Expression for an electric field is given by E = 4000x²i V/m. The electric flux through the cube of side 20 cm when placed in electric field (as shown in the figure) is __________ V cm. Numerical answer type. Work it out, then open the answer.

Answer: 640. The cube lies from x = 0 to x = 0.2 m; only the faces normal to x give flux. At x = 0: E = 0. At x = 0.2: E = 4000(0.2)² = 160 V/m. φ = 160×(0.2)² = 6.4 V m = 640 V cm.

Expression for an electric field is given by E = 4000x²i V/m. The… | JEE Main 2023 (31 Jan, Shift 1)

Q66JEE Main 2023 · 31 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

Expression for an electric field is given by $\vec{E} = 4000x^{2}\hat{i}\ \frac{\text{V}}{\text{m}}$ . The electric flux through the cube of side $20\text{ cm}$ when placed in electric field (as shown in the figure) is __________ $\text{V cm}$ .

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

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Answer: 640

The cube lies from $x = 0$ to $x = 0.2$ m; only the faces normal to $x$ give flux.
At $x = 0$ : $E = 0$ . At $x = 0.2$ : $E = 4000(0.2)^2 = 160$ V/m.
$\phi = 160\times(0.2)^2 = 6.4$ V m $= 640$ V cm.

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Electric Flux & Gauss's Law question from JEE Main 2023 (31 Jan, Shift 1)

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