Two equally charged spheres, each of mass 1.00 g, are placed 2.00 cm apart. When released, each accelerates at 225 m/s^2. What is the charge on each sphere?

• A. 50 nC
• B. 100 nC
• C. 200 nC
• D. 1 nC

Answer: (B)

The force pushing the spheres apart is given by Coulomb’s law, and since the charges are equal, each sphere experiences the same magnitude of force F = k q^2 / r^2. This force provides the acceleration: F = m a, where m is the mass of one sphere and a is its acceleration.

First compute the force from the given acceleration: F = m a = (0.001 kg)(225 m/s^2) = 0.225 N.

The separation is r = 2.00 cm = 0.02 m, so r^2 = 0.0004 m^2. Solve for q from F = k q^2 / r^2:

q^2 = F r^2 / k = (0.225 N)(0.0004 m^2) / (8.99×10^9 N·m^2/C^2).

Calculate: 0.225 × 0.0004 = 9.0×10^-5; divide by 8.99×10^9 gives q^2 ≈ 1.0×10^-14 C^2, so q ≈ 1×10^-7 C.

Thus each sphere carries about 100 nC of charge.

If the charge were smaller, the force would be too weak to produce the given acceleration; if larger, the acceleration would be larger. The value 100 nC matches the required acceleration of 225 m/s^2 for the 1.00 g mass at 2.00 cm apart.