Physics - Capacitor Question with Solution | TestHub

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 $50\text{ V}$ battery.

Let $C_s$ be the equivalent capacitance of the series branch containing $C_1,C_2,C_3$, and $C_4$.
$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}$
$\frac{1}{C_s}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{4}{10}\mu {\text{F}}^{-1}$
$C_s=2.5\mu \text{F}$

The total equivalent capacitance $C_{eq}$ of the circuit is the sum of the parallel capacitances $C_s$ and $C_5$:
$C_{eq}=C_s+C_5=2.5\mu \text{F}+2.5\mu \text{F}=5\mu \text{F}$

The voltage across the series branch is equal to the battery voltage, $V=50\text{ V}$.
The charge on the series combination is:
$Q_s=C_s\times V=2.5\mu \text{F}\times 50\text{ V}=125\mu \text{C}$
Since $C_1,C_2,C_3$, and $C_4$ are in series, they each carry the same charge $Q_s$.
Therefore, charge on $C_1,C_2,C_3,C_4=125\mu \text{C}$.

The voltage across capacitor $C_5$ is also $50\text{ V}$.
The charge on $C_5$ is:
$Q_5=C_5\times V=2.5\mu \text{F}\times 50\text{ V}=125\mu \text{C}$

Thus, the equivalent capacitance is $5\mu \text{F}$ and the charge on all capacitors is $125\mu \text{C}$.

Answer: $5\mu F$, $125\mu C$ on all capacitors