Question

Two large circular discs separated by a distance of $0.01\, m$ are connected to a battery via a switch as shown in the figure. Charged oil drops of density $900\, kg \,m ^{-3}$ are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of $200\, V$ across the discs. As a result, an oil drop of radius $8 \times 10^{-7} m$ stops moving vertically and floats between the discs. The number of electrons present in this oil drop is (neglect the buoyancy force, take acceleration due to gravity $=10 \,ms ^{-2}$ and charge on an electron $(e) =1.6 \times 10^{-19} C$ )

Answer 1

Electric field between the discs after closing the switch
$E=\frac{V}{d}=\frac{200}{0.01}=2 \times 10^{4} N / C$
When an oil drop stops and floats between the discs
$qE = mg$
$neE = mg$
$n=\frac{m g}{e E}=\frac{\rho \frac{4}{3} \pi r^{3} g}{e E}$
$n=900 \times \frac{4 \times 3.14 \times\left(8 \times 10^{-7}\right)^{3} \times 10}{3 \times 1.6 \times 10^{-19} \times 2 \times 10^{4}}$
$n=\frac{3 \times 4 \times 3.14 \times 2 \times 8}{100}=6$
The number of electrons present in this oil drop is $n =6$