Question

A parallel plate capacitor contains a mica sheet (thickness $={10}^{-3}m$) and a sheet of fibre (thickness $=0.5\times {10}^{-3}m$). The dielectric constant of mica is 8 and that of fibre is 2.5. Assuming that the fibre breaks down when subjected to an electric field of $6.4\times {10}^{6}V/m$, find the maximum safe voltage that can be applied to the capacitor.

• A. $2.08kV$
• B. $2.08V$
• C. $9.6kV$
• D. $9.6V$

Answer 1

The correct option is A $2.08kV$

Given,
$\begin{array}{cc}{d}_m& ={10}^{-3}m;d_f=0.5\times {10}^{-3}m\\ k_m& =8;k_f=2.5\\ & E_0=6.4\times {10}^{6}V{m}^{-1}\end{array}$

$\begin{array}{cc}{V}_{net}& =V_1+V_2\\ V_{net}& =E_1\times d_m+E_2\times d_f\\ & =\frac{E_0}{K_m}\times d_m+\frac{E_0}{K_f}\times d_f\end{array}$
$V_{net}=\frac{6.4\times {10}^6}{8}\times {10}^{-3}+\frac{6.4\times {10}^6}{2\cdot 5}\times 0.5\times {10}^{-3}$

On solving you get
$V_{net}=2.08\times {10}^{3}V$
$V_{net}=2.08kV$