चित्र में विध्यूत क्षेत्र अवयव $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0$ है, जिसमें $\alpha=800 \,N / C m ^{1 / 2}$ है। $(a)$ घन से गुजरने वाला फ्लक्स, तथा $(b)$ घन के भीतर आवेश परिकलित कीजिए। $a=0.1 \,m$ मानिए

$(a)$ since the electric field has only an $x$ component, for faces perpendicular to $x$ direction, the angle between $E$ and $\Delta S$ is $\pm \pi / 2 .$ Therefore, the flux $\phi= E . \Delta S$ is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric fleld at the left face $1 s$ $E_{L}=\alpha x^{1 / 2}=\alpha a^{1 / 2}$

$(x=a \text { at the left face })$ The magnitude of electric fleld at the right face is $E_{R}=\alpha x^{1 / 2}=\alpha(2 a)^{1 / 2}$

$(x=2 a$ at the right face). The corresponding fluxes are

$\phi_{L}= E _{L} \cdot \Delta S =\Delta S E _{L} \cdot \hat{ n }_{L}=E_{L} \Delta S \cos \theta$$=-E_{L} \Delta S,$ since $\theta=180^{\circ}$

$=-E_{L} a^{2}$

$\phi_{R}= E _{R} \cdot \Delta S =E_{R} \Delta S \cos \theta=E_{R} \Delta S,$ since $\theta=0^{\circ}$

$=E_{R} a^{2}$

Net flux through the cube 

$=\phi_{R}+\phi_{L}=E_{R} a^{2}-E_{L} a^{2}=a^{2}\left(E_{R}-E_{L}\right)$$=\alpha a^{2}\left[(2 a)^{1 / 2}-a^{1 / 2}\right]$

$=\alpha a^{5 / 2}(\sqrt{2}-1)$

$=800(0.1)^{5 / 2}(\sqrt{2}-1)$

$=1.05 \,N\, m ^{2} \,C ^{-1}$

$(b)$ We can use Gauss's law to find the total charge $q$ inside the cube.

We have $\phi=q / \varepsilon_{0}$ or $q=\phi \varepsilon_{0} .$

Therefore $q=1.05 \times 8.854 \times 10^{-12}\, C =9.27 \times 10^{-12} \,C$