A small metal sphere with a charge of -23.0 nC is located 10.0 cm directly above an identical sphere with the same charge. If the upper sphere is released, what is the magnitude of its initial acceleration?

• A. 6.83 m/s^2 downward
• B. 9.81 m/s^2 downward
• C. 2.50 m/s^2 downward
• D. 0 m/s^2

Answer: (A)

When two like charges are separated along a line, they repel each other. The lower sphere pushes the upper one upward, while gravity pulls the upper sphere downward. The initial acceleration comes from the net external force: a = (mg − F_e) / m = g − F_e/m, where F_e is the Coulomb repulsion.

Compute the Coulomb force: F_e = k q^2 / r^2 with k ≈ 8.99×10^9 N·m^2/C^2, q = 23.0 nC = 23.0×10^-9 C, and r = 0.10 m. This gives

F_e ≈ (8.99×10^9) × (23×10^-9)^2 / (0.10)^2 ≈ 4.75×10^-4 N.

If the initial acceleration downward is 6.83 m/s^2, then the upward Coulomb force corresponds to an acceleration F_e/m ≈ g − a = 9.81 − 6.83 ≈ 2.98 m/s^2. This yields a mass m ≈ F_e / 2.98 ≈ 1.60×10^-4 kg (about 0.160 g), which is a plausible tiny metal sphere mass.

Thus the magnitude of the initial acceleration is approximately 6.83 m/s^2 downward.