# Question 59646

Apr 7, 2017

$\frac{r}{\sqrt{\epsilon}} _ r$

#### Explanation:

Let two point charges ${q}_{1} \mathmr{and} {q}_{2}$ be located in air at a distance $r$ from each other.

Magnitude of Force between the two charges is given by Coulomb's Law
$| {\vec{F}}_{12 \text{ air}} | = {k}_{e} \frac{{q}_{1} {q}_{2}}{r} ^ 2$ ......(1)

where ${k}_{e} = \frac{1}{4 \pi {\epsilon}_{0}}$ is Coulomb's constant = 8.99×10^9 N m^2 C^-2, and the constant ε_0 is the permittivity of free space.

When the same charges are placed in a medium of relative permittivity ${\epsilon}_{r}$, the electric field between the charges is decreased relative to vacuum by this value. Hence, we have the relation for force as
|vecF_(12 " med")|= k_e/epsilon_r (q_1 q_2)/ r_"med"^2# ......(2)

To meet the given condtion equating (1) and (2) we get
${k}_{e} \frac{{q}_{1} {q}_{2}}{r} ^ 2 = {k}_{e} / {\epsilon}_{r} \frac{{q}_{1} {q}_{2}}{r} _ {\text{med}}^{2}$
$\implies {r}_{\text{med}}^{2} = {r}^{2} / {\epsilon}_{r}$
$\implies {r}_{\text{med}} = \frac{r}{\sqrt{\epsilon}} _ r$