Question

In Millikan's oil drop experiment an oil drop carrying a charge $Q$ is held stationary by a potential difference $2400V$ between the plates. To keep a drop of half the radius stationary the potential difference had to be made $600V$. What is the charge on the second drop:

• A. $\frac{Q}{4}$
• B. $\frac{Q}{2}$
• C. $Q$
• D. $\frac{3Q}{2}$

Answer 1

Answer: (B)

$\frac{Q}{2}$

Let the density of the oil drop $=d$

Volume of oil drop $=4/3\pi R^3$

Distance between the plates $=l$

Potential across plates $=4$

Electrostatic force on the oil drop $=F-e=QE=Q\times V/L$

Gravitational force on the oil drop

$\Rightarrow Q=4/3\pi R^{3}dgL/V$

$Q\alpha R^3/V$

$Q_2/Q_1=\frac{R_2^3.V_1}{R_1^3{V}_2}$

$Q_2=\frac{Q\times {\left(1/2\right)}^3}{2400/600}$

$Q_2=\frac{Q}{2}$

Hence, this is the answer.