JEE Challenger
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PhyHardAdv 2010 • Paper 2

Magnitude of Charge on Spherical Oil Drop in Electric Field

A tiny spherical oil drop carrying a net charge qq is balanced in still air with a vertical uniform electric field of strength 81π7×105 Vm−1\frac{81\pi}{7} \times 10^5\text{ Vm}^{-1}. When the field is switched off, the drop is observed to fall with terminal velocity 2×10−3 m s−12 \times 10^{-3}\text{ m s}^{-1}.

Given g=9.8 m s−2g = 9.8\text{ m s}^{-2}, viscosity of the air =1.8×10−5 Ns m−2= 1.8 \times 10^{-5}\text{ Ns m}^{-2} and the density of oil =900 kg m−3= 900\text{ kg m}^{-3}, the magnitude of qq is

Options

A

1.6×10−19 C1.6 \times 10^{-19}\text{ C}

B

3.2×10−19 C3.2 \times 10^{-19}\text{ C}

C

4.8×10−19 C4.8 \times 10^{-19}\text{ C}

D

8.0×10−19 C8.0 \times 10^{-19}\text{ C}

Correct

Topics & Concepts

Electric Charges and FieldsElectric FieldFluid Dynamics

Step-by-Step Solution

To find the magnitude of the net charge qq on the spherical oil drop, we analyze two states: when the electric field is applied, and when it is switched off.

1. Case 1: Electric Field is Switched OFF

When the electric field is switched off, the oil drop falls under gravity and reaches a terminal velocity vtv_t. At terminal velocity, the downward gravitational force is balanced by the upward viscous drag force given by Stokes' Law:

6πηrvt=mg6 \pi \eta r v_t = m g

where:

  • rr is the radius of the spherical oil drop,
  • η=1.8×10−5 N s m−2\eta = 1.8 \times 10^{-5} \text{ N s m}^{-2} is the coefficient of viscosity of air,
  • vt=2×10−3 m s−1v_t = 2 \times 10^{-3} \text{ m s}^{-1} is the terminal velocity,
  • m=43πr3ρm = \frac{4}{3} \pi r^3 \rho is the mass of the oil drop,
  • ρ=900 kg m−3\rho = 900 \text{ kg m}^{-3} is the density of the oil,
  • g=9.8 m s−2g = 9.8 \text{ m s}^{-2} is the acceleration due to gravity.

Equating the viscous force to the weight of the drop:
6πηrvt=43πr3ρg6 \pi \eta r v_t = \frac{4}{3} \pi r^3 \rho g

Solving for the radius squared (r2r^2):
r2=9ηvt2ρgr^2 = \frac{9 \eta v_t}{2 \rho g}

Substitute the given values into the expression for r2r^2:
r2=9×(1.8×10−5)×(2×10−3)2×900×9.8=32.4×10−817640=949×10−10 m2r^2 = \frac{9 \times (1.8 \times 10^{-5}) \times (2 \times 10^{-3})}{2 \times 900 \times 9.8} = \frac{32.4 \times 10^{-8}}{17640} = \frac{9}{49} \times 10^{-10} \text{ m}^2

Taking the square root gives the radius of the oil drop:
r=37×10−5 mr = \frac{3}{7} \times 10^{-5} \text{ m}

2. Case 2: Electric Field is Switched ON

When the vertical uniform electric field EE is switched on, the oil drop is balanced in still air. The upward electrostatic force balances the downward gravitational force:

∣q∣E=mg|q| E = m g

Using mg=6πηrvtm g = 6 \pi \eta r v_t, we substitute r=37×10−5 mr = \frac{3}{7} \times 10^{-5} \text{ m}:
mg=6π×(1.8×10−5)×(37×10−5)×(2×10−3)=64.8π7×10−13 Nm g = 6 \pi \times (1.8 \times 10^{-5}) \times \left(\frac{3}{7} \times 10^{-5}\right) \times (2 \times 10^{-3}) = \frac{64.8 \pi}{7} \times 10^{-13} \text{ N}

Given that the electric field strength is E=81π7×105 V m−1E = \frac{81 \pi}{7} \times 10^5 \text{ V m}^{-1}, we can now calculate the charge qq:

∣q∣=mgE=64.8π7×10−1381π7×105=64.881×10−18 C|q| = \frac{m g}{E} = \frac{\frac{64.8 \pi}{7} \times 10^{-13}}{\frac{81 \pi}{7} \times 10^5} = \frac{64.8}{81} \times 10^{-18} \text{ C}

64.881=0.8\frac{64.8}{81} = 0.8

∣q∣=0.8×10−18 C=8.0×10−19 C|q| = 0.8 \times 10^{-18} \text{ C} = 8.0 \times 10^{-19} \text{ C}

Conclusion

The magnitude of the charge qq is 8.0×10−19 C8.0 \times 10^{-19}\text{ C}.

Correct Option: D

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