Two metal plates (A, B) are kept horizontally with separation of (π12)cm, with plate A on the top. An atomizer jet sprays oil (density 1.5g/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=10m/s2)
(A)100 V and 580 V✓ Correct
(B)580 V and 100 V
(C)60 V and 400 V
(D)0 V and -200 V
Answer: (A)
Solution
A droplet stays suspended when the upward electric force balances its weight, qE=mg, with m=34πr3ρ and E=dΔV. Hence ΔV=3q4πr3ρgd. Substituting r=10−3m, ρ=1500kg/m3, d=π12×10−2m and q=5×10−9C gives ΔV=3×5×10−94π(10−3)3×1500×10×π12×10−2=480V. For the force on the positive droplets to point upward the field must point up, so the lower plate B is at the higher potential; the pair differing by 480 V with VB>VA is VA=100V and VB=580V.
JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option