The force of repulsion between two identical charges separated by a distance of 9cm in air is 144 dyne. find the magnitude of each charge.
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Answered by
0
Answer:
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Answered by
0
Explanation:
the distance between the charges initially be r.
Thus initial electrostatic force F
o
=
r
2
kQ
1
Q
2
New distance between the charges r
′
=3r
Charge of each particle is tripled Q
1
′
=3Q
1
and Q
2
′
=3Q
2
∴ New electrostatic force F
′
=
(3r)
2
k(3Q
1
)(3Q
2
)
=F
o
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