Question

The below figure shows a closed Gaussian surface in the shape of a cube of edge length $3.0\, m$. There exists an electric field given by $\vec{E}=[(2.0 x+4.0) \hat{i}+8.0 \hat{j}+3.0 \hat{k}] N / C$, where $x$ is in metres, in the region in which it lies. The net charge (in coulombs) enclosed by the cube is equal to $\alpha \varepsilon_{0}$. Find $\alpha$.

Answer 1

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

$\phi=6 \times 9=\frac{Q}{\varepsilon_{0}} $
$Q=54 \varepsilon_{0}$
Components of electric field which are constant do not contribute in net flux in or out.
$\frac{q_{\text {in }}}{\varepsilon_{0}}=54 $
$\Rightarrow q_{\text {in }}=54 \varepsilon_{0}$