The electric field in a region is given by E = (3/5E₀i + 4/5E₀j)N/C… | JEE Main 2021 (25 Feb, Shift 1)

Q80JEE Main 2021 · 25 FEB · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field in a region is given by $\vec{E} = \left(\frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}\right)\frac{\text{N}}{\text{C}}$ . The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y $-$ z plane) to that of the surface of area 0.3 m2 (parallel to x $-$ z plane) is a: b, where a = __________ [Here $\hat{i}\ \hat{j}$ and $\hat{k}$ are unit vectors along x, y and z-axes respectively.]

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

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Answer: 1

Y-Z surface: $\phi_1 = E_x A_1 = \dfrac{3}{5}E_0(0.2) = 0.12E_0$ .
X-Z surface: $\phi_2 = E_y A_2 = \dfrac{4}{5}E_0(0.3) = 0.24E_0$ .
$\phi_1 : \phi_2 = 1 : 2$ , so $a = 1$ .

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