A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ and $\varepsilon_2$, as shown in the figures. The distance between the plates is $d$ and the area of each plate is $A$. In the first configuration the two dielectrics are layers, each of thickness $d/2$, filling the whole plate area; in the second each fills half the plate area across the full gap $d$. If the capacitances in the first and second configurations are $C_1$ and $C_2$ respectively, then $\dfrac{C_1}{C_2}$ is:
Answer: (B) $\dfrac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$
Let $C_0 = \dfrac{\varepsilon_0A}{d}$.
**First configuration** (layers, in series): each layer is a capacitor of gap $d/2$, i.e. $2\varepsilon_1C_0$ and $2\varepsilon_2C_0$.
$$C_1 = \frac{(2\varepsilon_1C_0)(2\varepsilon_2C_0)}{2\varepsilon_1C_0 + 2\varepsilon_2C_0} = \frac{2\varepsilon_1\varepsilon_2C_0}{\varepsilon_1 + \varepsilon_2}$$
**Second configuration** (side by side, in parallel): each part has area $A/2$ and gap $d$.
$$C_2 = \frac{\varepsilon_1C_0}{2} + \frac{\varepsilon_2C_0}{2} = \frac{(\varepsilon_1 + \varepsilon_2)C_0}{2}$$
$$\frac{C_1}{C_2} = \frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$$
Solution by Sreeraj P, M.Sc Physics