Question

The electrostatic force of repulsion is 3.8×10^-8N in vacuum.if it is placed in a medium of dielectric constant 20. calculate the force in that medium .​

Answer 1

Answer:

3

Explanation:

By Coulomb's law, the force between two charges (q 1 ,q 2) in air is given by F= 4πϵ 0

r

2

q

1

q

2

where r= distance between two charges.

Now the force between two charges (q

1

,q

2

) in medium with dielectric constant K=4 becomes, F

′

=

4πKϵ

0

r

2

q

1

q

2

So, F

′

=

K

F

=

4

F