Question

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 $15pF$?
(given ${ϵ}_0=8.86\times {10}^{-12}{C}^2/N{m}^2)$.

• A. $3.8$
• B. $4.5$
• C. $6.2$
• D. $8.5$

Answer 1

The correct option is D $8.5$


$A={10}^{-4}{m}^2$
$E_{max}={10}^{6}V/m$
$C=15\mu F$
$C=\frac{k{ϵ}_{0}A}{d}$
$⟹\frac{Cd}{{ϵ}_{0}A}=k$

$d=\frac{V}{E_{max}}=\frac{500}{{10}^6}=500\times {10}^{-6}m$
$\therefore k=\frac{15\times {10}^{-12}\times 500\times {10}^{-6}}{8.86\times {10}^{-12}\times {10}^{-4}}$
$=\frac{15\times 5}{8.86}=8.465$

$k\approx 8.5$.