A parallel plate capacitor (plate length `L`, plate width `b`, separation between plates `d)` is connected to a battery of emf `V`. A dielectric slab of dielectric constant `K` and before capacitor. Determine the force by which dielectric is attrached by capacitor.

Explanation

Here capacitor is connected to battery and hence it is attracted by constant force, as shown in the following solution.
Let at some instant, `x` length of dielectric is inside capacitor.
`C_(1)=(Kxbin_(0))/(d), C_(2)=((L-x)bin_(0))/(d)`
`C_(1)` and `C_(2)` are in parallel
`C_(eq)=C_(1)+C_(2)=(bin_(0))/(d)[kx+(L-x)]`
`=(b in_(0))/(d)[(K-1)x+L]`
Energy stored in capacitor
`U=(1)/(2)C_(eq)V^(2)=(1)/(2)(bin_(0))/(d)[(K-1)x+L]V^(2)`
`F=-(dU)/(dx)=-((K-1)bin_(0)V^(2))/(2d)`
`-ve` sign shows that force `F` is attractive.
OR
Since force is constant, it can be determined as follows. when capacitor is empty i.e., no dielectric slab in it,
Energy stored `U_(1)=(1)/(2)CV^(2), C=(Lbin_(0))/(d)`
when dielectric slab is between plates,
`U_(2)=(1)/(2)KC V^(2)`
work done `W=U_(2)-U_(1)=(1)/(2)(K-1)CV^(2)` ...`(i)`
Since force is constant,
Work done by force =`FL` ...`(ii)`
`FL=(1)/(2)(K-1)CV^(2)=(1)/(2)(K-1)(Lbin_(0))/(d)V^(2)`
`F=((K-1)bin_(0)V^(2))/(2d)`