A capacitor P with capacitance $$10 \times 10^{-6} F$$ is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance $$20 \times 10^{-6} F$$. The charge on capacitor Q when equilibrium is established will be $$\alpha \times 10^{-5}C$$ (assume capacitor Q does not have any charge initially), the value of $$\alpha$$ is _________.

We are given capacitor P with capacitance $$C_P = 10 \times 10^{-6}$$ F charged to a potential difference of $$V_P = 6.0$$ V, and capacitor Q with capacitance $$C_Q = 20 \times 10^{-6}$$ F (initially uncharged). When P is disconnected from the battery and connected across Q, the charge on Q at equilibrium can be determined as follows.

Since the initial charge on P is given by $$Q = CV$$, substituting yields $$Q_{initial} = C_P \times V_P = 10 \times 10^{-6} \times 6.0 = 60 \times 10^{-6}\text{ C} = 60 \mu\text{C}$$.

After connection, the total charge is conserved because no external path allows charge to escape, and Q starts uncharged. Therefore $$Q_{total} = Q_P + Q_Q = 60 \mu\text{C}$$.

At equilibrium both capacitors share the same potential difference $$V_{common}$$, which follows from $$V_{common} = \frac{Q_{total}}{C_P + C_Q} = \frac{60 \times 10^{-6}}{(10 + 20) \times 10^{-6}} = \frac{60}{30} = 2\text{ V}$$.

Finally, the charge on Q is $$Q_Q = C_Q \times V_{common} = 20 \times 10^{-6} \times 2 = 40 \times 10^{-6}\text{ C} = 4 \times 10^{-5}\text{ C}$$. This shows that $$\alpha = 4$$.

The answer is 4.