Two metal plates (A, B) are kept horizontally with separation of ( 12 π ) cm, with plate A on the top. An atomizer jet sprays oil (density 1.5…

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

$$qE = mg$$qE=mg

The mass of the oil droplet is given by:

$$m = \dfrac{4}{3} \pi r^3 \rho$$m=34​πr3ρ

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

$m = \dfrac{4}{3} \pi (10^{-3})^3 \times 1500 = 2 \pi \times 10^{-6}$m=34​π(10−3)3×1500=2π×10−6 kg

The electric field $E$E required is:

$E = \dfrac{mg}{q} = \dfrac{2 \pi \times 10^{-6} \times 10}{5 \times 10^{-9}} = 4000 \pi$E=qmg​=5×10−92π×10−6×10​=4000π V/m

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

$\Delta V = E \times d = 4000 \pi \times \left(\dfrac{12}{\pi} \times 10^{-2}\right) = 480$ΔV=E×d=4000π×(π12​×10−2)=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.

$V_B - V_A = 480$VB​−VA​=480 V

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

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