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 done by the battery .

Explanation

When switch `S` is open:
`C_(eq)=(c)/(2)`
Charge on capacitor `Q_(i)=(CE)/(2)`
when swtich `S` is closed:
Since the second capacitor is connected between same point or `p.d`. Across it is zero, it is reduntant.
`C'_(eq)=C`
Charge on capacitors
`Q_(f)=CE`
(a) Extra charge supplied by battery, `DeltaQ=Q_(f)-Q-(f)=(CE)/(2)`
(b) work done by battery `W_(b)=DeltaQE=(CE^(2))/(2)`
(c) `U_(i)=(1)/(2)(c)/(2)E^(2)=(CE^(2))/(4),U_(f)=(1)/(2)CE^(2)`,
`DeltaU=U_(f)-U_(i)=(CE^(2))/(4)`
(d) `W_(b)=DeltaU+H`
`(CE^(2))/(2)=(CE^(2))/(4)+HimpliesH=(CE^(2))/(4)`
Alternatively for heat developed
heat developed
`H=(1)/(2)C(E-(E)/(2))^(2)+(1)/(2)C(0-(E)/(2))^(2)`
`=(CE^(2))/(8)+(CE^(2))/(8)=(CE^(2))/(4)`
(The polarity of plates of first capacitor remains same.)