Q 12-02-189JEE MainJEE Main 2019 (12 Jan, Shift 2)Easy
A parallel plate capacitor with plates of area $1\ \text{m}^2$ each, are at a separation of $0.1$ m. If the electric field between the plates is $100$ N/C, the magnitude of charge on each plate is: (Take $\varepsilon_0 = 8.85\times10^{-12}\ \dfrac{\text{C}^2}{\text{N m}^2}$)
Answer: (A) $8.85\times10^{-10}$ C
$E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0A}$, so
$$Q = \varepsilon_0EA = 8.85\times10^{-12}\times100\times1 = 8.85\times10^{-10}\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics