JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2025
- Easy
- Numerical
The electric field in a region is given by \(\overrightarrow{\mathrm{E}}=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times 10^3 \mathrm{~N} / \mathrm{C}\). 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 ________ \(\mathrm{cm}^2\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 15
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times 10^3 \cdot \mathrm{~A} \hat{\mathrm{j}} \\ & 6=4 \times 10^3 \mathrm{~A} \\ & \mathrm{~A}=1.5 \times 10^{-3} \mathrm{~m}^2 \\ & =15 \mathrm{~cm}^2\end{aligned}\)
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