Question

A drop of oil of radius ${10}^{-6}$ m carries a charge of four times that of electron. If the density of oil is $2000$$kg/m^3$, the potential difference which must be applied across the plates in Millikan’s experiment in order that the drop may float, when the distance between the plates is $5mm$ apart, is

• A. $620V$
• B. $641V$
• C. $541V$
• D. $341V$

Answer 1

The correct option is B $641V$

Charge on drop $=4\times 1.6\times {10}^{-19}$C

$=6.4\times {10}^{-19}$C

Radius of drop $={10}^{-6}m$

Mass of the drop $=$ volume $\times$ density

$=\frac{4}{3}\pi {10}^{-18}\times 2000$

$=8.3775\times {10}^{-15}kgs$

For equilibrium, weight $=$ electric force

$\Rightarrow 8.3775\times {10}^{-15}\times g=\frac{v}{5\times {10}^{-3}}\times 6.4\times {10}^{19}$

$\Rightarrow 5\times \frac{8.3775\times 10\times g}{6.4}=v$

$\Rightarrow v=641.04volts$