Question

An oil drop of radius $2\text{mm}$ with a density $3{\text{g cm}}^{-3}$ is held stationary under a constant electric field $3.55\times {10}^5{\text{V m}}^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Consider $g=9.81{\text{m s}}^{-2}$)

• A. $1.73\times {10}^{10}$
• B. $48.8\times {10}^{11}$
• C. $1.73\times {10}^{12}$
• D. $17.3\times {10}^{10}$

Answer 1

The correct option is A $1.73\times {10}^{10}$

Suppose the number of excess electrons is $n$.

Force due to electric field, $F_e=qE=\left(ne\right)E$
Weight of the oil drop, $W=mg$

Since drop is stationary, $F_e=W$ $\Rightarrow neE=mg$
$\Rightarrow n=\frac{mg}{eE}=\frac{\rho \times \frac{4}{3}\pi R^3\times g}{eE}$

$\Rightarrow n=\frac{(3\times 1000)\times \frac{4}{3}\times 3.14\times (2\times {10}^{-3}{)}^3\times 9.8}{1.6\times {10}^{-19}\times 3.55\times {10}^5}$

$\Rightarrow n=\frac{984704\times {10}^5}{5.68}=1.73\times {10}^{10}$

$\Rightarrow n=1.73\times {10}^{10}$