 2d, so the uniform field magnitude is V/d, directed toward the right plate. A particle of mass m therefore has constant acceleration $a=qV/(md)$.
Starting from rest and moving distance d, it obeys $d=at^2/2$. Thus $t=d\sqrt{2m/(qV)}$.
For fixed m and q, the time is proportional to **$d/\sqrt V$, C**.