A parallel plate capacitor with air medium between the plates has a capacitance of \(10 \mu \mathrm{F}\). The area of capacitor is divided into two equal halves and filled with two media (as shown in figure) having dielectric constant \(K_1=2\) and \(\mathrm{K}_2=4\). The capacitance of the system will be

\(\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=10 \mu \mathrm{F}\)
After dividing the area into two equal halves, the resultant capacitance is calculated as,
\(\mathrm{C}_{\text {eq }} =\mathrm{C}_1+\mathrm{C}_2 \)
\( =\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}+\frac{\mathrm{K}_1 \varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}=\frac{\frac{2 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}}+\frac{\frac{4 \varepsilon_0 \mathrm{~A}}{2}}{\mathrm{~d}} \)
\( \therefore \mathrm{C}_{\text {eq }} =\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}+\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{3 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=3 \times 10\) \(=30 \mu \mathrm{F}\)