A tiny 0.0250 - microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plates that carry equal but opposite charges on their facing surfaces. The plates are both circular with a radius of 6.50cm. (k = 1/4pi (epsilon) 0 = 9.0 x 10^9 N x m^2/C^2, e = 1.6 x 10^-19 C) a) How much excess charge must be on each plate to hold the oil drop steady? (b) Which plate must be positive, the upper one or the lower one?

(a) 12 * 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg = 0.025 * 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 * 10⁻² m

k = 1/4πε₀ = 9.0 * 10⁹ Nm²/C²

electric charge, e = 1.6 * 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε = Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F = W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 * 10⁻⁶ * 9.8 * (6.5 * 10⁻²) ²) : (4 * 9 * 10⁹ * 15 * 1.6 * 10⁻¹⁹) = 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.