A parallel plate capacitor is filled equally | vedclass

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(B) In the first configuration,the two dielectrics are in series. The capacitance of each part is $C_a = \frac{2 \varepsilon_1 \varepsilon_0 A}{d}$ an

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$\frac{4 \varepsilon_1 \varepsilon_2}{\left(\varepsilon_1+\varepsilon_2\right)^2}$

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(B) In the first configuration,the two dielectrics are in series. The capacitance of each part is $C_a = \frac{2 \varepsilon_1 \varepsilon_0 A}{d}$ and $C_b = \frac{2 \varepsilon_2 \varepsilon_0 A}{d}$.Since they are in series,the equivalent capacitance $C_1$ is given by:$C_1 = \frac{C_a C_b}{C_a + C_b} = \frac{(\frac{2 \varepsilon_1 \varepsilon_0 A}{d})(\frac{2 \varepsilon_2 \varepsilon_0 A}{d})}{\frac{2 \varepsilon_1 \varepsilon_0 A}{d} + \frac{2 \varepsilon_2 \varepsilon_0 A}{d}} = \frac{2 \varepsilon_1 \varepsilon_2 \varepsilon_0 A}{d(\varepsilon_1 + \varepsilon_2)}$.In the second configuration,the two dielectrics are in parallel. The capacitance of each part is $C_c = \frac{\varepsilon_1 \varepsilon_0 (A/2)}{d}$ and $C_d = \frac{\varepsilon_2 \varepsilon_0 (A/2)}{d}$.Since they are in parallel,the equivalent capacitance $C_2$ is given by:$C_2 = C_c + C_d = \frac{\varepsilon_0 A}{2d} (\varepsilon_1 + \varepsilon_2)$.Now,the ratio $\frac{C_1}{C_2}$ is:$\frac{C_1}{C_2} = \frac{2 \varepsilon_1 \varepsilon_2 \varepsilon_0 A}{d(\varepsilon_1 + \varepsilon_2)} 	imes \frac{2d}{\varepsilon_0 A (\varepsilon_1 + \varepsilon_2)} = \frac{4 \varepsilon_1 \varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$.

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