Question

Voltage rating of a parallel plate capacitor is $500\,V$. 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 $15\, pF$ ? (given $\in_0 = 8.86 \times 10^{-12} C^2 / Nm^2$)

• A. 3.8
• B. 4.5
• C. 6.2
• D. 8.5

Answer 1

Answer: (D)

8.5

Answer 2

$A = 10^{-4} m^{2}$
$E_{max } = 10^{6} V/m$
$C = 15 \mu F$
$C = \frac{k\varepsilon_{0}A}{d}$
$\frac{Cd}{\varepsilon_{0}A} = k$
$k = \frac{15 \times10^{-12} \times500 \times10^{-6}}{8.86 \times10^{-12} \times10^{4}}$
$= \frac{15 \times5}{ 8.86 } = 8.465$
$k \approx8.5$