JEE Main & Advanced
Physics
Electrostatics & Capacitance
Question Bank
Charge and Coulombs Law
question_answer
Two charges \[{{q}_{1}}\]and \[{{q}_{2}}\]are placed in vacuum at a distance \[d\] and the force acting between them is \[F\]. If a medium of dielectric constant 4 is introduced around them, the force now will be [MP PMT 1994]
A) \[4F\]
B) \[2F\]
C) \[\frac{F}{2}\]
D) \[\frac{F}{4}\]
Correct Answer:
D
Solution :
In the presence of medium force becomes \[\frac{1}{K}times\].