Two horizontal metal plates A and B are separated by (12 ÷ π) cm, with plate A above plate B. An atomizer sprays positively charged oil droplets horizontally between the plates. Each droplet has radius 1 mm, density 1.5 g·cm⁻³ and charge 5 nC. Neglect air resistance and take g = 10 m·s⁻². Which pair of plate potentials keeps the droplets from descending?
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Question

Two horizontal metal plates A and B are separated by (12 ÷ π) cm, with plate A above plate B. An atomizer sprays positively charged oil droplets horizontally between the plates. Each droplet has radius 1 mm, density 1.5 g·cm⁻³ and charge 5 nC. Neglect air resistance and take g = 10 m·s⁻². Which pair of plate potentials keeps the droplets from descending?

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G1 Physics · Accepted Answer

For the droplet to have no vertical acceleration, the electric force must balance its weight: qE = mg. The droplet mass is m = ρ(4πr³ ÷ 3). Convert the density: ρ = 1.5 g·cm⁻³ = 1500 kg·m⁻³. Radius: r = 1 mm = 1×10⁻³ m. Therefore, m = 1500×(4π ÷ 3)(10⁻³)³. The required electric field magnitude is E = mg ÷ q. The plate separation is d = (12 ÷ π) cm = 0.12 ÷ π m. Thus the required potential difference magnitude is ΔV = Ed = [mg ÷ q]d. Substitute q = 5×10⁻⁹ C: ΔV = {1500(4π ÷ 3)(10⁻³)³×10 ÷ [5×10⁻…