The capacitance of a parallel plate capacitor is `C_(0)` when the region between the plates has air. This region is now filled with a dielectric slab of dielectric constant `K`. The capacitor is connected to a cell of emf `V` and slab is taken out. Find (a) charge flown through the cell (b) energy absorbed by the cell and (c) change in energy stored in the capacitor.

Explanation

Capacitance: `C'=KC_(0)`
Charge on capacitor: `Q'=C'V=KC_(0)V`
Capacitance: `C_(0)`
Charge on capacitor: `Q=C_(0)V`
`QltQ'`
(a) charge flown through cell `=Q'-Q=(K-1)C_(0)V`
(b) work done on cell `=DeltaQV=(Q'-Q)V=(K-1)C_(0)V^(2)`
(c) charge in stored energy
`DeltaU=U'-U=(1)/(2)C'V^(2)-(1)/(2)C_(0)V^(2)`
`=(1)/(2)(K-1)C_(0)V^(2)`