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

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

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

Solution:

In balance condition

$\Rightarrow QE = mg \Rightarrow Q \frac{V}{d} = (\frac{4}{3} {πr}^{3} ρ) g$

$\Rightarrow Q \propto \frac{r^{3}}{V} \Rightarrow \frac{Q_{1}}{Q_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} \times \frac{V_{2}}{V_{1}}$

$\Rightarrow \frac{Q}{Q_{2}} = {(\frac{r}{r / 2})}^{3} \times \frac{600}{2400} = 2 \Rightarrow Q_{2} = Q / 2$

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