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QJEEM23Ph10ea: JEE Main 2023 Physics - Class 12 JEEM23Ph10ea Solution: A cubical volume is bounded by the surfaces . The electric…

A cubical volume is bounded by the surfaces x=0,x=a,y=0,y=a,z=0,z=a\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 E→=E0xi^\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} x \hat{i} . Where E0=4×104 NC−1 m−1\mathrm{E}_{0}=4 \times 10^{4} ~\mathrm{NC}^{-1} \mathrm{~m}^{-1} . If a=2 cm\mathrm{a}=2 \mathrm{~cm} , the charge contained in the cubical volume is Q×10−14C\mathrm{Q} \times 10^{-14} \mathrm{C} . The value of Q\mathrm{Q} is ________________.

(Take ϵ0=9×10−12 C2/Nm2\epsilon_{0}=9 \times 10^{-12} ~\mathrm{C}^{2} / \mathrm{Nm}^{2} )

Step-by-Step Verified Solution

JEE Main 2023 (Online) 1st February Evening Shift Physics - Electrostatics Question 85 English Explanation

amp;E→=E0xi^amp;ϕnet=ϕABCD=E0a⋅a2amp;qen∈0=E0a3amp;qen=E0∈0a3amp;=4×104×9×10−12×8×10−6amp;=288×10−14Camp;∴Q=288\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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