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 $$E_0 = 4 \times 10^4$$ NC$$^{-1}$$ 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 $$\epsilon_0 = 9 \times 10^{-12}$$ C$$^2$$ N$$^{-1}$$m$$^{-2}$$)

A cubical volume is bounded by $$x = 0, x = a, y = 0, y = a, z = 0, z = a$$ and it is placed in an electric field $$\vec{E} = E_0 x \hat{i}$$ with $$E_0 = 4 \times 10^4$$ NC$$^{-1}$$m$$^{-1}$$. Here $$a = 2$$ cm. We need to find the charge $$Q$$ enclosed in units of $$10^{-14}$$ C.

Gauss's law states that the enclosed charge equals $$\epsilon_0 \oint \vec{E} \cdot d\vec{A}$$. Since $$\vec{E} = E_0 x \hat{i}$$, only the faces perpendicular to $$\hat{i}$$ give nonzero flux. At $$x = 0$$ the field is zero, so its contribution is zero. At $$x = a$$ the field is $$\vec{E} = E_0 a \hat{i}$$ and the outward normal is $$\hat{i}$$, with area $$a^2$$.

The resulting flux through the face at $$x = a$$ is $$\Phi = E_0 a \cdot a^2 = E_0 a^3$$.

The enclosed charge becomes $$Q = \epsilon_0 E_0 a^3$$. Substituting $$\epsilon_0 = 9 \times 10^{-12}$$, $$E_0 = 4 \times 10^4$$, and $$a = 2 \times 10^{-2}$$ gives
$$Q = (9 \times 10^{-12}) \times (4 \times 10^4) \times (2 \times 10^{-2})^3$$
$$= 9 \times 10^{-12} \times 4 \times 10^4 \times 8 \times 10^{-6}$$
$$= (9 \times 4 \times 8) \times 10^{-14}$$
$$= 288 \times 10^{-14} \text{ C}$$.

Therefore, $$Q = 288$$.