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 1

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.

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  1. Answer:

    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

    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;

    [tex]F = \frac{Kq_1q_2}{r^2} \\\\q_1q_2 = \frac{Fr^2}{K} \\\\q_1q_2 = \frac{120*0.5^2}{8.99*10^9} \\\\q_1q_2 = 3.337*10^{-9} \ C^2[/tex] ————–  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

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