12) 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) 2 J
B) 2005
D) 250 J
Please solve with calculations
Answers
Answered by
4
Answer:
charge to be moved in an electric field=5C=q
work done=20J
we have to find potential difference b/w 2 point.
here, we use the formula that
workdone=(charge×potentialdifference) b/w 2 point.
so, potential difference b//w 2 point
V=
charge
workdone
=
5
20
=4V.
Explanation:
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