Question

Two small objects, each with a charge Q, exert a force F on each other. We replace one of the objects with another whose charge is 4Q. (a.) The original magnitude of the force on the Q charge was F. What is the force now? (b.) What is the magnitude of the force on the 4Q charge? (c.) Next, we move the charges to be three times as far apart as they were. What is the force on the Q charge now? (d.) If we change the signs of both charges to negative, how would that affect your answers to the previous questions? Briefly explain.

Answer 1

Answer:

a) 4*F₀ b) 4*F₀ c) (4/9)*F₀ d) F is the same as for a) b) and c).

Explanation:

a) Assuming that we can treat to both objects as point charges, the electrostatic force between them must obey Coulomb´s Law, as follows:

F₀ = k*Q²/r² (1)

a) If we replace one of the objects with another, whose charge is 4Q, the new force on charge Q is as follows:

F= k*4Q*Q / r² = 4*(k*Q²/r²)

From (1) we know that (k*Q²/r²) = F₀

⇒ F = 4*F₀

b) The force on the 4Q charge, according to Newton´s 3rd Law, must be equal in magnitude to the one on the charge Q, as follows:

F₄q = 4*F₀

c) If we move the charges to be three times apart as they were, the new magnitude of the force will be as follows:

F = k*Q*4Q / (3*r)² = (4/9)*(k*Q²/r²)

From (1) we know that (k*Q²/r²) = F₀

⇒ F = (4/9)*F₀

d) If we change the signs of both charges, the force between them  will remain to be repulsive (as now both charges will be negative), so the new forces will be the same as for the positive charges.