Q64 / 75 · JEE Main 2026 (02 April Shift 2) · Physics
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$^2$)
(Neglect the air resistance to the droplets and take $g = 10$ m/s$^2$)
- Correct answerA$100 \, V$ and $580 \, V$
- B$580 \, V$ and $100 \, V$
- C$60 \, V$ and $400 \, V$
- D$0 \, V$ and $-200 \, V$
Solution
For the 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 $580 \, V$
$qE = mg$
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 $580 \, V$
Hence, the correct answer is option: 'A'.