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 $\frac{\text{electronic charge on the moon}}{\text{electronic charge on the earth}}$ to be

• A. $ 1 $
• B. $0$
• C. $ \frac{g_{E}}{g_{M}}$
• D. $ \frac{g_{M}}{g_{E}}$

Answer 1

Answer: (A)

According to Millikan's oil drop experiment, electronic charge is given by $q=\frac{6 \pi n r\left(v_{1}+v_{2}\right)}{E}$ which is independent of $g$.
So, $\frac{\text { electronic charge on the moon }}{\text { electronic charge on the earth }}=1$