A cubical volume is bounded by the surfaces $$x = 0, x = a, y = 0, y = a, z = 0, z = a$$. The electric field in the region is given by $$\vec{E} = E_0 x\hat{i}$$. Where $$E_0 = 4 \times 10^4 \text{ NC}^{-1} \text{ m}^{-1}$$. If $$a = 2$$ cm, the charge contained in the cubical volume is $$Q \times 10^{-14}$$ C. The value of $$Q$$ is ______.
(Take $$\epsilon_0 = 9 \times 10^{-12} \text{ C}^2 \text{ N}^{-1}\text{m}^{-2}$$)

We consider a cube bounded by $$x=0$$ to $$x=a$$, $$y=0$$ to $$y=a$$, and $$z=0$$ to $$z=a$$ in an electric field $$\vec{E}=E_0 x \hat{i}$$, where $$E_0 = 4 \times 10^4$$ NC$$^{-1}$$m$$^{-1}$$, $$a = 2$$ cm $$= 0.02$$ m, and $$\epsilon_0 = 9 \times 10^{-12}$$ C$$^2$$N$$^{-1}$$m$$^{-2}$$.

By Gauss’s law, the enclosed charge is given by $$Q_{enc} = \epsilon_0 \oint \vec{E} \cdot d\vec{A}$$. Since $$\vec{E}=E_0 x \hat{i}$$ has no $$\hat{j}$$ or $$\hat{k}$$ components, only the faces perpendicular to the $$x$$-axis contribute to the flux.

At $$x=0$$ the field vanishes, so the flux through this face is zero. At $$x=a$$ the field is $$\vec{E}=E_0 a \hat{i}$$, and with an outward normal $$+\hat{i}$$ and area $$a^2$$, the flux becomes $$E_0 a \cdot a^2 = E_0 a^3$$.

Substituting into Gauss’s law gives $$Q = \epsilon_0 \cdot E_0 \cdot a^3 = 9 \times 10^{-12} \times 4 \times 10^4 \times (0.02)^3\,$$.

$$(0.02)^3 = 8 \times 10^{-6}$$ $$m^3$$

$$Q = 9 \times 10^{-12} \times 4 \times 10^4 \times 8 \times 10^{-6}$$

$$Q = 288 \times 10^{-14}$$ C

Therefore, $$Q = 288 \times 10^{-14}$$ C.