Physics Circle home Student Login

Home›Chapters›Electrostatics PYQs›Motion of Charges in an Electric Field›Question 3

JEE Main 2026 (2 Apr, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2026 (2 Apr, Shift 2)

Q3JEE Main 2026 · 2 APR · SHIFT 2Difficult

Motion of Charges in an Electric Field

Two metal plates (A, B) are kept horizontally with separation of (12π)\left(\frac{12}{\pi}\right) cm, with plate A on the top. An atomizer jet sprays oil (density 1.5 g/cm3) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials VA and VB are required on plates A and B respectively in order to ensure the droplets do not descend. The values of VA and VB are ______.

(Neglect the air resistance to the droplets and take g = 10 m/s2)

Show answer and solution

Answer: (a)

  1. Drop mass: m=1500×43π(10−3)3=2π×10−6 kgm = 1500 \times \frac{4}{3}\pi(10^{-3})^3 = 2\pi \times 10^{-6}\text{ kg}, so mg=2π×10−5 Nmg = 2\pi \times 10^{-5}\text{ N}.
  2. Balance: qE=mg⇒E=2π×10−55×10−9=4000π V/mqE = mg \Rightarrow E = \frac{2\pi \times 10^{-5}}{5 \times 10^{-9}} = 4000\pi\text{ V/m}, directed upward.
  3. VB−VA=Ed=4000π×0.12π=480 VV_B - V_A = Ed = 4000\pi \times \frac{0.12}{\pi} = 480\text{ V} (lower plate B at higher potential).
  4. Only option (a) gives VB−VA=580−100=480 VV_B - V_A = 580 - 100 = 480\text{ V}.
  5. So the answer is (a) 100 V and 580 V.
Report a mistake

More questions on Motion of Charges in an Electric Field