The electric field 20.0 cm from the surface of a copper ball with radius 4.2 cm is directed toward the ball’s center and has magnitude 425 N/C. How much charge is on the surface of the ball?
From Gauss law
Q=(425*(0.2+0.042)2/(9*109)
Q=2.7655*10-9 C
Since the direction of electric field is towards the center of the ball ,therefore
Q= - 2.7655*10-9 C
The electric field 20.0 cm from the surface of a copper ball with radius 4.2 cm...
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