Answer:
The smallest charge of the dial is 2.46 10-5 C
Explanation:
The Coulomb force is responsible for the electroactive repulsion, the equation that describes it is
F = k q1 q2 / r²
Where K is the Coulomb constant that is worth 8.99109 N m² / C², q is the electric charge of each sphere and r is the distance between them.
They also give us the condition that the sum of the charge is 5.23 10-5 C
Qt = q1 + q2 = 5.23 10⁻⁵ C
Let's replace in the Coulomb equation, let's clear and calculate
F = k (Qt -q2) q2 / r²
F = k q22 / r² - k Qt q2 / r²
1.0 = 8.99 10⁹ q2² /2.04² - 8.99 10⁹ 5.23 10⁻⁵ q2 / 2.04²
1.0 = 2.16 10⁹ q2²2 - 11.30 10⁴ q2
0 = 2.16 109 q2² - 11.30 10⁴ q2 -1.0 (* 1/2.16 109)
0 = q2² - 1.05 10⁻⁵ q2 - 0.463 10⁻⁹
Let's solve the second degree equation for q2
q2 = 1.05 10⁻⁵ ±√[(1.05 10⁻⁵)² - 4 1 (-0.463 10⁻⁹)] / 2
q2 = 1.05 10⁻⁵ ±√ [1.10 10⁻¹⁰ + 18.52 10⁻¹⁰] / 2
q2 = {1.05 10⁻⁵ ± 4.43 10⁻⁵} / 2
The solutions are
q2 ’= 2.74 10-5 C
q2 ’’ = -1.69 10-5 C
As the problem tells us that the spheres are positively charged, the correct solution is 2.74 10-5 C, let's see the charge of the other sphere
Qt = q1 + q2 ’
q1 = Qt -q2 ’
q1 = 5.23 10-5 - 2.74 10-5
q1 = 2.46 10-5 C
The smallest charge of the dial is 2.46 10-5 C
