A capacitor `C_(1)` is charged to a p.d.V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor `C_(2)`. The final p.d. across the combination is

Explanation

Intially when capacitor 1 is connected to the battery, then charge it acquires is , From Eq.25-1
`q_0=C_1V_0=(3.55 times 10^-6 F)(6.30V)`
`=22.365 times 10^-6 C`
When switch S in Fig.25-11 is closed and capacitor 1 begins to charge capacitor 2, the electric potential and charge on capacitor 1 decrease and those on capacitor 2 increase until
`V_1=V_2` (equilibrium)
From Eq.25-1 we can rewrite this as
`q_1/C_1=q_2/C_2` (equilibrium)
Because the total charge cannot magically change, the total after the transfer must be
`q_1+q_2=q_0` (charge conservation)
Thus `q_2=q_0-q_1`
We can now rewrite thus second equilibrium equation as
`q_1/C_1=(q_0-q_1)/C_2`
Solving this for `q_1` and substituting given data, we find
`q_1=6.35 mu C`
The rest of the initial charge `(q_0=22.365 mu C)` must be on capacitor 2.
`q_2=16.0 mu C`