Physics, asked by AymanRajput, 6 hours ago

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 ​

Answers

Answered by mjstule123456789
4

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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