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Electrostatics question

2023 · 1 Feb · Shift 2 · Q74
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Electrostatics question

2023 · 1 Feb · Shift 2 · Q74

JEE MainPhysicsElectrostaticsNumerical+4 / −1
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}x=0,x=a,y=0,y=a,z=0,z=a. The electric field in the region is given by E→=E0xi^\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} x \hat{i}E=E0​xi^. Where E0=4×104 NC−1 m−1\mathrm{E}_{0}=4 \times 10^{4} ~\mathrm{NC}^{-1} \mathrm{~m}^{-1}E0​=4×104 NC−1 m−1. If a=2 cm\mathrm{a}=2 \mathrm{~cm}a=2 cm, the charge contained in the cubical volume is Q×10−14C\mathrm{Q} \times 10^{-14} \mathrm{C}Q×10−14C. The value of Q\mathrm{Q}Q is ‾\underline{\hspace{2cm}}​. (Take ϵ0=9×10−12 C2/Nm2\epsilon_{0}=9 \times 10^{-12} ~\mathrm{C}^{2} / \mathrm{Nm}^{2}ϵ0​=9×10−12 C2/Nm2)
Numerical answer
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