Two identical metallic spheres, having unequal opposite charges are placed at a distance of 5 metres apart in air. After bringin

Question

Two identical metallic spheres, having unequal opposite charges are placed at a distance

of 5 metres apart in air. After bringing them in contact with each other, they are again

placed at the same distance apart. Now the force of repulsion between them is 20 N.

Calculate the final charge on each of them.​

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Dulcie 5 months 2021-08-23T11:28:45+00:00 1 Answers 3 views 0

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    2021-08-23T11:30:30+00:00

    Answer:

    Charge on the sphere is  2.48*10^{-9} C

    Explanation:

    distance apart r = 5 m

    force of repulsion F = 20 N

    The spheres had opposite unequal opposite charges, this means that on bringing them into contact, the sphere with the greater charge will have its charge cancel out the other charge on the other sphere. The resultant charge will then be evenly distributed between the two spheres. The result is that both spheres will now have like, equal amount of charge on them.

    applying Coulumb’s law,

    F = \frac{kQ^{2} }{r^{2} }

    where

    k = 9*10^{9}  m^{3} s^{-4} A^{-2}

    substituting values into the equation, we have

    20 = \frac{9*10^{9}*Q^{2} }{5^{2} }

    Q^{2}  = \frac{20*25}{9*10^{9} } = 6.17*10^{-18}

    Q = \sqrt{6.17*10^{-18} } = 2.48*10^{-9} C

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