First slide

Electrostatic potential

Question

An oil drop having charge 2e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m³, the radius of the drop will be

Easy

A

2.0 x 10^-6 m
B

1.7 x 10^-6 m
C

1.4 x 10^-6m
D

1.1 x 10^-6m

Solution

In equilibrium, QE = mg
$\Rightarrow Q \cdot \frac{V}{d} = mg = (\frac{4}{3} {πr}^{3} ρ) g \Rightarrow 2 \times 1.6 \times 10^{- 19} \times \frac{12000}{2 \times 10^{- 2}} = \frac{4}{3} {πr}^{3} \times 900 \times 10 \Rightarrow r = 1.7 \times 10^{- 6} m$

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