A parallel plate capacitor consists of two metal plates of area `A` and separation `d`. A slab of thickness `t` and electric constant `K` is inserted between the plates with its faces parallel to the plates and having the same surface area as that of the plates. Find the capacitance of the system.
IF `K=2`, for what value of `t//d` will the capacitance of the system be `3//2` times that of the air capacitor? Calculate the energy in the two cases and account for the energy change.

Answer

`C=(in_(0)A)/(d-t+(t)/(k)),(t)/(d)=(2)/(3),(U_(i))/(U_(F))=(3)/(2)`,
`Delta U=-(q^(2)d)/(6in_(0)A)`