A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness $\left(\frac{1}{4}\right)d$ and placed between plates of parallel plate capacitor, where d is the separation of the plates. The ratio of capacitances of capacitor with dielectric and with air as medium $\left(\frac{c}{c_0}\right)$ is
Answer & explanation
Correct answer: option 2
The correct answer is option (2) : $\frac{4K}{3K+1}$
When air was medium between plates of capacitor,
$C_0=\frac{Aε_0}{d}$
When dielectric was filled
$V=\frac{E_0}{K}\times \frac{d}{4}+E_0\times \frac{3d}{4}$ and $E_0=\frac{Q}{Aε_0}$
$⇒V=\frac{Qd}{4Aε_0}\left[\frac{1}{K}+3\right]$
$⇒V=\frac{Qd}{4Aε_0}\left[\frac{1+3K}{K}\right]$ and $ Q=V\times \frac{4Aε_0K}{d(1+3K)}$ also $Q=CV$
$⇒C=\frac{4Aε_0K}{d(1+3K)}$ and $C_0=\frac{Aε_0}{d}$
$⇒\frac{C}{C_0}=\frac{4K}{1+3K}$