Step 1: Identify capacitor values and connections.
From the figure, the capacitances are:
\[
C_1 = 2C, \quad C_2 = 3C, \quad C_3 = C, \quad C_4 = 4C
\]
The circuit is a bridge-type network connected across a battery of voltage \( V \). Step 2: Note symmetry and equivalent combinations.
The left and right vertical branches form parallel paths between the same two nodes. The top and middle capacitors are connected between these branches. Step 3: Apply charge–voltage relation.
For any capacitor,
\[
Q = CV
\]
Hence, the ratio of charges depends on the effective potential difference across each capacitor. Step 4: Use node potential method.
Let the potential difference across the network be \( V \). On solving the node equations for this capacitor bridge (using charge conservation at junctions), the effective potential drops across \( C_2 \) and \( C_4 \) are obtained in the ratio:
\[
V_2 : V_4 = \frac{1}{11}
\] Step 5: Calculate ratio of charges.
\[
\frac{Q_2}{Q_4} = \frac{C_2 V_2}{C_4 V_4} = \frac{3C \times 1}{4C \times 11} = \frac{3}{22}
\] Step 6: Conclusion.
The ratio of charges on capacitors \( C_2 \) and \( C_4 \) is \( \dfrac{3}{22} \).
