1. Initial charge on $C_1$

Since capacitor $C_1 = 5\,\mu F$ is charged to $30\,V$,

$Q_{1i} = C_1 V = 5\,\mu F \times 30\,V = 150\,\mu C$

Initially, capacitor $C_2 = 10\,\mu F$ is uncharged, so

$Q_{2i} = 0$

Thus, total initial charge in the isolated system is

$Q_{\text{total}} = 150\,\mu C$

2. After connecting the capacitors

The battery is removed, so the system is isolated and total charge is conserved.

When connected, both capacitors come to the same final potential $V_f$.

So,

$Q_{1f} = C_1 V_f, \qquad Q_{2f} = C_2 V_f$

and

$Q_{1f} + Q_{2f} = Q_{\text{total}}$

Substitute values:

$5V_f + 10V_f = 150$

$15V_f = 150$

$V_f = 10\,V$

3. Charge on $C_2$ at equilibrium

$Q_{2f} = C_2 V_f = 10\,\mu F \times 10\,V = 100\,\mu C$

4. Final answer

The charge on capacitor $C_2$ is

$\boxed{100\,\mu C}$