Question

if one of the charges is double changes and distance reduced is half​

Answer 1

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.

Answer 2

Answer:

16F is the answer ☺☺☺☺☺☺