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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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olatunjiolorundare20

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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