when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the sph

Question

when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.2525 N. What were the initial charges on the spheres

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Diễm Thu 3 weeks 2021-08-23T23:35:07+00:00 1 Answers 0 views 0

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    2021-08-23T23:36:08+00:00

    Answer:

    q1 = 7.6uC , -2.3 uC

    q2 = 7.6uC , -2.3 uC

    ( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

    Explanation:

    Solution:-

    – We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

    – To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb’s Law, which states:

                                  F = k\frac{|q_1|.|q_2|}{r^2}

    Where,

                         k: The coulomb’s constant = 8.99*10^9

    – Coulomb’s law assume the objects as point charges with separation or ( r ) from center to center.  

    – We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

    – Therefore, the force of attraction between the spheres would be:

                                 \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1Eq 1

    Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q’ ). This is the point when the re-distribution is complete ( current stops in the wire).

    – We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                                 q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

    – Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                              \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

    We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                             -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                             

                              q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

     

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