Question 290 of 311

A cube is placed inside an electric field, $\vec{E}=150{y}^{2}\hat{j}$ The side of the cube is $0.5m$ and is placed in the field as shown in the given figure. The charge inside the cube is:

A$8.3\times {10}^{-11}C$B$3.8\times {10}^{-11}C$C$3.8\times {10}^{-12}C$D$8.3\times {10}^{-12}C$

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Answer: (A) $8.3\times {10}^{-11}C$

By gauss law $\oint \vec{E}\cdot \vec{ds}=\dfrac{{q}_{\text{in}}}{{\epsilon }_{0}}$
${\epsilon }_{0}(150)(0.5{)}^{2}\times (0.5{)}^{2}={q}_{in}$
${q}_{in}=\dfrac{8.85\times {10}^{-12}\times 150}{16}=8.3\times {10}^{-11}C$