Electrostatic Potential and Capacitance Part 1

Given, area of dielectrics with dielectric constant $K_1$ and $K_2$, $A_1 = A_2 = \frac{A}{2}$ Thickness of dielectrics with dielectric constant $K_1$ and $K_2$, $d_1 = d_2 = \frac{d}{2}$ Other half of the capacitor is still filled with air, so distance betweenplates in other half of capacitor will be d but the area of plates for airsection will be $\frac{A}{2}$. Calculating the capacitance of section of capacitor filled with air. $C_a = \frac{\varepsilon_0 \left(\frac{A}{2}\right)}{d} = \frac{\varepsilon_0 A}{2d}$ Now, the capacitance of section of capacitor filled with dielectricconstant $K_1$ can be calculated as $C_1 = \frac{K_1 \varepsilon_0 A_1}{d_1} = \frac{K_1 \varepsilon_0 \left(\frac{A}{2}\right)}{\frac{d}{2}} = \frac{K_1 \varepsilon_0 A}{d}$ Similarly, the capacitance of section of capacitor filled withdielectric constant $K_2$ can be calculated as $C_2 = \frac{K_2 \varepsilon_0 A_2}{d_2} = \frac{K_2 \varepsilon_0 \left(\frac{A}{2}\right)}{\frac{d}{2}} = \frac{K_2 \varepsilon_0 A}{d}$ Now, the dielectric sections are in series with each other, so equivalent capacitance of combination can be calculated as $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{K_1 \varepsilon_0 A} + \frac{d}{K_2 \varepsilon_0 A}$ $\Rightarrow C_{eq} = \frac{\varepsilon_0 A}{d} \left[\frac{K_1 K_2}{K_1 + K_2}\right]$ Now, $C_{eq}$ is in parallel with the section of capacitor filled with air. Total capacitance of combination can be calculated as $C_T = C_a + C_{eq} = \frac{\varepsilon_0 A}{2d} + \frac{\varepsilon_0 A}{d} \left[\frac{K_1 K_2}{K_1 + K_2}\right]$ $\Rightarrow C_T = \frac{\varepsilon_0 A}{d} \left[\frac{1}{2} + \frac{K_1 K_2}{K_1 + K_2}\right]$