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12 April, 01:54

The electric field 10.0 cm from the surface of a copper ball of radius 5.00 cm is directed toward the ballâs center and has magnitude 9.00/times 10^2~/text{N/C}9.00Ã10 â2 ââ N/C. How much charge is on the surface of the ball?

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  1. 12 April, 02:19
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    Answer:0.01C

    Explanation:E=Kq/r^2

    E=electric potential=f/q=given as 9*10^9N/C

    K=constant=9*10^9Nm^2/C

    So q=Er^2/K

    =9*10^9*{0.1}^2/9*10^9=0.01C

    Note r is the distance from. the field and the copper.
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