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$$^{-2}$$)

The electric field is  $$\vec{E}=E_0x\,\hat{i}$$

Using Gauss's law,  $$\Phi=\frac{Q}{\epsilon_0}$$

The field is along the $$x$$-direction. Hence, there is no flux through the faces at $$x=0$$, $$y=0$$, $$y=a$$, $$z=0$$ and $$z=a$$.

Only the face at $$x=a$$ contributes to the flux.

At $$x=a$$,  $$E=E_0a$$

Area of the face is  $$A=a^2$$

Therefore, the total electric flux is  $$\Phi=E_0a\times a^2=E_0a^3$$

By Gauss's law,  $$Q=\epsilon_0E_0a^3$$

Given,  $$\epsilon_0=9\times10^{-12}\,C^2N^{-1}m^{-2}$$

$$E_0=4\times10^4\,NC^{-1}m^{-1}$$

$$a=2\,cm=2\times10^{-2}\,m$$

Therefore,  $$Q=(9\times10^{-12})(4\times10^4)(2\times10^{-2})^3$$

$$Q=(9\times10^{-12})(4\times10^4)(8\times10^{-6})$$

$$Q=288\times10^{-14}\,C$$

Hence,  $${Q=288}$$

Therefore, the correct answer is $$288$$.