The space between the plates of a parallel pl | vedclass

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(C) The capacitor is divided into two capacitors in series,each with plate separation $d' = d/2 = 0.885 \ mm = 0.885 	imes 10^{-3} \ m$ and area $A =

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) The capacitor is divided into two capacitors in series,each with plate separation $d' = d/2 = 0.885 \ mm = 0.885 	imes 10^{-3} \ m$ and area $A = 4 \ cm^2 = 4 	imes 10^{-4} \ m^2$.The capacitance of the first part is $C_1 = \frac{k_1 \varepsilon_0 A}{d'} = \frac{5 	imes 8.85 	imes 10^{-12} 	imes 4 	imes 10^{-4}}{0.885 	imes 10^{-3}} = 20 \ pF$.The capacitance of the second part is $C_2 = \frac{k_2 \varepsilon_0 A}{d'} = \frac{3 	imes 8.85 	imes 10^{-12} 	imes 4 	imes 10^{-4}}{0.885 	imes 10^{-3}} = 12 \ pF$.Since these are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \implies C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{20 	imes 12}{20 + 12} = \frac{240}{32} = 7.5 \ pF$.This capacitor is connected in parallel with another capacitor of $7.5 \ pF$. The total effective capacitance is $C_{total} = C_{eq} + 7.5 \ pF = 7.5 + 7.5 = 15 \ pF$.

$List-I$	$List-II$
$(P)$ The capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected, is	$(1)$ $3 C_0$
$(Q)$ The capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$, is	$(2)$ $C_0 / 2$
$(R)$ The capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$, is	$(3)$ $C_0 / 3$
$(S)$ The capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$, is	$(4)$ $2 C_0 / 3$
	$(5)$ $2 C_0$

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