A parallel plate of capacitance `C_(0)` is connected to a power supply `V_(0)`. A dielectric slab of dielectric constant `K=5` is now inserted into the gap between the plates. (a) find the extra charge flown through the power supply and work done by the supply. (b) find change in energy stored in capacitor and heat produced in connecting wires.

Explanation

(a) Here, potential of capacitor remains same
`Q_(1) = C_(0)V_(0),Q_(2)=KC_(0)V_(0)`
Extra charged supplied by battery
`Delta Q=Q_(2)-Q_(1)=(K-1)C_(0)V_(0)=4C_(0)V_(0)`
Work done by supply
`W_(b)=Delta QV_(0)=4C_(0)V_(0)^(2)`
(b) `U_(i)=(1)/(2)C_(0)V_(0)^(2),U_(f)=(1)/(2)KC_(0)V_(0)^(2)`
`DeltaU=U_(f)-U_(i)=(1)/(2)(K-1)C_(0)V_(0)^(2)=2C_(0)V_(0)^(2)`
`W_(b)=Delta U+H`,
Where H : heat produced in connecting wires,
`H=W_(b)-Delta U=4C_(0)V_(0)^(2)-2C_(0)V_(0)^(2)=2C_(0)V_(0)^(2)`