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?

Answer

`q_(1)=(C_(1)^(2)V(C_(2)+C_(3)))/(C_(1)C_(2)+C_(2)C_(3)+C_(1)C_(3),)q_(2)=q_(3)=(C_(2)C_(3)C_(1)CV)/(C_(1)C_(2)+C_(1)C_(2)+C_(2)C_(3))`.

Explanation

When switch is left
.
Appling loop rule:
`(q_(0) - q)/(C_(1)) - (q)/(C_(3)) - (q)/(C_(2)) = 0`
or `(q_(0))/(C_(1)) -
When switch is thrown right According to loop rule,
`(q_(0)-q)/C_(1)=q(1/C_(2)+1/C_(3))=2((C_(2)+C_(3))/(C_(2)C_(3)))`
`:. q=((C_(2)C_(3)q_(0))/(C_(1)C_(2)+C_(1)C_(3)+C_(2)C_(3)))`
`:. `q_(1)=q_(0)-q=C_(1)V-((C_(2)C_(3)q_(0))/(C_(1)C_(2)+C_(1)C_(3)+C_(2)C_(3)))`
`=(C_(1)^(2)V(C_(2)+C_(3)))/(C_(1)C_(2)+C_(2)C_(3)+C_(1)C_(3))`
Also, `q_(2)=q_(3)=q=(C_(2)C_(3)q_(0))/(C_(1)C_(2)+C_(2)C_(3)+C_(1)V_(3))`
`=(C_(2)C_(3)C_(1)V)/(C_(1)C_(2)+C_(1)C_(3)+C_(2)C_(3))`.