Physics · Electrostatics

JEE Main 2026 — 2 April, Evening Shift — Question 12

Two metal plates (A,B) are kept horizontally with separation of (12π)cm\left(\frac{12}{\pi}\right)\mathrm{cm}, with plate A on the top. An atomizer jet sprays oil (density 1.5g/cm31.5\mathrm{g/cm}^3) droplets of radius 1mm1\mathrm{mm} horizontally. All oil droplets carry a charge 5nC5\mathrm{nC}. The potentials VAV_A and VBV_B required on plates A and B respectively in order to ensure the droplets do not descend are (Neglect air resistance, g=10m/s2\mathrm{g}=10\mathrm{m/s}^2)

  1. Option A:

    100V and 580V

    Correct
  2. Option B:

    580V and 100V

  3. Option C:

    60V and 400V

  4. Option D:

    0V and -200V

Answer: A

Step-by-step solution

Electric force = weight: qΔVd=43πr3ρgq\frac{\Delta V}{d} = \frac43\pi r^3 \rho g solving gives ΔV=480V\Delta V = 480\mathrm{V}, with plate A lower potential thus VA=100,VB=580V_A=100, V_B=580

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Electric Charge and Coulomb's Law
Two metal plates (A,B) are kept horizontally with separation of (12/π… | JEE Main 2026 PYQ with Solution · DhiX AI