Question

A potential difference of $300V$ is applied between the plates of a plane capacitor spaced $1cm$ apart. A plane parallel glass plate with a thickness of $0.5cm$ and a plane parallel paraffin plate with a thickness of $0.5cm$ are placed in the space between the capacitor plates find:
(i) intensity of electric field in each layer
(ii) the drop of potential in each layer
(iii) the surface charge density of the charge on the capacitor plates.
Given that: $k_{glass}=6,k_{paraffin}=2$

Answer 1

$C{V}_1=3C{V}_2$ ....................(1)

$V_1+V_2=300$ ...................(2)

$V_1=75V$

$V_2=225V$

1. $E_1=\frac{v_1}{d_1}=\frac{75\times 100}{0.5}=1.5\times {10}^4$ V/m

$E_2=\frac{v_2}{d_2}=\frac{225\times 100}{0.5}=4.5\times {10}^4$ V/m

2. $V_1=75V$

$V_2=225V$

3. $Q=\frac{C_1{C}_2}{C_1+C_2}V=\frac{3}{4}CV=\frac{3}{4}\left(\frac{2{\epsilon }_{0}A}{d}\right)\times 300$

$\frac{Q}{A}=\frac{3\times 2\times 300\times 8.85\times {10}^{-12}}{4\times 0.5\times {10}^{-2}}=8\times {10}^{-7}C/m^2$