Question

A parallel plate capacitor contains a mica sheet ( thickness ${10}^{-3}m$) and a sheet of fiber (thickness $0.5\times {10}^{-3}m$). The dielectric constant of mica is $8$ and that of the fiber is $2.5$. Assuming that the fiber breaks down when subjected to an electric field of $6.4\times {10}^{6}V{m}^{-1}$, find the maximum safe voltage (in kV) that can be applied to the capacitor.

Answer 1

There are two capacitors and they are connected in series.
If $V_1$ be the potential across mica capacitor and $V_2$ for fiber capacitor, then
$V=V_1+V_2=E_1{d}_1+E_2{d}_2$
$=\frac{E_0{d}_1}{K_1}+\frac{E_0{d}_1}{K_1}$ $(E\propto \frac{1}{K})$
$=\frac{6.4\times {10}^6\times {10}^{-3}}{8}+\frac{6.4\times {10}^6\times 0.5\times {10}^{-3}}{2.5}=2.08\times {10}^{3}V\sim 2kV$