JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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/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 $^2$ )

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

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

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Solution

For the oil droplets to not descend, the upward electric force must balance the downward gravitational force.

$qE = mg$

The mass of the oil droplet is given by:

$m = \dfrac{4}{3} \pi r^3 \rho$

Substituting the given values $r = 10^{-3}$ m and $\rho = 1500$ kg/m $^3$ :

$m = \dfrac{4}{3} \pi (10^{-3})^3 \times 1500 = 2 \pi \times 10^{-6}$ kg

The electric field $E$ required is:

$E = \dfrac{mg}{q} = \dfrac{2 \pi \times 10^{-6} \times 10}{5 \times 10^{-9}} = 4000 \pi$ V/m

The potential difference $\Delta V$ between the plates is:

$\Delta V = E \times d = 4000 \pi \times \left(\dfrac{12}{\pi} \times 10^{-2}\right) = 480$ V

Since the charge on the droplet is positive, the electric field must be directed upwards to provide an upward force. Electric field lines point from higher potential to lower potential, so the bottom plate B must be at a higher potential than the top plate A.

$V_B - V_A = 480$ V

Checking the given options, $V_A = 100$ V and $V_B = 580$ V satisfies this condition.

Answer: $100 \, V$ and $580 \, V$

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Solution

For the oil droplets to not descend, the upward electric force must balance the downward gravitational force.

$qE = mg$

The mass of the oil droplet is given by:

$m = \dfrac{4}{3} \pi r^3 \rho$

Substituting the given values $r = 10^{-3}$ m and $\rho = 1500$ kg/m $^3$ :

$m = \dfrac{4}{3} \pi (10^{-3})^3 \times 1500 = 2 \pi \times 10^{-6}$ kg

The electric field $E$ required is:

$E = \dfrac{mg}{q} = \dfrac{2 \pi \times 10^{-6} \times 10}{5 \times 10^{-9}} = 4000 \pi$ V/m

The potential difference $\Delta V$ between the plates is:

$\Delta V = E \times d = 4000 \pi \times \left(\dfrac{12}{\pi} \times 10^{-2}\right) = 480$ V

$V_B - V_A = 480$ V

Checking the given options, $V_A = 100$ V and $V_B = 580$ V satisfies this condition.

Answer: $100 \, V$ and $580 \, V$

JEE Main 2026 — Physics Electromagnetism

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JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism