JEE PYQ: Capacitance Question 25
Question 25 - 2019 (08 Apr Shift 1)
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 if the capacitance is 15 pF?
(given $\epsilon_0 = 8.86 \times 10^{-12}$ C$^2$ m$^2$)
(1) 3.8
(2) 8.5
(3) 4.5
(4) 6.2
Show Answer
Answer: (2)
Solution
$C = \frac{k\epsilon_0 A}{d}$. $E = \frac{V}{d}$, so $d = \frac{500}{10^6} = 5 \times 10^{-4}$ m. $C = \frac{k\epsilon_0 AE}{V}$. $15 \times 10^{-12} = \frac{k \times 8.86 \times 10^{-12} \times 10^{-4} \times 10^6}{500}$. $k = 8.5$.
BETA
