the force of interaction between two point charges is f.what will be the force of interaction if magnitude of each charge is doubled and distance between them is halved.
Answers
Answered by
1
Explanation:
The force between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Hence, if distance between charges is halved (charges remaining kept constant), the force between the two charges is quadrupled.
Answered by
7
Answer:
By Coulomb's law, the force between two charges q
1
and q
2
is F=
r
2
kq
1
q
2
where r= separation between charges.
When charges are doubled and their distance is halved , the force of interaction becomes F
′
=
(r/2)
2
k(2q
1
)(2q
2
)
=16F
Thus, the value of n will be 16.
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