A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness left(frac13right)^mathrmrd$\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

### Related Formula C = fracepsilon_0 Ad - t + fractK$$C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$ Alternatively, treat as two capacitors in series: C_eq = fracC_1 C_2C_1 + C_2$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$ ### Core Logic Initial capacitance (vacuum): C = fracA epsilon_0d$C = \frac{A \epsilon_{0}}{d}$.

The system can be modelled as two capacitors in series: 1. A vacuum capacitor of thickness d - fracd3 = frac2d3$d - \frac{d}{3} = \frac{2d}{3}$ 2. A dielectric capacitor of thickness fracd3$\frac{d}{3}$ and dielectric constant K$K$.

### Step 1: Capacitors in Series The capacitances are: C_1 = fracepsilon_0 Aleft(frac2d3right) = frac32 left(fracepsilon_0 Adright) = frac32 C$$C_1 = \frac{\epsilon_0 A}{\left(\frac{2d}{3}\right)} = \frac{3}{2} \left(\frac{\epsilon_0 A}{d}\right) = \frac{3}{2} C$$ C_2 = fracK epsilon_0 Aleft(fracd3right) = 3K left(fracepsilon_0 Adright) = 3KC$$C_2 = \frac{K \epsilon_0 A}{\left(\frac{d}{3}\right)} = 3K \left(\frac{\epsilon_0 A}{d}\right) = 3KC$$ Now, equivalent capacitance in series: C_texteq = fracC_1 C_2C_1 + C_2 = fracleft(frac32 Cright) times (3KC)frac32 C + 3KC$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\left(\frac{3}{2} C\right) \times (3KC)}{\frac{3}{2} C + 3KC}$$ C_texteq = fracfrac92KC^2frac32C(1 + 2K)$$C_{\text{eq}} = \frac{\frac{9}{2}KC^2}{\frac{3}{2}C(1 + 2K)}$$ C_texteq = frac3KC2K + 1$$C_{\text{eq}} = \frac{3KC}{2K + 1}$$ ### Pattern Recognition Partial dielectric filling of thickness t$t$: Use formula C_textnew = fracepsilon_0 Ad - t + t/K$C_{\text{new}} = \frac{\epsilon_0 A}{d - t + t/K}$. Plugging t = d/3$t = d/3$ directly gives fracepsilon_0 Ad - d/3 + d/3K = fracepsilon_0 Afrac2d3 + fracd3K = frac3K epsilon_0 A2Kd + d = frac3KC2K+1$\frac{\epsilon_0 A}{d - d/3 + d/3K} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{3K}} = \frac{3K \epsilon_0 A}{2Kd + d} = \frac{3KC}{2K+1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics