If $g_{E}$ and $g_{M}$ are the accelerations due to gravity on the surface 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
$\frac{e l e c t r o n i c c h a r g e o n t h e m o o n}{e l e c t r o n i c c h a r g e o n t h e e a r t h}$ to be

g_{M} / g_{E}

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g_{E} / g_{M}

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Solution

The correct option is B $1$
It is obvious that by charge conservation law, electronic charge must be independent of the acceleration due to gravity. Hence it will remain constant irrespective of where the experiment is performed.

Hence $\frac{e l e c t r o n i c c h a r g e o n m o o n}{e l e c t r o n i c c h a r g e o n e a r t h} = 1$

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Similar questions

g_{E}

g_{M}

\frac{electronic charge on the moon}{electronic charge on the earth}

_{E}

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(g_{m} = \frac{g_{e}}{6})

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ρ_{e}

ρ_{m}

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