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QJEEM23Ph3a29: JEE Main 2023 Physics - Class 12 JEEM23Ph3a29 Solution: Expression for an electric field is given by . The electric…

Expression for an electric field is given by E→=4000x2i^Vm\overrightarrow{\mathrm{E}}=4000 x^{2} \hat{i} \frac{\mathrm{V}}{\mathrm{m}} . The electric flux through the cube of side 20 cm20 \mathrm{~cm} when placed in electric field (as shown in the figure) is __________ V cm\mathrm{V} \mathrm{~cm} .

JEE Main 2023 (Online) 31st January Morning Shift Physics - Electrostatics Question 80 English

Step-by-Step Verified Solution

The flux will be only from DEFGD E F G surface as on the surface OABCO A B C field is 0 and for rest of the surface, area vector is perpendicular to field.

JEE Main 2023 (Online) 31st January Morning Shift Physics - Electrostatics Question 80 English Explanation

amp; So ϕ=EAamp;=4000×(⋅2)2×⋅2×⋅2amp;=325Vmamp;=325×100 V cmamp;=640Vcm\begin{aligned}& \text { So } \phi=E A \\\\& =4000 \times(\cdot 2)^{2} \times \cdot 2 \times \cdot 2 \\\\& =\frac{32}{5} \mathrm{Vm} \\\\& =\frac{32}{5} \times 100 \mathrm{~V} \mathrm{~cm} \\\\& =640 \mathrm{Vcm}\end{aligned}

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