Q 12-02-176JEE MainJEE Main 2019 (8 Apr, Shift 1)Easy
Voltage rating of a parallel plate capacitor is $500\ \text{V}$. Its dielectric can withstand a maximum electric field of $10^6\ \text{V/m}$. The plate area is $10^{-4}\ \text{m}^2$. What is the dielectric constant if the capacitance is $15\ \text{pF}$? (given $\varepsilon_0 = 8.86\times10^{-12}\ \text{C}^2/\text{N m}^2$)
Answer: (B) $8.5$
Plate separation: $d = \dfrac{V}{E_{\max}} = \dfrac{500}{10^6} = 5\times10^{-4}\ \text{m}$.
$$K = \frac{Cd}{\varepsilon_0A} = \frac{15\times10^{-12}\times5\times10^{-4}}{8.86\times10^{-12}\times10^{-4}} \approx 8.5$$
Solution by Sreeraj P, M.Sc Physics