The electric field in a region is given by E = (2i + 4j + 6k) × 10³ N/C. The flux of the field through a rectangular surface parallel to x - z plane is 6.0 N m²C⁻¹. The area of the surface is _____________ cm². Numerical answer type. Work it out, then open the answer.

Answer: 15. Surface parallel to the x-z plane has area vector along j, so φ = Ey A. A = (6.0)/(4×10³) = 1.5×10⁻³ m² A = 1.5×10⁻³×10⁴ = 15 cm²

The electric field in a region is given by E = (2i + 4j + 6k) × 10³… | JEE Main 2025 (7 Apr, Shift 2)

Q16JEE Main 2025 · 7 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

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

Numerical answer type. Work it out, then open the answer.

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Answer: 15

Surface parallel to the x-z plane has area vector along $\hat j$ , so $\phi = E_y A$ .
$A = \dfrac{6.0}{4\times10^3} = 1.5\times10^{-3}\text{ m}^2$
$A = 1.5\times10^{-3}\times10^4 = 15\text{ cm}^2$

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