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Electrostatics question

2023 · 31 Jan · Shift 1 · Q70
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Electrostatics question

2023 · 31 Jan · Shift 1 · Q70

JEE MainPhysicsElectrostaticsNumerical+4 / −1
Expression for an electric field is given by E→=4000x2i^Vm\overrightarrow{\mathrm{E}}=4000 x^{2} \hat{i} \frac{\mathrm{V}}{\mathrm{m}}E=4000x2i^mV​. The electric flux through the cube of side 20 cm20 \mathrm{~cm}20 cm when placed in electric field (as shown in the figure) is ‾\underline{\hspace{2cm}}​V cm\mathrm{V} \mathrm{~cm}V cm. JEE Main 2023 (Online) 31st January Morning Shift Physics - Electrostatics Question 99 English
Numerical answer
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