Question

Three identical conducting spheres are located at the vertices of an equilateral triangle ABC. Initially the charge the charge of the sphere at point A is q A =0 and the spheres at B and C carry the same charge q A =q B =q. It is known that the sphere B exerts an electrostatic force on C which has a magnitude F=4N. Suppose an engineer connects a very thin conducting wire between spheres A and B. Then she removes the wire and connects it between spheres A and C. After these operations, what is the magnitude of the force of interaction in Newtons between B and C?

Answer 1

F(BC) =kq²/a².

q´(B)=q´(A) =(q+0)/2=q/2,
q´´(A) =q´(C) = {q´(A)+c(C)}/2 = 3q/4,

F´(BC) =k q´(B)• q´(C)/a² =
=(3/8)• kq²/ a²=3•4/8=3.2 = 1.5 N