In a Millikan's oil drop method, a charged drop of oil has mass m and charge q. It is made stationary by applying an electric field E. Find the strength of the electric field. If the result of Millikans experiment with different drops is $6.48 \times 10^{- 19} C, 12.82 \times 10^{- 19} C, 19.3 \times 10^{- 19} C, 8.02 \times 10^{- 19} C, 25.62 \times 10^{- 19} C, 16.04 \times 10^{- 19} C,$ find the elementary charge.

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Solution

If suspended charged oil drop has mass $m$ and charge $q$ , then its weight $m g$ is exactly equal to the electric force applied $q E$ , where $E$ is the electric field.
i.e., $m g = q E$
$⟹ E = \frac{m g}{q}$
Also, irrespective of the charge on different oil drops, the elementary charge is always a multiple of the charge of an electron ( $- 1.6 \times 10^{- 19}$ ).

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