A cube is placed inside an electric field, ${ }^{\hat{\mathrm{E}}-=150 \mathrm{y}^2 \mathrm{j}}$. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is :
Select the correct option:
A
$\beta .8 \times 10^{-11} \mathrm{C}$
B
$8.3 \times 10^{-11} \mathrm{C}$
C
$3.8 \times 10^{-12} \mathrm{C}$
D
$8.3 \times 10^{-12} \mathrm{C}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
As electric field is in $y$-direction so electric flux is only due to top and bottom surface Bottom surface $y=0$
$$
\Rightarrow E=0 \Rightarrow \varphi=0
$$
Top surface $\mathrm{y}=0.5 \mathrm{~m}$
$$
\Rightarrow E=150(.5)^2=\frac{150}{4}
$$
Now flux $\phi=\mathrm{EA}=\frac{150}{4}(0.5)^2=\frac{150}{16}$
By Gauss's law
$$
\phi=\frac{Q_{i n}}{\epsilon_0}
$$
$$
\begin{aligned}
& \frac{150}{16}=\frac{Q_n}{\epsilon_0} \\
& Q_{\text {in }}=\frac{150}{16} \times 8.85 \times 10^{-12}=8.3 \times 10^{-11} \mathrm{C}
\end{aligned}
$$
Option (2)
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