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)

$\frac{4F}{9}$

Answer 2

$F' = \frac{1}{4\pi\varepsilon_{0}} \frac{q_{1}q_{2}}{\left(r-t+t\sqrt{k}\right)^{2}}$ $\frac{F}{F'} = \frac{\left[r-\left(\frac{r}{2}\right)+\left(\frac{r}{2}\right)\sqrt{4}\right]^{2}}{r^{2}}$ $= \frac{\left(\frac{3r}{2}\right)^{2}}{r^{2}}$ $= \frac{\frac{9}{4}r^{2}}{r^{2}}$ $\frac{F}{F'} =\frac{9}{4}$ $F' = \frac{4}{9} F$