An oil droplet with $15$excess electrons is observed between two parallel electrodes spaced $12mm$ apart. The droplet hangs motionless if the upper electrode is$25V$ more positive than the lower electrode. The density of the oil is $860kg/m^3$. What is the radius of the droplet?

We need to find out the radius of oil droplet placed between two parallel electrodes .

Number of excess electrons; n=$15$

Separation between electrode; l$=12mm=12\times {10}^{-3}m$

The droplet hangs motionless if the upper electrode potential; V=

The density of the oil ;d $=860Kg/m^3$

The electric force is balanced by the mass of the object. So,

$$\begin{gathered} F=mg=qE \\ q=\frac{mgl}{V}.....\left(1\right);\{E=\frac{V}{l}\} \end{gathered}$$

Also , charges come in integral multiples of electronic units ;

q = ne $.....\left(2\right)$

Now, equate (1) & (2)

$\Rightarrow$$ne=\frac{mgl}{V}.....\left(3\right)$ ,density ;d= mass/ volume

{ d = m/ vol.}

$\Rightarrow$ m= d $\times$vol. = d$\times \frac{4}{3}\pi r^3......\left(4\right)$

Putting (4) in (3) ,

$$\begin{gathered} \frac{d\times \frac{4}{3}\times \pi r^3\times gl}{V}=\frac{860\times \frac{4}{3}\times \pi r^3\times 9.8\times 12\times {10}^{-3}}{25} \\ \Rightarrow r=\sqrt[3]{\frac{15\times 1.6\times {10}^{-19}\times 25\times 3}{4\times 860\times 3.14\times 9.8\times 12\times {10}^{-3}}} \end{gathered}$$

Thus, the radius of oil droplet is$0.1123m$