Three plates A, B, C each have an area of 50 | vedclass

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(B) The system consists of two capacitors connected in parallel.Each capacitor has capacitance $C = \frac{\varepsilon_0 A}{d}$.Since there are two suc

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$2.1 	imes 10^{-9} \, J$

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(B) The system consists of two capacitors connected in parallel.Each capacitor has capacitance $C = \frac{\varepsilon_0 A}{d}$.Since there are two such capacitors in parallel,the total capacitance is $C_{eq} = 2 	imes \frac{\varepsilon_0 A}{d} = \frac{2 \varepsilon_0 A}{d}$.The energy stored in the system is given by $U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \left( \frac{2 \varepsilon_0 A}{d} ight) V^2 = \frac{\varepsilon_0 A V^2}{d}$.Given: $\varepsilon_0 = 8.854 	imes 10^{-12} \, F/m$,$A = 50 	imes 10^{-4} \, m^2$,$d = 3 	imes 10^{-3} \, m$,$V = 12 \, V$.$U = \frac{8.854 	imes 10^{-12} 	imes 50 	imes 10^{-4} 	imes (12)^2}{3 	imes 10^{-3}}$.$U = \frac{8.854 	imes 50 	imes 144 	imes 10^{-16}}{3 	imes 10^{-3}} = \frac{63748.8 	imes 10^{-16}}{3 	imes 10^{-3}} \approx 2.125 	imes 10^{-9} \, J$.Thus,the energy stored is approximately $2.1 	imes 10^{-9} \, J$.

Three plates $A, B, C$ each have an area of $50 \, cm^2$ and the distance between them is $3 \, mm$. Find the energy stored in the system when the plates are fully charged.

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