The capacitors are connected in parallel and series combinations.
Step 1: Identify the Equivalent Capacitance
The 4 \(\mu\)F and 8 \(\mu\)F capacitors are in series. The equivalent capacitance is:
\[
\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}
\]
\[
C_{\text{eq}} = \frac{8}{3} \mu F
\]
This C\(_eq\) is now in parallel with the 4 \(\mu\)F capacitor:
\[
C_{\text{parallel}} = 4 + \frac{8}{3} = \frac{12}{3} + \frac{8}{3} = \frac{20}{3} \mu F
\]
Step 2: Find Charge Stored
The total charge stored in the system is:
\[
Q = C_{\text{total}} V = \frac{20}{3} \times 63
\]
\[
Q = 420 \mu C
\]
Since the 5 \(\mu\)F capacitor is in series with the parallel combination, it gets the same charge:
\[
V_{\text{5\(\mu\)F}} = \frac{Q}{C} = \frac{420}{5} = 48V
\]
Thus, the potential difference across the 5 \(\mu\)F capacitor is 48 V.
