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Two charges \small q_1, q_2 are placed in vacuum at a distance \small d and the force acting between them is \small F. If medium of dielectric constant 4 is introduced around them, the force now will be-

Option: 1

4F


Option: 2

2F


Option: 3

F/2


Option: 4

F/4


Answers (1)

best_answer

\text { We know that, force is given by, }

\vec{F}=\frac{1}{4 \pi c_0 k} \frac{q_1 q_2}{d^2}                    -(1)

In equation (1),  k is dielectric constant which is 4 in question statement. Then,

\vec{F}^{\prime} \propto \frac{1}{k}

F^{\prime} \propto \frac{1}{4} \quad \text { force will be } F / 4

 

Posted by

Riya

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