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← JEE Advanced 2020 · Paper 2
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JEE Advanced 2020 · Paper 2, Physics question 3

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−3900\ \mathrm{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 8×10−78\times10^{-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 m s−2= 10\ \mathrm{m\,s^{-2}} and charge on an electron (e)=1.6×10−19(e) = 1.6\times10^{-19} C)

Two horizontal circular discs 0.01 m apart, connected through a switch to a 200 V battery. Oil drops fall through a small hole in the centre of the top disc into the gap.
Answer and worked solution

Official answer: 6

The drop floats when the electric force balances its weight (buoyancy neglected): qE=mgqE=mg.

Mass.
m=900×43π(8×10−7)3=1.93×10−15 kg.m=900\times\tfrac43\pi(8\times10^{-7})^3=1.93\times10^{-15}\text{ kg}.

Field. E=200/0.01=2×104 V m−1E=200/0.01=2\times10^4\ \mathrm{V\,m^{-1}}.

Charge.
q=mgE=1.93×10−142×104=9.65×10−19 C=9.65×10−191.6×10−19e≈6e.q=\frac{mg}E=\frac{1.93\times10^{-14}}{2\times10^4}=9.65\times10^{-19}\text{ C}=\frac{9.65\times10^{-19}}{1.6\times10^{-19}}e\approx6e.
Answer: 6

Topics:Electric field
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