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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

  • Option 1)

    4F\;

  • Option 2)

    \; 2F\;

  • Option 3)

    \; \frac{F}{2}\,

  • Option 4)

    \, \frac{F}{4}

 

Answers (1)

best_answer

As we learned

When dielectric insert between the charges -

F_{med}=\frac{F_{air}}{K}=\frac{1q_{1}q_{2}}{4\pi \varepsilon _{0}kr^{2}}

- wherein

 

 In the presence of medium force becomes \frac{1}{K}times

 


Option 1)

4F\;

Option 2)

\; 2F\;

Option 3)

\; \frac{F}{2}\,

Option 4)

\, \frac{F}{4}

Posted by

prateek

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