Spheres A and B have identical sizes and carry identical charges $q$, separated by a distance $r$. A

Question

Spheres A and B have identical sizes and carry identical charges $q$, separated by a distance $r$. A third uncharged sphere C of the same size touches A, then touches B, and is finally removed. What is the new force of repulsion between A and B in terms of the original force $F$?

Professor PC Thomas Classes · Answer

When C touches A, they share charge equally: $q_A' = q/2$, $q_C = q/2$. When C touches B (charge $q$), they share total charge $(q/2 + q) = 3q/2$ equally: $q_B' = 3q/4$. New force $F' = k(q_A')(q_B')/r^2 = k(q/2)(3q/4)/r^2 = \frac{3}{8} k q^2/r^2 = \frac{3}{8} F$.