A parallel plate capacitor having plate area `A` and separation `4d` is connected to a battery of emf `V_(0)`. A diellectric slab of thickness `2d` and dielectric constant `2.0` is inserted into the gap. (a) find the increase in energy strored in capacitor. (b) Now battery is removed and dielectric slab taken out, find further increase in energy.

Explanation

(a) `C_(1)=(Ain_(0))/(4d),V_(1)=V_(0)`
`U_(1)=(1)/(2)C_(1)V_(1)^(2)=(1)/(2)(Ain_(0))/(4d)V_(0)^(2)=(Ain_(0)V_(0)^(2))/(8d)`
`C_(2)=(Ain_(0))/(4d-t+(t)/(K))=(Ain_(0))/(4d-2d+(2d)/(2))=(Ain_(0))/(3d),V_(2)=V_(0)`
`U_(2)=(1)/(2)C_(2)V_(2)^(2)=(1)/(2)(Ain_(0))/(3d).V_(0)^(2)=(Ain_(0)V_(0)^(2))/(6d)`
`DeltaU=U_(2)-U_(1)=((1)/(6)-(1)/(8))(Ain_(0)V_(0)^(2))/(d)=(Ain_(0)V_(0)^(2))/(24d)`
`Q=C_(2)V_(2)=(Ain_(0)V_(0))/(3d)`
(c) `C_(3)=C_(1)=(Ain_(0))/(4d)`
`U_(3)=(Q^(2))/(2C_(3))=((Ain_(0)V_(0))/(3d))^(2)(1)/(2.(Ain_(0))/(4d))=(Ain_(0)V_(0)^(2))/(9d)`
`DeltaU'=U_(3)-U_(2)=((2)/(9)-(1)/(6))(Ain_(0)V_(0)^(2))/(d)`
`=(Ain_(0)V_(0)^(2))/(18d)`