Physics Q39 – JEE-MAIN 2026

Home Physics Electric Charges and FieldsQuestion #6

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}

)

Clue 1 of 3: Equilibrium Condition

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

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate the mass of the oil droplet✦ Active

First, convert all given values to SI units. The radius $r = 1 mm = 1 \times 10^{- 3} m$ . The density $ρ = 1.5 {g/cm}^{3} = 1.5 \times 10^{3} {kg/m}^{3}$ . The volume of a spherical droplet is $V = \frac{4}{3} π r^{3}$ . The mass of the droplet is $m = ρ V$ .

V = \frac{4}{3} π (1 \times 10^{- 3})^{3} = \frac{4}{3} π \times 10^{- 9} m^{3}

m = (1.5 \times 10^{3} {kg/m}^{3}) \times (\frac{4}{3} π \times 10^{- 9} m^{3}) = 2 π \times 10^{- 6} kg

Step 2: Determine the required electric field strength○ Expand

For the droplets to not descend, the upward electric force ( $F_{e}$ ) must balance the downward gravitational force ( $F_{g}$ ). The gravitational force is $F_{g} = m g$ . The electric force is $F_{e} = q E$ , where $q = 5 nC = 5 \times 10^{- 9} C$ is the charge on the droplet. Thus, $q E = m g$ .

F_{g} = (2 π \times 10^{- 6} kg) \times (10 {m/s}^{2}) = 2 π \times 10^{- 5} N

E = \frac{F_{g}}{q} = \frac{2 π \times 10^{- 5} N}{5 \times 10^{- 9} C} = \frac{2 π}{5} \times 10^{4} V/m

💡 Teacher's Secret Hint

Remember that the electric field must be directed upwards to exert an upward force on a positive charge.

Step 3: Calculate the potential difference between the plates○ Expand

The separation between the plates is $d = (\frac{12}{π}) cm = (\frac{12}{π}) \times 10^{- 2} m$ . For a uniform electric field, the potential difference $Δ V$ is given by $E d$ . Since plate A is on top and plate B is on the bottom, and the electric field is upwards (from B to A), plate B must be at a higher potential than plate A ( $V_{B} > V_{A}$ ). Therefore, $Δ V = V_{B} - V_{A}$ .

V_{B} - V_{A} = E d = (\frac{2 π}{5} \times 10^{4} V/m) \times (\frac{12}{π} \times 10^{- 2} m)

V_{B} - V_{A} = \frac{2 \times 12}{5} \times 10^{4 - 2} = \frac{24}{5} \times 10^{2} = 4.8 \times 100 = 480 V

Checking the options, only option 1 ( $V_{A} = 100 V$ , $V_{B} = 580 V$ ) gives $V_{B} - V_{A} = 580 - 100 = 480 V$ . This matches our calculated potential difference.

💡 Teacher's Secret Hint

Pay close attention to the direction of the electric field and the relative potentials of the plates. The electric field points from higher to lower potential.

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Physics Question 39 – JEE-MAIN 2026

Clue 1 of 3: Equilibrium Condition

🧠 Full Solution Path