Question

If $g_{E}$ and $g_{M}$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if 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. $\frac{g_{M}}{g_{E}}$
• B. 1
• C. 0
• D. $\frac{g_{E}}{g_{M}}$

Answer 1

Answer: (B)

Since electronic charge $\left(e=1.6 \times 10^{-19} C \right)$ is a universal constant. It does not depend on the value of acceleration due to gravity.
$\therefore $ Electronic charge on the moon = electronic charge on the earth
or $\frac{\text { electronic charge on themoon }}{\text { electronic charge on the earth }}=1$