A cube is placed inside an electric field, $\overrightarrow{\mathrm{E}}=150 \mathrm{y}^{2} \hat{\mathrm{j}}$. The side of the cube is $0.5 \mathrm{~m}$ and is placed in the field as shown in the given figure. The charge inside the cube is:
Correct Option: , 2
As electric field is in y-direction so electric flux is
only due to top and bottom surface
Bottom surface $y=0$
$\Rightarrow \mathrm{E}=0 \Rightarrow \phi=0$
Now flux $\phi=\mathrm{EA}=\frac{150}{4}(.5)^{2}=\frac{150}{16}$
By Gauss's law $\phi=\frac{Q_{\text {in }}}{\epsilon_{0}}$
$\frac{150}{16}=\frac{\mathrm{Q}_{\text {in }}}{\epsilon_{0}}$
$\mathrm{Q}_{\mathrm{in}}=\frac{150}{16} \times 8.85 \times 10^{-12}=8.3 \times 10^{-11} \mathrm{C}$
Option (2)
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