The electric field in a region is given by vec E =(2 hat i +4 hat j +6 hat k ) × 10 3 N / C . The flux of the field through a rectangular surface parallel to x-z plane is 6.0 Nm 2 C -1 . The area of the surface is cm 2 .

Physics · Electrostatics

JEE Main 2025 — 7 April, Evening Shift — Question 64

The electric field in a region is given by $\vec{E}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^{3} \mathrm{~N} / \mathrm{C}$ .

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 $\qquad$ $\mathrm{cm}^{2}$ .

Numerical answer — enter this value.

$\vec{E}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^{3} \mathrm{~N} / \mathrm{C}$

$\phi=\vec{E} \cdot \vec{S}$ $=[(2 \hat{i}+4 \hat{j}+6 k)-(a \hat{j})] \times 10^{3}$

$6=A 4 \times 10^{3}$ $\begin{aligned} A & =\frac{6}{4 \times 10^{3}} \mathrm{~ms} & =15 \mathrm{~cm} \end{aligned}$

Answer key and solution verified before publishing.

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