Q. If $g_{E}$ and $g_{M}$ are the acceleration due to gravity on the surfaces of the earth and the moon respectively and it Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio $\frac{electronic\ charge\ on\ the\ moon\ }{electronic\ charge\ on\ the\ earth\ }$ to be
A) 1
B) Zero
C) $g_{E}{/g}_{M}$
D) $g_{M}{/g}_{E}$
Explanation
(a) According to Millikan’s oil drop experiment, electronic charge is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = \frac{6\pi\eta r\left( v_{1} + v_{2} \right)}{E}$
Which is independent of g.
So, $\frac{electronic\ charge\ on\ the\ moon\ }{electronic\ charge\ on\ the\ earth} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.