Three plates A, B and C each of area 0.1 m^2 | vedclass

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(B) The system consists of three plates $A, B,$ and $C$. Based on the circuit diagram, plates $A$ and $C$ are connected together to one terminal of th

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$10^{-1} \ \mu J$

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(B) The system consists of three plates $A, B,$ and $C$. Based on the circuit diagram, plates $A$ and $C$ are connected together to one terminal of the battery, and plate $B$ is connected to the other terminal. This configuration forms two capacitors in parallel, each with plates separated by distance $d = 0.885 \ mm = 0.885 	imes 10^{-3} \ m$.The capacitance of each capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Substituting the values: $C = \frac{8.854 	imes 10^{-12} 	imes 0.1}{0.885 	imes 10^{-3}} = 10^{-9} \ F = 1 \ nF$.Since the two capacitors are in parallel, the equivalent capacitance is $C_{eq} = C + C = 2 \ nF = 2 	imes 10^{-9} \ F$.The energy stored in the system is $U = \frac{1}{2} C_{eq} V^2$.Substituting $C_{eq} = 2 	imes 10^{-9} \ F$ and $V = 10 \ V$:$U = \frac{1}{2} 	imes (2 	imes 10^{-9}) 	imes (10)^2 = 10^{-9} 	imes 100 = 10^{-7} \ J$.Converting to microjoules $(\mu J)$:$U = 10^{-7} \ J = 10^{-1} 	imes 10^{-6} \ J = 10^{-1} \ \mu J$.

Three plates $A, B$ and $C$ each of area $0.1 \ m^2$ are separated by $0.885 \ mm$ from each other as shown in the figure. A $10 \ V$ battery is used to charge the system. The energy stored in the system is

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