The value of the final charge on the sphere on the right is; 2.1855 × 10^(-6) C
The initial charge q1 on the sphere is;
-2.365 × 10^(-6) C or +6.736 × 10^(-6) C
We are told that the spheres are identical and as such;
Final charge = (q1 + q2)/2
Where q1 and q2 are charges on both sphere's.
Now, formula for force between two charges is;
F = k*q1*q2/r²
Where;
k is coulombs constant = 9 × 10^(9) C
r is distance of separation
We are given;
F = 72.1 mN = 0.0721 N
r = 1.41 m
Thus;
(9 × 10^(9) × q1 * q2)/(1.41²) = -0.0721
Thus;
q1*q2 = (0.0721 × 1.41^(2))/(9 × 10^(9))
q1*q2 = -15.93 × 10^(-12) C² - - - (eq 1)
Formula for final force will be;
F_f = k*(q1 + q2)²/4r²
F_f = (9 × 10^(9) × (q1 + q2)²)/(4 × 1.41²)
Since they repel each other with a force of 21.63 mN = 0.02163 N, then;
(9 × 10^(9) × (q1 + q2)²)/(4 × 1.41²) = 0.02163
(9 × 10^(9) × (q1 + q2)²) = 0.02163 × 4 × 1.41²
(q1 + q2)² = (0.02163 × 4 × 1.41²)/9 × 10^(9)
(q1 + q2)² = 19.11 × 10^(-12) C
q1 + q2 = 4.371 × 10^(-6) C - - - (eq 2)
Making q1 the subject in eq 1 gives;
q1 = (-15.93 × 10^(-12))/q2
Putting that in eq 2 gives;
((-15.93 × 10^(-12))/q2) + q2 = 4.371 × 10^(-6)
Multiply through by q2 to get;
(q2)² - (4.371 × 10^(-6)) - (15.93 × 10^(-12)) = 0
Solving using quadratic equation calculator gives;
q2 = +6.736 × 10^(-6) C or -2.365 × 10^(-6) C
q1 = -2.365 × 10^(-6) C or +6.736 × 10^(-6) C
Final charge = (4.371 × 10^(-6))/2
Final charge = 2.1855 × 10^(-6) C
Read more about force between charges at; https://brainly.com/question/17692887