JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

Two metal plates ( $A, B$ ) are kept horizontally with separation of $(\frac{12}{π})$ cm, with plate $A$ on the top. An atomizer jet sprays oil (density $1.5$ g/cm $^{3}$ ) 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/s $^{2}$ )

A
$100$ V and $580$ V
B
$580$ V and $100$ V
C
$60$ V and $400$ V
D
$0$ V and $- 200$ V

Solution

For the droplet to hover, the electric force on the positive charge must point upward, so the field must point upward (from $B$ to $A$ ), meaning $V_{B} > V_{A}$ .

Equilibrium: $q E = m g$ , with $E = \frac{Δ V}{d}$ and $m = \frac{4}{3} π r^{3} ρ$ .

Here $r = 1 0^{- 3}$ m, $ρ = 1500$ kg/m $^{3}$ , $q = 5 \times 1 0^{- 9}$ C, $d = \frac{12}{π} \times 1 0^{- 2}$ m.

$Δ V = \frac{m g d}{q} = \frac{( 4/3 ) π ( 1 0 ^{- 9} ) ( 1500 ) ( 10 ) \cdot ( 12/ π ) \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 9}} = \frac{4 \cdot 1500 \cdot 10 \cdot 12 \times 1 0 ^{- 11}}{3 \cdot 5 \times 1 0 ^{- 9}} = 480$ V.

Among the given options, only $V_{A} = 100$ V, $V_{B} = 580$ V satisfies $V_{B} - V_{A} = 480$ V.

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