Voltage rating of a parallel plate capacitor is $500 \mathrm{~V}$. Its dielectric can withstand a maximum electric field of $10^{6}$ $\mathrm{V} / \mathrm{m}$. The plate area is $10^{-4} \mathrm{~m}^{2}$. What is the dielectric constant if the capacitance is $15 \mathrm{pF}$ ?
(given $\epsilon_{0}=8.86 \times 10^{-12} \mathrm{C}^{2} \mathrm{~m}^{2}$ )
Correct Option: , 2
(2) Capacitance of a capacitor with a dielectric of dielectric constant $\mathrm{k}$ is given by
$C=\frac{k \in_{0} A}{d}$
$\because E=\frac{V}{d}$ $\therefore C=\frac{k \in_{0} A E}{V}$
$15 \times 10^{-12}=\frac{k \times 8.86 \times 10^{-12} \times 10^{-4} \times 10^{6}}{500}$
$k=8.5$
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