Question

The plates of parallel plate capacitor are charged up to 100 V. A 2 mm thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the plate is

Answer 1

Let $V_1$ be the potential difference across capacitor in the absence of dielectric medium.

$V_1=E_{1}d$

($E_1$ = electric field in vaccum between plates, d = distance between plates)

$=\frac{\sigma }{{\epsilon }_{0}d}$

Let $V_2$ be the potential difference in the presence of dielectric medium.

$V_2=E_1(d-t)+E_{2}t=\frac{\sigma }{{\epsilon }_0}(d-t)+\frac{\sigma }{{\epsilon }_{0}k}t$

($E_2$ = electric filed in dielectric)

Let $V_3$ be the potential difference with dielectric medium and increased distance between plates by d'


$V_3=E_1\left[\right(d+d^{\prime })-t]+E_{2}t$

$=\frac{\sigma }{{\epsilon }_0}\left(\right(d+d^{\prime })-t+\frac{t}{k})$

According to the given condition,
$V_1=V_3$

$\frac{\sigma }{{\epsilon }_0}d=\frac{\sigma }{{\epsilon }_0}\left(\right(d+d^{\prime })-t+\frac{t}{k})$

$k=\frac{t}{t-d^{\prime }}$

$=\frac{2mm}{2mm-1.6mm}$

= 5

hence, the dielectric constant of slab is 5