To solve this problem, we need to understand how the electric field $E$ varies with the insertion of dielectric slabs.$\\$Given:$\\$• Two dielectric slabs with dielectric constants $K_1$ and $K_2$ are inserted between the plates.$\\$• $K_1 < K_2 \\$The electric field $E$ in a dielectric medium is given by:$\\ E = \frac{E_0}{K} \\$where $E_0$ is the electric field without the dielectric.$\\$Since $K_1 < K_2,$ the electric field in the region with $K_1$ will be greater than in the region with $K_2. \\$Let's analyze the situation:$\\$• In the region with dielectric constant $K_1,$ the electric field is $E_1 = \frac{E_0}{K_1}. \\$• In the region with dielectric constant $K_2,$ the electric field is $E_2 = \frac{E_0}{K_2}. \\$Since $K_1 < K_2,$ it follows that $E_1 > E_2. \\$The correct graph should show:$\\$• A higher electric field in the region with $K_1. \\$• A lower electric field in the region with $K_2. \\$Option 3 correctly represents this variation:$\\$• The electric field is higher in the first region ($K_1$) and lower in the second region ($K_2$).$\\$Therefore, the correct option is Option 3.