Question

The force of repulsion between two identical positive charges when kept with a separation 'r' in air is 'F' Half the gap between the two charges is filled by a dielectric slab of dielectric constant = 4. Then the new force of repulsion between those two charges becomes

• A. $\frac{F}{3}$
• B. $\frac{F}{2}$
• C. $\frac{F}{4}$
• D. $\frac{4F}{9}$

Answer 1

Answer: (D)

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(r-t+t\sqrt{k}\right)}^2}$
$\frac{F}{F^{\prime }}=\frac{{\left[r-\left(\frac{r}{2}\right)+\left(\frac{r}{2}\right)\sqrt{4}\right]}^2}{r^2}$
$=\frac{{\left(\frac{3r}{2}\right)}^2}{r^2}$
$=\frac{\frac{9}{4}{r}^2}{r^2}$
$\frac{F}{F^{\prime }}=\frac{9}{4}$
$F^{\prime }=\frac{4}{9}F$