Physics - Electrostatics Question with Solution | TestHub

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

For a dielectric of dielectric constant K between the charges, the effective separation in air r_effis given by

$\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{\mathrm{eff}}^2}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{K{r}^2}$

$\Rightarrow r_{\mathrm{eff}}=\sqrt{K}r$

$\therefore F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(\frac{r}{2}+\frac{\sqrt{K}r}{2}\right)}^2}$

$\Rightarrow \frac{F^{\prime }}{F}=\frac{1}{{\left(\frac{1}{2}+\frac{\sqrt{4}}{2}\right)}^2}=\frac{4}{9}$