Question

In Millikan's oil drop experiment, a charged drop of mass $1.8\times {10}^{-14}kg$ is stationary between the plates. The distance between the plates is 0.90cm and potential difference between them is 2.0 kV. The number of electrons on the drop is

• A. 500
• B. 50
• C. 5
• D. 0

Answer 1

The correct option is C 5

Given that, particle is at rest and so net force is zero.
From the free body diagram we can see that
$F_e=F_g$

$mg=q\frac{V}{d}=\left(ne\right)\frac{V}{d}$

where $q=ne$ (n is Number of electrons)

Substituting the given values of-

$m=1.4\times {10}^{-14}kg$
$V=2\times {10}^3$ volts
$d=9\times {10}^{-3}m$
e = $1.602\times {10}^{-19}$ Coulomb

We get , n = \dfrac{mg d}{e V}=$\frac{(1.8\times {10}^{-14})\left(9.8\right)(9\times {10}^{-3})}{(1.602\times {10}^{-19})(2\times {10}^3)}=5$

So, the answer is option (C).