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JEE MAIN 2025
07-04-2025 S2
Question
The electric field in a region is given by $\overrightarrow{\mathrm{E}}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^3 \mathrm{~N} / \mathrm{C}$. The flux of the field through a rectangular surface parallel to $x-z$ plane is $6.0 \mathrm{Nm}^2 \mathrm{C}^{-1}$. The area of the surface is $\_\_\_\_$ $\mathrm{cm}^2$.
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Question Tags
JEE Main
Physics
Easy
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