Question

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. $k \sqrt{d}$
• B. $1 \cdot 5 d \sqrt{k}$
• C. $2 d \sqrt{k}$
• D. $d \sqrt{k}$

Answer 1

Answer: (D)

$d \sqrt{k}$

Answer 2

F=(4πε0​)1​kd2q1​q2​​( in medium)
FAir ​=4πε0​1​d′2q1​q2​​
F=FAir​
4πε0​kd2q1​q2​​=4πε0​d′2q1​q2​​
d′=dk​