In the circuit shown in the figure, if the po | vedclass

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(A) The equivalent capacitance $C_{eq}$ of the system is calculated as follows: The capacitors $C_2$ and $C_3$ are in parallel, and this combination i

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$2.4 	imes 10^{-3} \, C ; 1.2 	imes 10^{-3} \, C$

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(A) The equivalent capacitance $C_{eq}$ of the system is calculated as follows: The capacitors $C_2$ and $C_3$ are in parallel, and this combination is in series with $C_1$.$C_{23} = C_2 + C_3 = 4 \, \mu F + 2 \, \mu F = 6 \, \mu F$.Now, $C_1$ and $C_{23}$ are in series:$C_{eq} = \frac{C_1 	imes C_{23}}{C_1 + C_{23}} = \frac{3 \, \mu F 	imes 6 \, \mu F}{3 \, \mu F + 6 \, \mu F} = \frac{18}{9} \, \mu F = 2 \, \mu F$.The total charge $q$ drawn from the source is:$q = C_{eq} 	imes V = 2 \, \mu F 	imes 1800 \, V = 3600 \, \mu C = 3.6 	imes 10^{-3} \, C$.This charge $q$ flows through $C_1$. The potential drop across $C_1$ is:$V_1 = \frac{q}{C_1} = \frac{3600 \, \mu C}{3 \, \mu F} = 1200 \, V$.The potential drop across the parallel combination of $C_2$ and $C_3$ is:$V_{23} = V_{total} - V_1 = 1800 \, V - 1200 \, V = 600 \, V$.Since $C_2$ and $C_3$ are in parallel, the potential drop across each is $600 \, V$.The charge on $C_2$ is $q_2 = C_2 	imes V_{23} = 4 \, \mu F 	imes 600 \, V = 2400 \, \mu C = 2.4 	imes 10^{-3} \, C$.The charge on $C_3$ is $q_3 = C_3 	imes V_{23} = 2 \, \mu F 	imes 600 \, V = 1200 \, \mu C = 1.2 	imes 10^{-3} \, C$.

In the circuit shown in the figure, if the point $R$ is earthed and point $P$ is given a potential of $+1800 \, V$, then the charges on $C_2$ and $C_3$ are respectively:

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