A capacitor of capacitance C_1 = 10 mu F is c | vedclass

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(A) Given:$C_1 = 10 \mu F = 10^{-5} 	ext{ F}$$V_1 = 9 	ext{ V}$Initial charge on capacitor $C_1$:$q_1 = C_1 V_1 = 10^{-5} 	imes 9 = 9 	imes 10^{-5

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$6.0 	imes 10^{-5} 	ext{ C}$

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(A) Given:$C_1 = 10 \mu F = 10^{-5} 	ext{ F}$$V_1 = 9 	ext{ V}$Initial charge on capacitor $C_1$:$q_1 = C_1 V_1 = 10^{-5} 	imes 9 = 9 	imes 10^{-5} 	ext{ C}$When the uncharged capacitor $C_2 = 20 \mu F = 2 	imes 10^{-5} 	ext{ F}$ is connected in parallel to the charged capacitor $C_1$,charge flows from $C_1$ to $C_2$ until both capacitors reach a common potential $V$.The common potential $V$ is given by:$V = \frac{	ext{Total Charge}}{	ext{Total Capacitance}} = \frac{q_1}{C_1 + C_2}$$V = \frac{9 	imes 10^{-5}}{10^{-5} + 2 	imes 10^{-5}} = \frac{9 	imes 10^{-5}}{3 	imes 10^{-5}} = 3 	ext{ V}$The charge on capacitor $C_2$ at equilibrium is:$q_2 = C_2 V = (2 	imes 10^{-5} 	ext{ F}) 	imes (3 	ext{ V}) = 6 	imes 10^{-5} 	ext{ C}$

$A$ capacitor of capacitance $C_1 = 10 \mu F$ is charged using a $9 \text{ V}$ battery. It is then removed from the battery and connected to another uncharged capacitor $C_2 = 20 \mu F$ as shown in the figure. The charge on $C_2$ after equilibrium is reached is:

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