2 If a charge of 5 C is moved against an electric field of 10 NC through a distance of 5 m, the P.E gained by charge is:
A) 25 J
C)2J
B) 200 J
D) 250 J
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U=q∆V
Given,
q=5C,
E=10,
d=5m
we know,E*d=V
V=10*5=50
U=5*50=250J
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