Question

A parallel plate capacitor with plates of width $4$ cm, length $10$ cm and separation between plates is $4$ cm, is connected across a battery of emf $12$ Volt. A dielectric slab of dielectric constant $K=7$ is slowly introduced between the plates. Force exerted on dielectric slab by the field is

• A. $114{\in }_0$
• B. $432{\in }_0$
• C. $7{\in }_0$
• D. $326{\in }_0$

Answer 1

The correct option is B $432{\in }_0$

Let, dielectric moves through a distance $dx$ inside the plates. This increases capacitance to $C+dC$.
As potential difference remains constant at V, the battery supply more charge to capacitor.
and
$dQ=dC.V$
So,work done by battery $d{W}_b=VdQ=dC.V^2$
and work done by force
$d{W}_F=-Fdx$ (F = force on slab)
$dC.V^2-Fdx=\frac{1}{2}\left(dC\right)V^2$
$Fdx=\frac{1}{2}\left(dC\right)V^2$
We know that,
$C=C_1+C_2=\frac{{\in }_0(l-x)w}{d}+\frac{K{\in }_{0}xw}{d}$
$dC=\frac{{\in }_{0}w}{d}(K-1)dx$
$\Rightarrow Fdx=\frac{1}{2}\frac{{\in }_{0}w}{d}(K-1)dx{V}^2$
$\Rightarrow F=\frac{1}{2}\frac{{\in }_{0}w(K-1)}{d}{V}^2$

$\Rightarrow F=\frac{{\in }_0\times 4\times {10}^{-2}\times (12{)}^2\times (7-1)}{2\times 4\times {10}^{-2}}$
$\Rightarrow F=432{\in }_0$