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17 September, 04:05

Two small, positively charged spheres have a combined charge of 5.23 x 10^-5 C. If each sphere is repelled from the other by an electrostatic force of 1.00 N when the spheres are 2.04 m apart, what is the charge on the sphere with the smaller charge?

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  1. 17 September, 04:34
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    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
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