NVAEasyJEE Main 2023 · 31 January, Shift 1Gauss's Law Applications

Physics Question from JEE Main 2023 · 31 January, Shift 1

Expression for an electric field is given by E=4000x2i^V/m\vec{E} = 4000x^2 \hat{i} \, \text{V/m}. The electric flux through the cube of side 20cm20 \, \text{cm} when placed in the electric field (as shown in the figure) is ..... Vcm\text{Vcm}.

A cube of side 20 cm is shown with x, y, z axes through one corner labeled (0,0,0). The rightward edge is along x, vertical edge along y, and slanted base edge along z.

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