To solve this problem, we need to calculate the total charge on the capacitors \(C_2\) and \(C_3\) after they are connected to the charged capacitor \(C_1\). Here are the steps to solve the problem:

Step 1: Calculate the initial charge on \(C_1\).

The charge \(Q_1\) on the capacitor \(C_1\) can be determined using the formula:

\(Q = CV\)

Where:

• \(C = 1.0 \, \mu \text{F}\)
• \(V = 60 \, \text{V}\)

Thus,

\(Q_1 = 1.0 \times 60 = 60 \, \mu \text{C}\)

Step 2: Determine the equivalent capacitance when \(C_1\) is connected to \(C_2\) and \(C_3\).

The capacitors \(C_2\) and \(C_3\) are connected in series, so the equivalent capacitance \(C_{\text{eq}}\) is given by:

\(\frac{1}{C_{\text{eq}}} = \frac{1}{C_2} + \frac{1}{C_3}\)

Substituting the given values:

\(\frac{1}{C_{\text{eq}}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\)

Thus, \(C_{\text{eq}} = 2 \, \mu \text{F}\)

Step 3: Calculate the final charge distribution.

When \(C_1\) is connected to the equivalent capacitance \(C_{\text{eq}}\), the total charge is conserved. The initial charge on \(C_1\) is now distributed over \(C_1\) and the series combination \(C_2\) and \(C_3\).

The new combined capacitance becomes:

\(C_{\text{total}} = C_1 + C_{\text{eq}} = 1 + 2 = 3 \, \mu \text{F}\)

The voltage across the capacitors is:

\(V_{\text{final}} = \frac{Q_1}{C_{\text{total}}} = \frac{60}{3} = 20 \, \text{V}\)

Hence, the charge on the series combination \((Q_2 = Q_3)\) is:

\(Q_2 = C_{\text{eq}} \times V_{\text{final}} = 2 \times 20 = 40 \, \mu \text{C}\)

Conclusion: The sum of the final charges on \(C_2\) and \(C_3\) is \(40 \, \mu \text{C}\).