Jee Main
Electrostatics
JEE Main 2023 (Online) 31st January Morning Shift
INTEGER+4 / -12023
Expression for an electric field is given by \(\overrightarrow{\mathrm{E}}=4000 x^{2} \hat{i} \frac{\mathrm{V}}{\mathrm{m}}\). The electric flux through the cube of side \(20 \mathrm{~cm}\) when placed in electric field (as shown in the figure)
is __________ \(\mathrm{V} \mathrm{~cm}\).
