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(D) The capacitance in air is $C_1 = \frac{\varepsilon_0 S}{d}$.The capacitor can be modeled as two parts in parallel. One part has area $S/2$ filled

Vedclass · Accepted Answer

$7/3$

Vedclass · Answer

(D) The capacitance in air is $C_1 = \frac{\varepsilon_0 S}{d}$.The capacitor can be modeled as two parts in parallel. One part has area $S/2$ filled with dielectric $\varepsilon_1$,and the other part has area $S/2$ containing two dielectrics in series,each of thickness $d/2$ with permittivities $\varepsilon_1$ and $\varepsilon_2$.For the parallel branch with dielectric $\varepsilon_1$: $C_A = \frac{\varepsilon_1 \varepsilon_0 (S/2)}{d} = \frac{2 \varepsilon_0 S}{2d} = \frac{\varepsilon_0 S}{d} = C_1$.For the other branch,we have two capacitors in series,$C_B$ and $C_C$,each with area $S/2$ and thickness $d/2$:$C_B = \frac{\varepsilon_1 \varepsilon_0 (S/2)}{d/2} = \varepsilon_1 \frac{\varepsilon_0 S}{d} = 2 C_1$$C_C = \frac{\varepsilon_2 \varepsilon_0 (S/2)}{d/2} = \varepsilon_2 \frac{\varepsilon_0 S}{d} = 4 C_1$The equivalent capacitance of this series branch is $C_{series} = \frac{C_B C_C}{C_B + C_C} = \frac{(2 C_1)(4 C_1)}{2 C_1 + 4 C_1} = \frac{8 C_1^2}{6 C_1} = \frac{4}{3} C_1$.The total capacitance $C_2 = C_A + C_{series} = C_1 + \frac{4}{3} C_1 = \frac{7}{3} C_1$.Therefore,the ratio $\frac{C_2}{C_1} = \frac{7}{3}$.

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