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 us filled by a dielectric slab of dielectric constant \( =4 \), Then the new force of repulsion between those two charges becomes

Given, force of repulsion between two identical positive charges with separation, \( r=F \).
Half of gap is filled with dielectric constant \( =4 \).
Now \( F=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \)
where \( q_{1} \) and \( q_{2} \) are two positive charges. For identical charges, \( q_{1}=q-2=q \)
\(F=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q^{2}}{r^{2}} \rightarrow(1)\)
When the gap is half filled with dielectric constant then
\(F^{\prime}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q^{2}}{(r-t+t \sqrt{k})^{2}}\)
where \( k \) is dielectric constant \( =4 ; t \) is thickness of dielectric \( =r / 2 \). Therefore,
\(F^{\prime}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q^{2}}{\left(r-\frac{r}{2}+\frac{r}{2} \sqrt{4}\right)^{2}}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q^{2}}{\left(r-\frac{r}{2}+r\right)^{2}} \)
\(=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q^{2}}{(2 r-r / 2)^{2}}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q^{2}}{\left(\frac{3}{2} r\right)^{2}}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q^{2}}{r^{2}} \times \frac{4}{9}\)
\(\text {Thus } F^{\prime}=\frac{4}{9} F \quad \quad(U \operatorname{sing} \text { Eq. (1)) }\)