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(B) The capacitor can be modeled as two capacitors in series: one with a dielectric $(C_{1})$ and one with air $(C_{2})$.Given: $A = 100\, m^{2}$,$d =

Vedclass · Accepted Answer

$161$

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(B) The capacitor can be modeled as two capacitors in series: one with a dielectric $(C_{1})$ and one with air $(C_{2})$.Given: $A = 100\, m^{2}$,$d = 10\, m$,$t = 5\, m$,$K = 10$,$\varepsilon_{0} = 8.85 	imes 10^{-12} F \cdot m^{-1}$.The thickness of the dielectric is $t = 5\, m$ and the thickness of the air gap is $d - t = 10 - 5 = 5\, m$.$C_{1} = \frac{K \varepsilon_{0} A}{t} = \frac{10 	imes \varepsilon_{0} 	imes 100}{5} = 200 \varepsilon_{0}$.$C_{2} = \frac{\varepsilon_{0} A}{d - t} = \frac{\varepsilon_{0} 	imes 100}{5} = 20 \varepsilon_{0}$.Since they are in series,the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{(200 \varepsilon_{0})(20 \varepsilon_{0})}{200 \varepsilon_{0} + 20 \varepsilon_{0}} = \frac{4000 \varepsilon_{0}^{2}}{220 \varepsilon_{0}} = \frac{400}{22} \varepsilon_{0} = \frac{200}{11} \varepsilon_{0}$.Substituting $\varepsilon_{0} = 8.85 	imes 10^{-12} F \cdot m^{-1}$:$C_{eq} = \frac{200}{11} 	imes 8.85 	imes 10^{-12} F = 18.1818 	imes 8.85 	imes 10^{-12} F \approx 160.909 	imes 10^{-12} F$.Since $1\, pF = 10^{-12} F$,$C_{eq} \approx 160.909\, pF$.Rounding to the nearest integer,$x = 161$.

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