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  • A cubical volume is bounded by the surfaces <img alt="\mathrm{x=0, x=\mathrm{a}, y=0, y=\mathrm{a}, z=0, z=\mathrm{a}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%3D0%2C%20x

A cubical volume is bounded by the surfaces \mathrm{x=0, x=\mathrm{a}, y=0, y=\mathrm{a}, z=0, z=\mathrm{a}}. The electric field in the region is given by \mathrm{\overrightarrow{\mathrm{E}}=\mathrm{E}_0 x \hat{\imath}}. Where \mathrm{\mathrm{E}_0=4 \times 10^4 \mathrm{NC}^{-1} \mathrm{~m}^{-1}}. If \mathrm{\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 \mathrm{\in_{0}=9\times 10^{-12}C^{2}/Nm^{2}} )

Option: 1

288


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\begin{aligned} & \phi=\mathrm{\overrightarrow{E}} \cdot \mathrm{\overrightarrow{A}} \\ & \mathrm{E}=\mathrm{E}_0 \mathrm{ai} \end{aligned}}

\mathrm{ \phi=\mathrm{E}_0 \mathrm{a} \cdot \mathrm{a}^2=\mathrm{E}_0 \mathrm{a}^3 }
\mathrm{\mathrm{q}_{\text {enc. }} =\phi \varepsilon_0 }
\mathrm{\mathrm{q}_{\text {enc. }} =\mathrm{E}_0 \mathrm{a}^3 \varepsilon_0 }
           \mathrm{ =4 \times 10^4 \times 8 \times 10^{-6} \times 9 \times 10^{-12} }
\mathrm{\mathrm{q}_{\text {enc. }} =288 \times 10^{-14} \mathrm{C}}
Hence the value of \mathrm{\mathrm{Q} \: is \: 288 .}

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