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 .
Answers
Answered by
0
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
Similar questions