Two dielectrics of equal size are inserted inside a parallel plate capacitor as shown. With what factor the effective capacitance increases ?

$\frac{k_1+k_2}{2}$

Here two condensers are formed, which are in parallel .

Area of each = A/2 and thickness = d

Total capacitance c = $\frac{\varepsilon_0 k(A / 2)}{d}+\frac{\varepsilon_0 k_2(A / 2)}{d}=\frac{\varepsilon_0 A}{d} \times\left(\frac{k_1+k_2}{d}\right)$

∴ Total capacitance increases by the factor $\frac{k_1+k_2}{2}$