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(A) The circuit consists of a $4 \mu F$ capacitor in series with a parallel combination of $3 \mu F$ and $9 \mu F$ capacitors. First,calculate the equ

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$420$

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(A) The circuit consists of a $4 \mu F$ capacitor in series with a parallel combination of $3 \mu F$ and $9 \mu F$ capacitors. First,calculate the equivalent capacitance of the parallel part: $C_p = 3 \mu F + 9 \mu F = 12 \mu F$. Now,the circuit branch has a $4 \mu F$ capacitor in series with $C_p = 12 \mu F$ across an $8 \ V$ source. The total equivalent capacitance of this branch is $C_{eq} = \frac{4 	imes 12}{4 + 12} = \frac{48}{16} = 3 \mu F$. The total charge flowing through this branch is $q = C_{eq} 	imes V = 3 \mu F 	imes 8 \ V = 24 \mu C$. This charge $q$ passes through the $4 \mu F$ capacitor. The voltage across the parallel combination $C_p$ is $V_p = V - V_{4\mu F} = 8 \ V - \frac{24 \mu C}{4 \mu F} = 8 \ V - 6 \ V = 2 \ V$. The charge on the $9 \mu F$ capacitor is $q_{9\mu F} = C_{9\mu F} 	imes V_p = 9 \mu F 	imes 2 \ V = 18 \mu C$. The total charge $Q$ is the sum of charges on the $4 \mu F$ and $9 \mu F$ capacitors: $Q = 24 \mu C + 18 \mu C = 42 \mu C = 42 	imes 10^{-6} \ C$. The electric field at a distance $r = 30 \ m$ is $E = \frac{kQ}{r^2} = \frac{9 	imes 10^9 	imes 42 	imes 10^{-6}}{30^2} = \frac{9 	imes 10^9 	imes 42 	imes 10^{-6}}{900} = 420 \ N/C$.

$A$ combination of capacitors is set up as shown in the figure. The magnitude of the electric field,due to a point charge $Q$ (having a charge equal to the sum of the charges on the $4 \mu F$ and $9 \mu F$ capacitors),at a point distance $30 \ m$ from it,would equal ....... $N/C$.

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