Question

In a Millikans oil drop experiment, an oil drop of mass $0.64\times {10}^{-14}kg$, carrying a charge $1.6\times {10}^{-19}C$ remains stationary between two plates separated by a distance of 5 mm. Given $g=9.8m/s^2$; the voltage that must be applied between the plates being

• A. 980 V
• B. 1960 V
• C. 3920V
• D. 2880V

Answer 1

Answer: (B)

1960 V

Let the voltage applied by $V_0$,
According to equilibrium eqn.,

$\Rightarrow 0.64\times {10}^{-14}\times 9.8=1.6\times {10}^{-16}\times \frac{V_0}{5}$
$\Rightarrow \frac{0.64\times 5\times 9.8\times {10}^2}{1.6}=V_0$
$\Rightarrow V_0=1960volts$

So, the answer is option (B).