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

qE = mg ….(i)

q = ne

V = Ed ⇒ E = $\frac{V}{d}$

from equation (i)

ne (v/d) = mg

n = mgd/eV

= $900 \times \frac{4\pi}{3} \times \frac{8 \times 8 \times 8 \times 10^{-21} \times 10 \times 0.01}{1.6 \times 10^{-19} \times 200}$

n = 6 (approx)