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Electrostatics Part 2

Section: Physics

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Question : 94 of 100

Marks: +1, -0

An oil drop of radius

2\,\mathrm{mm}

3\,\mathrm{g}\,\mathrm{cm}^{-3}

is held stationary under a constant electric field

3.55 \times 10^{5}\,\mathrm{Vm}^{-1}

in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Take,

g=9.81\,\mathrm{m}/\mathrm{s}^{2}

[18 Mar 2021 Shift 1]

$48.8 \times 10^{11}$
$1.73 \times 10^{10}$
$17.3 \times 10^{10}$
$1.73 \times 10^{12}$

Solution:

Given, the radius of oil drop,

r=2\,\mathrm{mm}=0.002\,\mathrm{m}

The density of oil drop,

\rho=3\,\mathrm{g}/\mathrm{cm}^{3}=3\times 10^{3}\,\mathrm{kg}/\mathrm{m}^{3}

The constant electric field,

E=3.55\times 10^{5}\,\mathrm{Vm}^{-1}

The constant electric field,

E=3.55\times 10^{5}\,\mathrm{Vm}^{-1}

Under stationary condition of oil drop,

m g = q E

\Rightarrow\;\; \rho V g = q E

[ \because m = \rho V ]

\Rightarrow\;\; \rho\left(\;\frac{4}{3}\pi r^{3}\right) g = n e E

[ \because V = \frac{4}{3}\pi r^{3}

and

q = n e ]

n = \frac{\rho\left(\frac{4}{3}\pi r^{3}\right) g}{e E}

\Rightarrow\;\; n = \frac{3 \times 10^{3} \times \left(\frac{4}{3}\pi (0.002)^{3}\right) 9.81}{1.6 \times 10^{-19} \times (3.55 \times 10^{5})}

\Rightarrow\;\; n = 1.73 \times 10^{10}

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