Two small equal spheres carrying unlike and unequal charges are placed 8 cm apart and attract each other with a force of 4 x 10^-5 N.After they have

been connected for a moment by a thin conducting wire , they repel each

other with a force of 2 x 10^-5 N.Calculate their original charges.

let the charge on them be q1 and -q2

4X 10^-5 = 9X10¹³. Q1.Q2/64------------------------1

NOW , when they r connected, charge on each would be q1-q2/2 as the spheres are similar.

2X10^-5 = 9X10¹³(Q1-Q2)²/4X64---------------------2

Solve these two eqns.

F_a =-4 x 10 ^-5 N  F_r =2 x 10^-5 N    q1 q2 b the two charges                                            F _a =(k q1 q2)/(8 x 10 ^-2  )²  since it is force of attraction ,the force is designated a minus sign, as the charges posses positive and negative charges and their product yields a  force (k q1 q2) with a negative sign  either q1 or q2 is having - sign so. 

after the two spheres are connected with a wire on each sphere is thes same with same sign...it is equal to (q1 +q2)/2. so the expression for the F_r = K[(q1+ q2)/2]² /(64 x 10 ^-4 )

repulsion is when spheres  have the same sign of electric charge...

2 x 10 ^-5 =(9 x 10⁹/4 x 64 x 10^-4 )(q1 + q2)²

(q1+q2)² = 56.88 x 10  ^-18 C2   ; (q1+ q2)= + or - 7.54 x 10^ -9 C-------------------------------------------1

q1 q2=- 28.44 x 10^ -18 C²

(q1 -q2)² =(q1 +q2)²  - 4 q1 q2  ;(q1-q2)² =170.64 x 10 ^-18 ; (q1 - q2)=+ or - 13.06 x 10^-9 C_____2

solving the eqn for q1 and q2 from 1and 2 ;u get q1= 10.30 x 10^-9 C or -10.30x 10^-9

q2 = - 2.76 or  +2.76  x 10 ^-9 C  either way it is possible check the answers in the original eqn for forces...