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  4. If two charges q 1 and q 2 are separated with distance ' d ' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?

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Q. If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?

JEE MainJEE Main 2023Electric Charges and Fields

Solution:

$ F =\frac{1}{\left(4 \pi \varepsilon_0\right)} \frac{ q _1 q _2}{ kd ^2}(\text { in medium) } $
$ F _{\text {Air }}=\frac{1}{4 \pi \varepsilon_0} \frac{ q _1 q _2}{ d ^{\prime 2}} $
$ F = F _{ Air }$
$ \frac{ q _1 q _2}{4 \pi \varepsilon_0 kd ^2}=\frac{ q _1 q _2}{4 \pi \varepsilon_0 d ^{\prime^2}} $
$ d ^{\prime}= d \sqrt{ k }$