When switch `S` is thrown to the left in figure, the plates of capacitor 1 acquire a potential difference `V_0`. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges `q_1, q_2` and `q_3` on the capacitors?

Explanation

`q_("total")=C_1V_0`
After switch is thrown towards right `C_23` and `C_1` are in parallel. The common potential is
`V=("Total charge")/("Total capacity")=(C_1V_0)/(C_1+((C_2C_3)/(C_2+C_3)))`
this is the same result as given in the answer.
`q_(C_2)+q_(C_3)=C_23 V`
`((C_2C_3)/(C_2+C_3))[(C_1V_0)/(C_1+(C_2C_3)/(C_2+C_3))]`