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 $\overrightarrow{E}=\left[\right(2.0x+4.0)\hat{i}+8.0\overrightarrow{j}+3.0\hat{k}]N/C$, 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

Flux in x direction:
$E_x=2x+4,S=(3{)}^2=9m^2$
At $x=0,$
$E_x=4N{C}^{-1}$
$\Rightarrow {\varphi }_{x=0}=4\left(9\right)=36N{m}^2{C}^{-1}$

At $x=-3,$
$E_x=-2N{C}^{-1}$
$\Rightarrow {\varphi }_{x=-3}=-2(-9)=+18N{m}^2{C}^{-1}$
${\varphi }_x=36+18=54N{m}^2{C}^{-1}$

Flux in y direction:
$E_y=8\hat{j}$
At y = 0,
${\varphi }_{y=0}=8\hat{j}(-9)\hat{j}=-72N{m}^2{C}^{-1}$
At y=3,
${\varphi }_{y=3}=8\hat{j}\left(9\right)\hat{j}=72N{m}^2{C}^{-1}$
${\varphi }_y=-72+72=0$


Flux in z direction:
$E_z=3\hat{k}$
At z = 0,
${\varphi }_{z=0}=8\hat{k}(-9)\hat{k}=-27N{m}^2{C}^{-1}$
At z=3,
${\varphi }_{z=3}=8\hat{k}\left(9\right)\hat{k}=27N{m}^2{C}^{-1}$
${\varphi }_z=27-27=0$

Net flux $\left(\varphi \right)={\varphi }_x+{\varphi }_y+{\varphi }_z=54N{m}^2{C}^{-1}$

We know that
$\varphi =\frac{q}{{\epsilon }_0}=54$
$q=54{\epsilon }_0$
Hence, value of $\alpha$ is 54