JEE Main 2026 · 2 Apr · Shift 2 · Physics Q39

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PhysicsSingle correct+4 / -1Question 39 of 75

Two metal plates ( $A$ , $B$ ) are kept horizontally with separation of $\left(\dfrac{12}\pi\right)$ 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 $V_A$ and $V_B$ are required on plates A and B respectively in order to ensure the droplets do not descend. The values of $V_A$ and $V_B$ are ________.
(Neglect the air resistance to the droplets and take $g=10$ m/s²)

100 V and 580 V

580 V and 100 V

60 V and 400 V

0 V and −200 V

Answer and worked solution

Official answer: A

Weight of a droplet. $m=\frac43\pi(10^{-3})^3(1500)=2\pi\times10^{-6}$ kg, so $mg=2\pi\times10^{-5}$ N.

Field and voltage. $qE=mg$ gives $E=\dfrac{2\pi\times10^{-5}}{5\times10^{-9}}=4\pi\times10^3$ V/m, and
$\Delta V=Ed=4\pi\times10^3\times\frac{0.12}\pi=480\text{ V}.$

Polarity. The force on the positive charge must be upward, so E points up: the lower plate B is 480 V higher. Only (A) has $V_B-V_A=480$ V.

(A)

Topics:Electric field Electric potential

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JEE Main 2026 · 2 Apr · Shift 2, Physics question 39