An electric field is given by (6i^+5j^+3k^) N C^(-1). The electric flux through a surface area 30i^ m^2 lying in YZ-plane(in SI unit) is:

Correct answer: (C) 180. Given: E→=(6i^+5j^+3k^) N C^(-1) A→=30i^ m^2 ϕ=E→·A→ ⇒ϕ=(6i^+5j^+3k^)·(30i^)=180 SI unit

Q75 / 90 · JEE Main 2024 (29 Jan Shift 2) · Physics

An electric field is given by $(6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N C^{- 1}$ . The electric flux through a surface area $30 \hat{i} m^{2}$ lying in YZ-plane(in SI unit) is:

Given:

$\vec{E} = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) N C^{- 1}$

$\vec{A} = 30 \hat{i} m^{2}$

$ϕ = \vec{E} \cdot \vec{A}$

$\Rightarrow ϕ = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i}) = 180 SI unit$

Hence, the correct answer is option: 'C'.