Physics, asked by apurbasangma7, 1 month ago

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.​

Answers

Answered by ankitjaydebpal
0

Answer:

okkkkkkkkkkkkkkkkkkkkkk

Answered by vijayarajv639
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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