if charge of 5C is moved against electric field of 10 NC-1 through distance of 5m the P.E gained by charge is
Answers
Answered by
10
Answer:
P.E gained = 250 Joules
Explanation:
P.E= force × displacement
= q ×E × d
= 5 × 10 × 5
=250 Joules
Answered by
0
Answer:
= - 55 j
Explanation:
del U = delQ+delW
= del q +f ×Displacement
=5+10×5
= - 55 j
Similar questions