Voltage rating of a parallel plate capacitor is $500V$. Its dielectric can withstand a maximum electric field of  ${10}^{6}V/m$. The plate area is ${10}^{-4}{m}^2$. What is the dielectric constant, if the capacitance is $15pF$? (Take, ${\epsilon }_0=8.86\times {10}^{-12}{C}^2/N-m^2$)

As we know, capacitance of a capacitor filled with dielectric medium
$C=\frac{{\epsilon }_{o}KA}{d}$
And potential difference between plates is
$E=\frac{V}{d}\Rightarrow d=\frac{V}{E}$
So by combining both Eqs(i) and (ii) we get
$K=\frac{CV}{{\epsilon }_{0}AE}$ …(iii)
Given,  $C=15pF=15\times {10}^{-12}F$
$V=500\mathrm{V},E={10}^6{\mathrm{Vm}}^{-1}$
$A={10}^{-4}{m}^2$ and  ${\epsilon }_0=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$
Substituting values in Eq.(iii), we get
$K=\frac{15\times {10}^{-12}\times 500}{8.85\times {10}^{-12}\times {10}^{-4}\times {10}^6}=8.47=8.5$