question_answer

Voltage rating of a parallel plate capacitor is 500V. 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 \[{{\in }_{0}}=8.86\times {{10}^{-12}}{{C}^{2}}/N{{m}^{2}}\]) [JEE Main 8-4-2019 Morning]

A) 3.8

B) 4.5

C) 6.2

D) 8.5

Correct Answer: D

Solution :

[d]

\[A={{10}^{-4}}{{m}^{2}}\]

\[{{E}_{\max }}={{10}^{6}}V/m\]

\[C=15\mu F\]

\[C=\frac{k{{\varepsilon }_{0}}A}{d}\]

\[\frac{Cd}{{{\varepsilon }_{0}}A}=k\]

\[k=\frac{15\times {{10}^{-12}}\times 500\times {{10}^{-6}}}{8.86\times {{10}^{-12}}\times {{10}^{4}}}\]

\[=\frac{15\times 5}{8.86}=8.465\]

\[k\approx 8.5\]