Q 12-01-058JEE MainJEE Main 2024 (29 Jan, Shift 2)Easy
An electric field is given by $(6\hat i + 5\hat j + 3\hat k)\ \text{N C}^{-1}$. The electric flux through a surface area $30\hat i\ \text{m}^2$ lying in the YZ-plane (in SI unit) is:
Answer: (C) 180
$$\Phi = \vec E\cdot\vec A = (6\hat i + 5\hat j + 3\hat k)\cdot30\hat i = 180\ \text{N m}^2\,\text{C}^{-1}$$
Solution by Sreeraj P, M.Sc Physics