First slide

Electrostatic potential

Question

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

Easy

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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