Question

Two small conducting point charges, separated by 0.5 m, carry a total charge of 170 mu or micro CC. They repel one another with a force of 120 N. For the universal constant k use the value 8.99 times 109 N m2/C2.

Find the charge on the larger of the two point charges.

Answer 1

the two point charges are equal

q₁ = 147.353 x 10⁻⁶ C or 22.646 x 10⁻⁶ C

q₂ = 22.647 x 10⁻⁶ C or 147.354 x 10⁻⁶ C

Step-by-step explanation:

Given;

distance between the two charges, r = 0.5 m

let the first charge = q₁

let the second charge = q₂

q₁ + q₂ = 170μ = 170 x 10⁻⁶ C ---------- equation (1)

Force of repulsion between the two charges, q₁ and q₂ = 120 N

Coulomb's constant, K = 8.99 x 10⁹ N m²/C²

Apply coulomb's law, for force of attraction or repulsion between two point charges;

`F = (Kq_1q_2)/(r^2) \\\\q_1q_2 = (Fr^2)/(K) \\\\q_1q_2 = (120*0.5^2)/(8.99*10^9) \\\\q_1q_2 = 3.337*10^(-9) \ C^2` -------------- equation (2)

From equation (1); q₂ = 170 x 10⁻⁶ C - q₁

Substitute in the value of q₂ in equation (2)

q₁(170 x 10⁻⁶ - q₁) = 3.337 x 10⁻⁹

170 x 10⁻⁶q₁ - q₁² = 3.337 x 10⁻⁹

170 x 10⁻⁶q₁ = q₁² + 3.337 x 10⁻⁹

0 = q₁² - 170 x 10⁻⁶q₁ + 3.337 x 10⁻⁹ (this is quadratic equation)

q₁ = 147.353 x 10⁻⁶ C or 22.646 x 10⁻⁶ C

q₂ = 170 x 10⁻⁶ C - q₁

q₂ = 170 x 10⁻⁶ C - 147.353 x 10⁻⁶ C or 170 x 10⁻⁶ C - 22.646 x 10⁻⁶ C

q₂ = 22.647 x 10⁻⁶ C or 147.354 x 10⁻⁶ C

Therefore, the two point charges are equal