What charges will flow after the shorting of the swich `Sw` in the circuit illustrated in Fig through sections 1 and 2 in the directions indicated by the arrown?

Explanation

Intially, charge on the capacitor `C_(1)` or `C_(2)`
`q = (xi C_(1) C_(2))/(C_(1) + C_(2))` , as they are in seies combination (Fig).
when the swich is closed, in the circuit `CDEFC` from `-Delta varphi = 0`, Fig
`xi = - (q_(2))/(C_(2)) = 0` or `q_(2) = C_(2) xi` ....(1)
And in the closed loop `BCFAB` from `Delta varphi = 0`
`(-q_(1))/(C_(1)) + (q_(2))/(C_(2)) - xi = 0` ...(2)

From (1) and (2) `q_(1) = 0`
Now, charge flown through section `1 = (q_(1) + q_(2)) = 0 = C_(2) xi`
and charge flows throgh section `2 = q_(1) - q = (xi C_(1) C_(2))/(C_(1) + C_(2))`