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Electrostatics question

2021 · 18 Mar · Shift 1 · Q47
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  5. /2021 · 18 Mar · Shift 1 · Q47

Electrostatics question

2021 · 18 Mar · Shift 1 · Q47

JEE MainPhysicsElectrostaticsMCQ+4 / −1
An oil drop of radius 2 mm with a density 3g cm −-− 3 is held stationary under a constant electric field 3.55 ×\times× 105 V m −-− 1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider g = 9.81 m/s2)
  1. A
    48.8 ×\times× 1011
  2. B
    1.73 ×\times× 1010
  3. C
    17.3 ×\times× 1010
  4. D
    1.73 ×\times× 1012
View written solutionFree

Correct answer: B

  1. Condition for suspension

Since the oil drop is held stationary, the electric force balances its weight:

qE=mgqE = mgqE=mg

So,

q=mgEq = \frac{mg}{E}q=Emg​

  1. Mass of the oil drop

Given:

  • Radius: r=2 mm=2×10−3 mr = 2\,\text{mm} = 2 \times 10^{-3}\,\text{m}r=2mm=2×10−3m
  • Density: ρ=3 g cm−3=3×103 kg m−3\rho = 3\,\text{g cm}^{-3} = 3 \times 10^3\,\text{kg m}^{-3}ρ=3g cm−3=3×103kg m−3

Volume of the drop:

V=43πr3=43π(2×10−3)3V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2 \times 10^{-3})^3V=34​πr3=34​π(2×10−3)3

(2×10−3)3=8×10−9(2 \times 10^{-3})^3 = 8 \times 10^{-9}(2×10−3)3=8×10−9

Hence,

V=43π×8×10−9=32π3×10−9 m3V = \frac{4}{3}\pi \times 8 \times 10^{-9} = \frac{32\pi}{3} \times 10^{-9}\,\text{m}^3V=34​π×8×10−9=332π​×10−9m3

Mass:

m=ρV=3×103×32π3×10−9m = \rho V = 3 \times 10^3 \times \frac{32\pi}{3} \times 10^{-9}m=ρV=3×103×332π​×10−9

m=32π×10−6 kgm = 32\pi \times 10^{-6}\,\text{kg}m=32π×10−6kg

Using π≈3.1416\pi \approx 3.1416π≈3.1416,

m≈32×3.1416×10−6=100.53×10−6=1.0053×10−4 kgm \approx 32 \times 3.1416 \times 10^{-6} = 100.53 \times 10^{-6} = 1.0053 \times 10^{-4}\,\text{kg}m≈32×3.1416×10−6=100.53×10−6=1.0053×10−4kg

  1. Weight of the drop

mg=1.0053×10−4×9.81mg = 1.0053 \times 10^{-4} \times 9.81mg=1.0053×10−4×9.81

mg≈9.86×10−4 Nmg \approx 9.86 \times 10^{-4}\,\text{N}mg≈9.86×10−4N

  1. Charge on the drop

Given electric field:

E=3.55×105 V m−1E = 3.55 \times 10^5\,\text{V m}^{-1}E=3.55×105V m−1

Therefore,

q=9.86×10−43.55×105q = \frac{9.86 \times 10^{-4}}{3.55 \times 10^5}q=3.55×1059.86×10−4​

q≈2.78×10−9 Cq \approx 2.78 \times 10^{-9}\,\text{C}q≈2.78×10−9C

  1. Number of excess electrons

If the drop has nnn excess electrons, then

q=neq = neq=ne

where e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C.

So,

n=qe=2.78×10−91.6×10−19n = \frac{q}{e} = \frac{2.78 \times 10^{-9}}{1.6 \times 10^{-19}}n=eq​=1.6×10−192.78×10−9​

n≈1.74×1010n \approx 1.74 \times 10^{10}n≈1.74×1010

  1. Match with options

This corresponds to:

1.73×1010\boxed{1.73 \times 10^{10}}1.73×1010​

So the correct option is B.

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