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 is the capacitance is $15pF$ ?
(given ${\in }_0=8.86\times 10^{-12}{C}^2/N{m}^2$)

• A. 3.8
• B. 4.5
• C. 6.2
• D. 8.5

Answer 1

Answer: (D)

$A=10^{-4}{m}^2$
$E_{max}=10^{6}V/m$
$C=15\mu F$
$C=\frac{k{\epsilon }_{0}A}{d}$
$\frac{Cd}{{\epsilon }_{0}A}=k$
$k=\frac{15\times 10^{-12}\times 500\times 10^{-6}}{8.86\times 10^{-12}\times 10^4}$
$=\frac{15\times 5}{8.86}=8.465$
$k\approx 8.5$