In the following figure, the capacitors have plate area `A=lxxb`, separation `'d"`. If the slab is displaced slightly, find the time period of the oscillation. Given mass of the dielectric `=m`. (Given `(md)/(epsilon_(0)b(k-1)epsilon^(2))=(1)/(2)`.)

Explanation

When the slab is displaced by a distance `x`, the work done is `=` change in stored energy `+` work done by battery
Initial capacitance of the capacitors
`C=(epsilon_(0)lb)/(2d)+(epsilon_(0)kbl)/(2d)`
Final capacitances
`C_(1f)=(epsilon_(0)b)/(d)(l//2+x)+(epsilon_(0)bk)/(d)(l//2-x)`
`C_(2f)=(epsilon_(0)b)/(d)(l//2-x)+(epsilon_(0)bk)/(d)(l//2+x)`
Change in energy stored `= -(1)/(2)epsilon^(2)[C_(1f)-C]+(1)/(2)epsilon^(2)[C_(2f)-C]`
Work done by battery `=epsilondq`
`=epsilonxxepsilon[(C_(1f)-C)-(C_(2f)-C)]`
`:. ` Work done `=(epsilon^(2)epsilon_(0)b)/(d)x(k-1)`
Net acceleration `a=(F)/(m)=(epsilon_(0)b)/(md)(k-1)epsilon^(2)`
Motion is oscillatory but not SHM.
and, `x=(1)/(2)at^(2)`
`:. t=sqrt((2x)/(a))`
`:.` Time period `=4t`
`=sqrt((32xmd)/(epsilon_(0)b(k-1)epsilon^(2)))`