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  4. In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 6.35 × 10-19 C. The number of excess electrons on the drop is

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Q. In a Millikan's oil drop experiment the charge on an oil drop is calculated to be $6.35 × 10^{-19}\, C$. The number of excess electrons on the drop is

VITEEEVITEEE 2011

Solution:

As we know, $Q = ne$
Number of electron $n=\frac{Q}{e}$
$\frac{6.35\times10^{-19}}{1.6\times10^{-19}}$
$=3.9=4$