When capacitors are connected in series, the charge \( Q \) on each capacitor is the same. The equivalent capacitance \( C_{\text{eq}} \) for capacitors \( C_1 \) and \( C_2 \) connected in series is given by:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Given:
\[ C_1 = 4 \, \mu F \]
\[ C_2 = 6 \, \mu F \]
\[ V = 10 \, V \]
First, calculate the equivalent capacitance \( C_{\text{eq}} \):
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} \]
\[ \frac{1}{C_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \]
\[ C_{\text{eq}} = \frac{12}{5} \, \mu F = 2.4 \, \mu F \]
Now, using the total voltage \( V \) across the combination, we can find the charge \( Q \) on the capacitors:
\[ Q = C_{\text{eq}} \times V \]
\[ Q = 2.4 \, \mu F \times 10 \, V \]
\[ Q = 24 \, \mu C \]
Since the charge on capacitors in series is the same, the charge on \( C_2 \) is:
\[ Q_{C_2} = 24 \, \mu C \]
Therefore, the charge on \( C_2 \) is \( 24 \, \mu C \).
