Q. The region between the parallel plates of a capacitor is filled with parallel layers of air and paper (of dielectric constant $4$). The space between the plates is $1 \,mm$ and the thickness of paper is $0.75\, mm$. The ratio of the voltages across air and paper is
Solution:
Here, $d = 1 \, mm, t = 0.75\, mm$
The given capacitor is equivalent to a two capacitors are connected in series. As in series charge on each capacitor is same,
$ \therefore \, Q_A = Q_P$
where the subscripts A and P refer air and paper respectively,
Thus,
$C_{A} V_{A} = C_{P} V_{P} $
$ \frac{V_{A}}{V_{P}} = \frac{C_{P}}{C_{A}} = \frac{\frac{e_{0}KA}{t}}{\frac{e_{0}A}{\left(d-t\right)}} = \frac{K\left(d-t\right)}{t} $
$ = \frac{4\left(0.25 mm\right)}{0.75 mm} = \frac{4}{3}$
