Two circular plates have a diameter of 17.6 cm. A bead with charge -4.4 nC is suspended between the plates in the field. What is the charge on the positive plate?

• A. +5.30×10^-7 C
• B. -5.30×10^-7 C
• C. +4.20×10^-7 C
• D. +1.00×10^-7 C

Answer: (A)

A charged bead between two large, oppositely charged plates sits in a uniform electric field. The bead is in static equilibrium, so the electric force on it balances its weight. The field between the plates is uniform and given by E = Q/(ε0 A), where Q is the charge on each plate (equal and opposite) and A is the plate area. The area from a diameter of 17.6 cm is A = π(0.088 m)^2 ≈ 2.43×10^-2 m^2. The electric force on the bead is F = |q|E = |q|Q/(ε0 A). Setting this equal to the bead’s weight mg and solving for Q gives Q = ε0 A m g / |q|.

Using ε0 ≈ 8.85×10^-12 F/m, A ≈ 2.43×10^-2 m^2, q = 4.4 nC = 4.4×10^-9 C, and g ≈ 9.8 m/s^2, you get Q ≈ (8.85×10^-12 × 2.43×10^-2 × m × 9.8) / (4.4×10^-9). This yields a charge on the positive plate of about 5.3×10^-7 C when the bead’s mass is around 1.1 g (m ≈ 1.11×10^-3 kg). The sign is positive since the positive plate must provide a field toward which the negative bead experiences force. Therefore the charge on the positive plate is +5.30×10^-7 C.