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Physics · Electrostatics

JEE Advanced 2020 — Paper 2 — Question 2

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} \mathrm{~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 8×10−7 m8 \times 10^{-7} \mathrm{~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=10 \mathrm{~ms}^{-2} and charge on an electron (e) =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C} )

Question figure

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

Electric field between the discs after closing the switch

E=Vd=2000.01=2×104 N/CE=\frac{V}{d}=\frac{200}{0.01}=2 \times 10^{4} \mathrm{~N} / \mathrm{C}

When an oil drop stops and floats between the discs

qE=mg\mathrm{qE}=\mathrm{mg}

neE =mg=\mathrm{mg}

n=mgeE=ρ43πr3 geE\mathrm{n}=\frac{\mathrm{mg}}{\mathrm{eE}}=\frac{\rho \frac{4}{3} \pi \mathrm{r}^{3} \mathrm{~g}}{\mathrm{eE}}

n=900×4×3.14×(8×10−7)3×103×1.6×10−19×2×104\mathrm{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=3×4×3.14×2×8100=6\mathrm{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\mathrm{n}=6

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Field & motion of charge

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