Question

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

• 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

$1.73\times {10}^{10}$

The explanation for the correct option:

Step 1: Given data with a diagram:

The radius of oil drop, $R=2mm$

The density of oil drops, $\rho =3gc{m}^{-3}=3000kg{m}^{-3}$

The intensity of the electric field, $E=3.55\times {10}^{5}V{m}^{-1}$

Step 2: Finding the number of excess electrons:

Using, $F=qE=\left(ne\right)E$……………..$\left(1\right)$

Also from the diagram we get,

$F=mg$……………..$\left(2\right)$

Comparing equation $\left(1\right)$ & $\left(2\right)$,

$\begin{array}{rcl}\left(ne\right)E& =& mg\\ & \Rightarrow & n=\frac{mg}{eE}\\ & \Rightarrow & n=\frac{\rho x{\displaystyle \frac{4}{3}}\pi R^{3}xg}{eE}\\ & \Rightarrow & n=\frac{3000x{\displaystyle \frac{4}{3}}x3.142x{\left(2x{10}^{-3}\right)}^{3}x9.81}{1.6x{10}^{-19}x3.55x{10}^5}\\ & \Rightarrow & n=\frac{984704x{10}^5}{5.68}\\ & \Rightarrow & n=1.73x{10}^{10}units\end{array}$

Hence, Option A is correct.