As shown in the figure,two parallel plate cap | vedclass

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(C) From the figure,the total distance $d = 5 \, mm$ is composed of $a + c + b$,where $c = 1 \, mm$ is the thickness of the conducting slab between th

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$5$

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(C) From the figure,the total distance $d = 5 \, mm$ is composed of $a + c + b$,where $c = 1 \, mm$ is the thickness of the conducting slab between the two capacitors.Thus,$a + b = d - c = 5 \, mm - 1 \, mm = 4 \, mm = 4 	imes 10^{-3} \, m$.The two capacitors are in series,with effective distance $d_{eff} = a + b = 4 	imes 10^{-3} \, m$.The equivalent capacitance $C_{eq}$ is given by $C_{eq} = \frac{\varepsilon_0 A}{d_{eff}}$.Given $A = 200 \, cm^2 = 200 	imes 10^{-4} \, m^2$.$C_{eq} = \frac{\varepsilon_0 	imes 200 	imes 10^{-4}}{4 	imes 10^{-3}} = \frac{200 	imes 10^{-1}}{4} \varepsilon_0 = 5 \varepsilon_0 \, F$.Comparing this with $x \varepsilon_0 \, F$,we get $x = 5$.

As shown in the figure,two parallel plate capacitors having equal plate area of $200 \, cm^2$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \varepsilon_0 \, F$. The value of $x$ is $..........$.

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