1. Electric Charges and Fields
medium

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?

A

$d \sqrt{ k }$

B

$k \sqrt{ d }$

C

$1.5 d \sqrt{ k }$

D

$2 d \sqrt{ k }$

(JEE MAIN-2023)

Solution

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

$F _{\text {Air }}=\frac{1}{4 \pi \varepsilon_0} \frac{ q _1 q _2}{ d ^{\prime 2}}$

$F = F _{ Air }$

$\frac{ q _1 q _2}{4 \pi \varepsilon_0 kd ^2}=\frac{ q _1 q _2}{4 \pi \varepsilon_0 d ^{\prime 2}}$

$d ^{\prime}= d \sqrt{ k }$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.