In an oscillating circuit consisting of a parallel- plate capacitor and an inductance coil with negligible active resistance the oscillations with energy `W` are sustained. The capacitor plates were slowly drawn aparto to increase the oscillation frequency `eta-` fold. What work was doen in the process ?

Explanation

In the oscillating circuit, let
`q=q_(m) cos omegat `
be the change on the condenser where
`omega^(2)=(1)/(LC)` and `C` is the intantaneous capacity of the condenser `(S=` area of plates `)`
`C=(epsilon_(0)S)/( y)`
`y=` distance between the plates. Since the oscillation frequency increases `eta` fold, the quancity
`omega^(2)=(y)/( epsilon_(0) SL)`
change `eta^(2)` fold and so does `y` i.e. changes from `y_(0)` initially to `eta^(2) y_(0)` finally. Now the `P.D.` across the condenser is
`V=(q_(m))/(C) cos omegat=(yq_(m))/( epsilon_(0)S) cos omegat`
and hence the electric field between the plates is
`E=(q_(m))/( epsilon_(0)S) cos omegat`
Thus, the charge on the plate being `q_(m)cos omegat` the force on the plate is
`E=(q_(m)^(2))/( epsilon_(0)S) cos^(2) omegat`
Since this force is always positive and the plate is pulled slowly we can use the average force
`bar(F)=(q_(m)^(2))/(2 epsilon_(0)S)`
and work done is `A=bar(F)( eta^(2)y_(0)-y_(0))=( eta^(2)-1)( q_(m)^(2)y_(0))/( 2 epsilon_(0)S)`
But `(q_(m)^(2)y_(0))/(2 epsilon_(0)S)=(q_(m)^(2))/(2 C_(0))=W` the initial stored energy. Thus.
`A=( eta^(2)-1) W`.