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 \[\frac{1}{K}times\].


You need to login to perform this action.
You will be redirected in 3 sec spinner