Consider the situation shown in figure (31-E23 ) . The switch S is open for a long time and then closed . (a) Find the charge flown through the battery when the switch S is closed (b)Find the work done by the battery .

Explanation

a. Since a switch is open for long time. The capacitor is in series. Hence equivalent capacity is `C_(eq)=C//2`. Initial charge on each capacitor is `q_(0)=Cepsilon//2`. When switch is closed, find charge on capacitor `C_(1)`, is `q_(1)=Cepsilon` and find charge in `C_(2)`, is `q_(2)=0`. Hence, when the switch is closed, charge flow through battery is
`Deltaq=(Cepsilon-(Cepsilon)/2)=(Cepsilon)/2`.
b. Work done by battery is `Deltaq.epsilon=(Cepsilon)/2xxepsilon=(Cepsilon^(2))/2`
c. Change in energy stored is
`1/2Cepsilon^(2)-1/2xx(C-2)xxepsilon^(2)=1/4Cepsilon^(2)`
d. Heat developwd is `W_("battery")-DeltaU=(Cepsilon^(2))/2-(Cepsilon^(2))/4=(Cepsilon^(2))/4`.