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Physics · 2025

JEE Main · 7 April 2025, Shift 2 · Q50

The electric field in a region is given by E =(2 i +4 j +6 k ) × 10^3 N / C. The flux of the field through a rectangular surface parallel to x-z…

The electric field in a region is given by $\displaystyle \overrightarrow{\mathrm{E}}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^3 \mathrm{~N} / \mathrm{C}$. The flux of the field through a rectangular surface parallel to $\displaystyle x-z$ plane is $\displaystyle 6.0 \mathrm{Nm}^2 \mathrm{C}^{-1}$. The area of the surface is $\displaystyle \_\_\_\_$ $\displaystyle \mathrm{cm}^2$.
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.