Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0765 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 = 0.100 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1.
q1__________?
q2__________?

Answer 1

Answer:

q₁ =± 1.30 10⁻⁶ C  and   q₂ = ± 1.28 10⁻⁶ C

Explanation:

We will solve this problem with Coulomb's law

    F = K q₁q₂ / r²

Where the Coulomb constant is value 8.99 10⁹ N m² / C²

Let's apply this equation to our problem

Case 1

    F1 = k q₁ q₂ / r₁²

Where r₁ = 0.440 m and F1 = 0.0765 N

Case 2

The charges are the same

    F2 = k q q / r₂²

With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N

When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres

    q₁ + q₂ = 2 q

Let's replace

    F2 = k ½ (q₁ + q₂) / r²

Let's write the two equations and solve the system of equations

    F1 = k q₁ q₂ / r²

    F2 = ½ k (q₁ + q₂) / r²    

    F1 r² / k = (q₁ q₂)

    F2 r² / k = (q₁ + q₂)/2

    q₁ = 2F2 r² / k - q₂

We substitute in the other equation

    F1 r² / k = (2F2 r² / k - q₂) q₂

    0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²

Let's solve the second degree equation

    F1 r² / K = 0.0765 0.440² / 8.99 10⁹

    F1 r² / K = 1.65 10⁻¹²

   (2F2 r² / k) =2  0.10 0.44² / 8.99 10⁹

    (2F2 r2 / k) = 4.30 10⁻¹²

    q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0

    q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}

    q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}

    q₂ = ± 1.2845 10⁻⁶ C

Now we calculate q1

    F1 = k q₁ q₂ / r²

    q₁ = F1 r² / (k q₂)

    q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)

    q₁ = 1.30 10⁻⁶ C