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  4. Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106 V/m. The plate area is 10-4 m2. What is the dielectric constant is the capacitance is 15 pF ? (given ∈0 = 8.86 × 10-12 C2 / Nm2)

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Q. 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 $\in_0 = 8.86 \times 10^{-12} C^2 / Nm^2$)

JEE MainJEE Main 2019Electrostatic Potential and Capacitance

Solution:

$A = 10^{-4} m^{2}$
$ 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 $