Electrostatics

JEE Main level, single correct, medium

56 of 311

Question 56 of 311

Two metal plates (A,B)(A, B) are kept horizontally with separation of (12π)\left(\dfrac{12}{\pi}\right) cm, with plate A on the top. An atomizer jet sprays oil (density 1.51.5 g/cm3^3) droplets of radius 11 mm horizontally. All oil droplets carry a charge 55 nC. The potentials VAV_A and VBV_B are required on plates A and B respectively in order to ensure the droplets do not descend. The values of VAV_A and VBV_B are _______.
(Neglect the air resistance to the droplets and take g=10g = 10 m/s2^2)

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Answer: (A) 100 V100 \, V and 580 V580 \, V

For the oil droplets to not descend, the upward electric force must balance the downward gravitational force.

qE=mgqE = mg

The mass of the oil droplet is given by:

m=43πr3ρm = \dfrac{4}{3} \pi r^3 \rho

Substituting the given values r=10−3r = 10^{-3} m and ρ=1500\rho = 1500 kg/m3^3:

m=43π(10−3)3×1500=2π×10−6m = \dfrac{4}{3} \pi (10^{-3})^3 \times 1500 = 2 \pi \times 10^{-6} kg

The electric field EE required is:

E=mgq=2π×10−6×105×10−9=4000πE = \dfrac{mg}{q} = \dfrac{2 \pi \times 10^{-6} \times 10}{5 \times 10^{-9}} = 4000 \pi V/m

The potential difference ΔV\Delta V between the plates is:

ΔV=E×d=4000π×(12π×10−2)=480\Delta V = E \times d = 4000 \pi \times \left(\dfrac{12}{\pi} \times 10^{-2}\right) = 480 V

Since the charge on the droplet is positive, the electric field must be directed upwards to provide an upward force. Electric field lines point from higher potential to lower potential, so the bottom plate B must be at a higher potential than the top plate A.

VB−VA=480V_B - V_A = 480 V

Checking the given options, VA=100V_A = 100 V and VB=580V_B = 580 V satisfies this condition.

Answer: 100 V100 \, V and 580 V580 \, V

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