Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.110 N when their center-to-center separation is 57.3 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0489 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other

  q₂ = ± 1.306 10⁻⁶ C

   q₂ =  ± 3.073 10⁻⁶ C

Explanation:

We must solve this exercise using Coulomb's law, let's write the equilibrium equation for each case

first case

     F₁ = k q₁ q₂ / r²         (1)

where F1 is the force on sphere 1, with a value of F₁ = 0.110 N; r is the distance between them r = 0.573 m

second case.

The spheres were joined with a conductive wire, so the charge of the two is equal, then the wire is removed and

       -F₂ = k q₃ q₃ / r²           (2)

In this case the force is expelling and is equal to F₂ = 0.0489 N

when the spheres are joined with the wire the charge of the spheres is distributed evenly between them two

         q₁ + q₂ = 2 q₃             (3)

we can see that there are three equations with three unknowns for which the system can be solved

let's substitute equation 3 in 2

         - F₂ = k (q₁ + q₂)² / r²

we join this equation with equation 1

          F₁ = k q₁ q₂ / r²

          -F₂ = k (q₁ + q₂)²/4r²

          q₁ = F₁ r² / k q₂

         - F₂ = k (F₁ r² / k q₂ + q₂)²/4r²

Let's replace the values and work out

         - 0.0489 = 9 109 (0.110 0.573² / (9 10⁹ q₂) + q₂)² / 0.573²

          - 0.0489 4 0.573²/9 10⁹ = (0.110 0.573² / (9 10⁹ q₂) + q₂)²

          - 7.1356 10⁻¹² = (4.013 10⁻¹² / q₂ + q₂)²

          - 7.1356 10⁻¹² = 1 / q₂² (4.013 10⁻¹² + q₂² )²

          - 7.1356 10⁻¹² q₂² = (16.104 10⁻²⁴ + 4.013 10⁻¹² q₂² + q₂⁴

          q₂⁴ + q₂² (4,013 10⁻¹² + 7,1356 10⁻¹²) + 16,104 10⁻²⁴ = 0

we change variables

        q’ = q₂²

          q'² + q'  11.1486 10⁻¹² + 16.104 10⁻²⁴ = 0

we solve this quadratic equation

          q '= [- 11.1486 10⁻¹² ±√ ((11.1486 10⁻¹² )² - 4 16.104 10⁻²⁴)] / 2

          q '= [- 11.1486 10 12 ±√ (59.8753 10⁻²⁴)] / 2

          q '= [- 11.1486 ± 7.7379] / 2 10⁻¹²

          q'1 = -1.70535 10⁻¹²

          q'2 = -9.44325 10⁻¹²

           q' = q₂²

           q₂ = √ q'

           q₂ = ± 1.306 10⁻⁶ C

           q₂ =  ± 3.073 10⁻⁶ C

since q 'is squared of these charges they can be positive or negative giving the same result.

a straight line sloped

i think the answer would be reserve.