Doing The Electric Dance

Initially, $q_A = 0$ and $q_B = q_C = q$ . If a thin conducting wire is connected between $A$ and $B$ then the charge $q$ of $B$ will be uniformly distributed at points $A$ and $B$ giving a new charge for points $A$ and $B$ , and that is: $q_A = q_B = \frac{q}{2}$ .

The thin conducting wire is then removed and is now connected at points $A$ and $C$ , following the same concept, i.e, Law of conservation of charge then the charge $\frac{q}{2}+q = \frac{3q}{2}$ will be equally distributed which implies $q_A = q_C = \frac{\frac{3q}{2}}{2} = \frac{3q}{4}$

By Coulomb's first law on electrostatic charges, which states that the magnitude of the force between two charges is directly proportional to the product of the charges, then the force of interaction at points $B$ and $C$ is now $(\frac{1}{2})(\frac{3}{4})(4 N) = \boxed{\frac{3}{2} N}$

At first the Electrostatic force between B and C is given by kQ^2/L^2 k = 1/ (4 * pi * epsilon) and L=length of the side of the triangle.... The value of force is given 4N . Now when sphere A(which is uncharged initially) is touched with B then the charge Q is equally shared between them because they are identical....Now again A is touched with another sphere C and charge on each of them becomes (Q/2 + Q)/2 = 3Q/4 ... Now the new force of attraction between B and C is 3/8 th of their original force 4N ... hence answer is = 1.5N.

Since all spheres have same dimensions, when connected by thin wire the charge on each will be equal to half the total sum of charges on the two. Using this get the final charges and solve with Coulomb's law.