Question

The below figure shows a closed Gaussian surface in the shape of a cube of edge length $3.0m$. There exists an electric field given by $\vec{E}=[(2.0x+4.0)\hat{i}+8.0\hat{j}+3.0\hat{k}]N/C$, where $x$ is in metres, in the region in which it lies. The net charge (in coulombs) enclosed by the cube is equal to $\alpha {\epsilon }_0$. Find $\alpha$.

Answer 1

Answer: 54

$\varphi =\int \vec{E}\cdot d\vec{s}$
Direction of field at $x=-3m$ is along negative $x$-axis.
Area vector is also along the same direction.

$\varphi =6\times 9=\frac{Q}{{\epsilon }_0}$
$Q=54{\epsilon }_0$
Components of electric field which are constant do not contribute in net flux in or out.
$\frac{q_{\text{in }}}{{\epsilon }_0}=54$
$\Rightarrow q_{\text{in }}=54{\epsilon }_0$