Five capacitors of capacitances \(C_1=C_2=C_3=C_4=10 \mu F\) and \(C_5=2.5 \mu F\) are connected as shown, along with a battery of 50 V .

(B) \(5 \mu F, 125 \mu C\) on all capacitors
From the given circuit diagram, the capacitors \(C_1, C_2, C_3\), and \(C_4\) are connected in series. This series combination is connected in parallel with the capacitor \(C_5\). The entire combination is connected across the 50 V battery.
Let \(C_s\) be the equivalent capacitance of the series branch containing \(C_1, C_2, C_3\), and \(C_4\).
\(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}\)
\(\frac{1}{C_s}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{4}{10} \mu F^{-1}\)
\(C_s=2.5 \mu F\)
The total equivalent capacitance \(C_{e q}\) of the circuit is the sum of the parallel capacitances \(C_8\) and \(C_5\) :
\(C_{e q}=C_s+C_5=2.5 \mu F+2.5 \mu F=5 \mu F\)
The voltage across the series branch is equal to the battery voltage, \(V=50 V\).
The charge on the series combination is:
\(Q_s=C_s \times V=2.5 \mu F \times 50 V=125 \mu C\)
Since \(C_1, C_2, C_3\), and \(C_4\) are in series, they each carry the same charge \(Q_s\).
Therefore, charge on \(C_1, C_2, C_3, C_4=125 \mu C\).
The voltage across capacitor \(C_5\) is also 50 V .
The charge on \(C_5\) is:
\(Q_5=C_5 \times V=2.5 \mu F \times 50 V=125 \mu C\)
Thus, the equivalent capacitance is \(5 \mu F\) and the charge on all capacitors is \(125 \mu C\).