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To charges q_{1}, q_{2} are placed in vacuum at a distance d and the force acting between them is F.  If medium of dielectric constant 4 is introduced around them, the force now will be-

Option: 1

4 \: \mathrm{F}


Option: 2

2\, \mathrm{F}


Option: 3

\mathrm{F/2}


Option: 4

\mathrm{F/4}


Answers (1)

best_answer

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 ans.

 

Posted by

shivangi.bhatnagar

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