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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