JEE Main 2025 Apr 7 Shift 2, Physics Q24: Gauss's Law

JEE Main2025Apr 7, Shift 2Physics

The electric field in a region is given by $E = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 1 0^{3}$ 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 $^{2}$ .

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Solution

A rectangular surface parallel to the x-z plane has its area vector along $\hat{j}$ (the normal direction). Only the $\hat{j}$ -component of $E$ contributes to the flux through it.

$Φ = E \cdot A = (4 \times 1 0^{3}) (A) = 6.0$ Nm $^{2}$ C $^{- 1}$ .

$A = \frac{6.0}{4 \times 1 0 ^{3}} = 1.5 \times 1 0^{- 3}$ m $^{2} = 15$ cm $^{2}$ .

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JEE Main 2025 Apr 7 Shift 2, Physics Q24: Gauss's Law

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