JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 23)
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 __________ $$\mathrm{V} \mathrm{~cm}$$.
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Answer
640
Explanation
The flux will be only from $D E F G$ surface as on the surface $O A B C$ field is 0 and for rest of the surface, area vector is perpendicular to field.
_31st_January_Morning_Shift_en_23_2.png)
$$ \begin{aligned} & \text { So } \phi=E A \\\\ & =4000 \times(\cdot 2)^{2} \times \cdot 2 \times \cdot 2 \\\\ & =\frac{32}{5} \mathrm{Vm} \\\\ & =\frac{32}{5} \times 100 \mathrm{~V} \mathrm{~cm} \\\\ & =640 \mathrm{Vcm} \end{aligned} $$
$$ \begin{aligned} & \text { So } \phi=E A \\\\ & =4000 \times(\cdot 2)^{2} \times \cdot 2 \times \cdot 2 \\\\ & =\frac{32}{5} \mathrm{Vm} \\\\ & =\frac{32}{5} \times 100 \mathrm{~V} \mathrm{~cm} \\\\ & =640 \mathrm{Vcm} \end{aligned} $$
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