A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant varepsilon_1$\varepsilon_1$ and varepsilon_2$\varepsilon_2$, as shown in figures. The distance between the plates is d$d$ and area of each plate is A$A$. If capacitance in first configuration and second configuration are C_1$C_1$ and C_2$C_2$ respectively, then fracC_1C_2$\frac{C_1}{C_2}$ is:

### Related Formula Capacitance with dielectric: C = fracvarepsilon_r varepsilon_0 Ad$$C = \frac{\varepsilon_r \varepsilon_0 A}{d}$$ Series Capacitors: C_texteq = fracC_a C_bC_a + C_b$$C_{\text{eq}} = \frac{C_a C_b}{C_a + C_b}$$ Parallel Capacitors: C_texteq = C_a + C_b$$C_{\text{eq}} = C_a + C_b$$ ### Core Logic Let C_0 = fracvarepsilon_0 Ad$C_0 = \frac{\varepsilon_0 A}{d}$ be the capacitance without any dielectric. - **First Configuration (Series connection)**: The dielectrics split the gap vertically, so the effective thickness of each slab is d/2$d/2$, and the area remains A$A$. C_a = fracvarepsilon_1 varepsilon_0 Ad/2 = 2varepsilon_1 C_0$$C_a = \frac{\varepsilon_1 \varepsilon_0 A}{d/2} = 2\varepsilon_1 C_0$$ C_b = fracvarepsilon_2 varepsilon_0 Ad/2 = 2varepsilon_2 C_0$$C_b = \frac{\varepsilon_2 \varepsilon_0 A}{d/2} = 2\varepsilon_2 C_0$$ Since they are in series: C_1 = fracC_a C_bC_a + C_b = frac(2varepsilon_1 C_0)(2varepsilon_2 C_0)2varepsilon_1 C_0 + 2varepsilon_2 C_0 = frac4varepsilon_1varepsilon_2 C_0^22C_0(varepsilon_1 + varepsilon_2) = frac2varepsilon_1varepsilon_2varepsilon_1 + varepsilon_2 C_0$$C_1 = \frac{C_a C_b}{C_a + C_b} = \frac{(2\varepsilon_1 C_0)(2\varepsilon_2 C_0)}{2\varepsilon_1 C_0 + 2\varepsilon_2 C_0} = \frac{4\varepsilon_1\varepsilon_2 C_0^2}{2C_0(\varepsilon_1 + \varepsilon_2)} = \frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2} C_0$$ - **Second Configuration (Parallel connection)**: The dielectrics split the area horizontally, so the effective area of each slab is A/2$A/2$, and the distance remains d$d$. C_c = fracvarepsilon_1 varepsilon_0 (A/2)d = fracvarepsilon_1 C_02$$C_c = \frac{\varepsilon_1 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_1 C_0}{2}$$ C_d = fracvarepsilon_2 varepsilon_0 (A/2)d = fracvarepsilon_2 C_02$$C_d = \frac{\varepsilon_2 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_2 C_0}{2}$$ Since they are in parallel: C_2 = C_c + C_d = (varepsilon_1 + varepsilon_2) fracC_02$$C_2 = C_c + C_d = (\varepsilon_1 + \varepsilon_2) \frac{C_0}{2}$$

### Step 1: Calculating the Ratio Now, compute fracC_1C_2$\frac{C_1}{C_2}$: fracC_1C_2 = fracleft(frac2varepsilon_1varepsilon_2varepsilon_1 + varepsilon_2right) C_0left(fracvarepsilon_1 + varepsilon_22right) C_0 = frac4varepsilon_1varepsilon_2(varepsilon_1 + varepsilon_2)^2$$\frac{C_1}{C_2} = \frac{\left(\frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2}\right) C_0}{\left(\frac{\varepsilon_1 + \varepsilon_2}{2}\right) C_0} = \frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$$ ### Pattern Recognition For dielectric-filled capacitors: splitting the gap (d/2$d/2$) leads to a series combination, while splitting the plate area (A/2$A/2$) leads to a parallel combination. Shortcut: C_textseries = textharmonic mean$C_{\text{series}} = \text{harmonic mean}$, C_textparallel = textarithmetic mean$C_{\text{parallel}} = \text{arithmetic mean}$. The ratio fracC_1C_2$\frac{C_1}{C_2}$ is always the ratio of the harmonic mean of the dielectric constants to their arithmetic mean! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance