An electric circuit shown in figure has a negligibly small active resistance. The left`-` hand capacitor was charged to a voltage `V_(0)` and then at the moment `t=0` the switch `Sw` was closed. Find the time dependence of the voltages in left and right capacitors.

Explanation

Initially `q_(1)=CV_(0)` and `q_(2)=0`. After the switch is closed change flows and we get
`q_(1)+q_(2)=CV_(0)`
`(q_(1))/(C)+L(dI)/( dt)-(q_(2))/( C)=0` `……..(1)`
Also `I=dot(q_(1))=-dot(q_(2)). `Thus
`Lddot(I)+ (2I)/( C) =0`
Hence `ddot(I)+omega_(0)^(2)I=0` `omega_(0)^(2)=(2)/(LC),`
The solution of this equation subject to
`I=0` at `t=0`
is `I=I_(0) sin omega_(0)t`
Integrating `q_(1)=A-(I_(0))/( omega_(0))cos omega_(0)t`
`q_(2)=B+(I_(0))/( omega_(0))cos omega_(0)t`
Finally substituting in `(1)`
`(A-B)/( C)-(2I_(0))/( omega_(0) C) cos omega_(0)t+ LI_(0) omega_(0) cos omega_(0) t=0`
Thus `A=B=(CV_(0))/( 2)` and
`(CV_(0))/(2)+(I_(0))/(omega_(0))=0`
so `q_(1)=(CV_(0))/(2)(1+ cos omega_(0)t)`
`q_(2)=(CV_(0))/(2)(1-cos omega_(0)t)`