JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 28)
A cubical volume is bounded by the surfaces $$\mathrm{x}=0, x=\mathrm{a}, y=0, y=\mathrm{a}, \mathrm{z}=0, z=\mathrm{a}$$. The electric field in the region is given by $$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} x \hat{i}$$. Where $$\mathrm{E}_{0}=4 \times 10^{4} ~\mathrm{NC}^{-1} \mathrm{~m}^{-1}$$. If $$\mathrm{a}=2 \mathrm{~cm}$$, the charge contained in the cubical volume is $$\mathrm{Q} \times 10^{-14} \mathrm{C}$$. The value of $$\mathrm{Q}$$ is ________________.
(Take $$\epsilon_{0}=9 \times 10^{-12} ~\mathrm{C}^{2} / \mathrm{Nm}^{2}$$)
Answer
288
Explanation
_1st_February_Evening_Shift_en_28_1.png)
$\begin{aligned} & \overrightarrow E = {E_0}x\widehat i \\\\ & \phi_{\mathrm{net}}=\phi_{\mathrm{ABCD}}=\mathrm{E}_0 \mathrm{a} \cdot \mathrm{a}^2 \\\\ & \frac{\mathrm{q}_{\mathrm{en}}}{\in_0}=\mathrm{E}_0 \mathrm{a}^3 \\\\ & \mathrm{q}_{\mathrm{en}}=\mathrm{E}_0 \in_0 \mathrm{a}^3 \\\\ & =4 \times 10^4 \times 9 \times 10^{-12} \times 8 \times 10^{-6} \\\\ & =288 \times 10^{-14} \mathrm{C} \\\\ & \therefore \mathrm{Q}=288\end{aligned}$
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