In Millikan’s oil drop experiment, an oil drop of mass $16 imes 10^{- 6}$ kg is balanced by an electric field of $10^{6}\ {Vm}^{- 1}$. The charge in coulomb on the drop is $(assuming\ g = 10\ {ms}^{- 2})$

Question

In Millikan’s oil drop experiment, an oil drop of mass $16 	imes 10^{- 6}$ kg is balanced by an electric field of $10^{6}\ {Vm}^{- 1}$. The charge in coulomb on the drop is $(assuming\ g = 10\ {ms}^{- 2})$

Dr. Abubacker · Accepted Answer

(c) Robert Millikan performed the experiment to determine the charge on an electron.
When a drop is suspended, its weight mg is exactly equal to the electric force applied $qE$, where E is electric field, q the charge, m the mass of drop and g the acceleration due to gravity.
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Hence, solving for q, we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = \frac{mg}{E}$
Given, $m = 16 	imes 10^{- 6}kg,\ \ \ \ \ \ \ \ \ \ g = 10{ms}^{- 2}$,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = 10^{6}$V–$m^{- 1}$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ q = \frac{16 	imes 10^{6} 	imes 10}{10^{6}} = 16 	imes 10^{- 11}$ C

Q. In Millikan’s oil drop experiment, an oil drop of mass $16 \times 10^{- 6}$ kg is balanced by an electric field of $10^{6}\ {Vm}^{- 1}$. The charge in coulomb on the drop is $(assuming\ g = 10\ {ms}^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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