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Electrostatics question

2020 · Shift 2 · Q38
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Electrostatics question

2020 · Shift 2 · Q38

JEE AdvancedPhysicsElectrostaticsNumerical+3 / −1
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 ‾\underline{\hspace{2cm}}​. (neglect the buoyancy force, take acceleration due to gravity = 10 ms−2 and charge on an electron (e) = 1.6 ×\times× 10–19 C) JEE Advanced 2020 Paper 2 Offline Physics - Electrostatics Question 41 English
Numerical answer
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Correct answer: 6

  1. Electric field between the discs

The discs are separated by d=0.01 md = 0.01\,\text{m}d=0.01m and the applied voltage is V=200 V.V = 200\,\text{V}.V=200V.

So the uniform electric field between the discs is E=Vd=2000.01=2×104 N/C.E = \frac{V}{d} = \frac{200}{0.01} = 2\times 10^4\,\text{N/C}.E=dV​=0.01200​=2×104N/C.

  1. Condition for the drop to float

When the oil drop stops moving vertically, the electric force balances its weight: qE=mg.qE = mg.qE=mg.

  1. Mass of the oil drop

Given:

  • density of oil: ρ=900 kg/m3\rho = 900\,\text{kg/m}^3ρ=900kg/m3
  • radius: r=8×10−7 mr = 8\times 10^{-7}\,\text{m}r=8×10−7m

Volume of the spherical drop: Vdrop=43πr3.V_{\text{drop}} = \frac{4}{3}\pi r^3.Vdrop​=34​πr3.

So mass is m=ρ⋅43πr3.m = \rho \cdot \frac{4}{3}\pi r^3.m=ρ⋅34​πr3.

Now, r3=(8×10−7)3=512×10−21=5.12×10−19 m3.r^3 = (8\times 10^{-7})^3 = 512\times 10^{-21} = 5.12\times 10^{-19}\,\text{m}^3.r3=(8×10−7)3=512×10−21=5.12×10−19m3.

Thus, m=900×43π×5.12×10−19.m = 900 \times \frac{4}{3}\pi \times 5.12\times 10^{-19}.m=900×34​π×5.12×10−19.

Since 900×43=1200,900\times \frac{4}{3} = 1200,900×34​=1200, we get m=1200π×5.12×10−19m = 1200\pi \times 5.12\times 10^{-19}m=1200π×5.12×10−19 m=6144π×10−19m = 6144\pi \times 10^{-19}m=6144π×10−19 m≈1.93×10−15 kg.m \approx 1.93\times 10^{-15}\,\text{kg}.m≈1.93×10−15kg.

  1. Weight of the drop

mg=1.93×10−15×10=1.93×10−14 N.mg = 1.93\times 10^{-15} \times 10 = 1.93\times 10^{-14}\,\text{N}.mg=1.93×10−15×10=1.93×10−14N.

  1. Charge on the drop

Using q=mgE,q = \frac{mg}{E},q=Emg​,

q=1.93×10−142×104q = \frac{1.93\times 10^{-14}}{2\times 10^4}q=2×1041.93×10−14​ q≈9.65×10−19 C.q \approx 9.65\times 10^{-19}\,\text{C}.q≈9.65×10−19C.

  1. Number of electrons

If the drop has charge equal to nnn electrons, q=ne,q = ne,q=ne, so n=qe=9.65×10−191.6×10−19≈6.03.n = \frac{q}{e} = \frac{9.65\times 10^{-19}}{1.6\times 10^{-19}} \approx 6.03.n=eq​=1.6×10−199.65×10−19​≈6.03.

Hence, n≈6.n \approx 6.n≈6.

Therefore, the number of electrons on the oil drop is 6.\boxed{6}.6​.

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