Expression for an electric field is given by $$\vec{E} = 4000 \; x^2 \; \hat{i}$$ V m$$^{-1}$$. The electric flux through the cube of side 20 cm when placed in electric field (as shown in the figure) is ______ V cm.

The electric field is uniform in direction (along $$+\hat{i}$$) but varies with the $$x$$-coordinate:
$$\vec E = 4000\,x^{2}\,\hat{i}\;{\text{V m}}^{-1}$$.

Place the cube so that its faces are perpendicular to the coordinate axes and the two faces perpendicular to the $$x$$-axis lie at $$x = 0$$ and $$x = a$$, where the edge length of the cube is
$$a = 20 \text{ cm} = 0.20 \text{ m}$$.

Electric flux through a closed surface is
$$\Phi = \oint_S \vec E \cdot d\vec A$$.

The field has no $$y$$ or $$z$$ components, so the four faces whose outward normals are $$\pm\hat{j}$$ or $$\pm\hat{k}$$ give zero contribution because $$\vec E \cdot d\vec A = 0$$ there.

Only the two faces perpendicular to the $$x$$-axis contribute.

Outward normal $$\hat n = +\hat{i}$$,
$$d\vec A = \hat{i}\,dA$$,
$$\vec E \cdot d\vec A = E_x\,dA = (4000\,a^{2})\,dA$$.

Total flux through this face:
$$\Phi_{x=a} = E_x(a)\,A = 4000\,a^{2}\,A$$, where $$A = a^{2}$$ is the area of one face.

Outward normal $$\hat n = -\hat{i}$$,
$$d\vec A = -\hat{i}\,dA$$,
$$\vec E \cdot d\vec A = -E_x(0)\,dA = 0$$ because $$E_x(0)=0$$.

Hence $$\Phi_{x=0} = 0$$.

Net outward flux through the entire cube:
$$\Phi = \Phi_{x=a} - \Phi_{x=0} = 4000\,a^{2}\,A$$.

Substitute $$a = 0.20 \text{ m}$$:
$$A = a^{2} = (0.20)^{2} = 0.040 \text{ m}^{2}$$,
$$E_x(a) = 4000\,(0.20)^{2} = 4000 \times 0.040 = 160 \text{ V m}^{-1}$$.

Therefore
$$\Phi = 160 \text{ V m}^{-1} \times 0.040 \text{ m}^{2} = 6.4 \text{ V m}.$$

The question asks for the answer in $$\text{V cm}$$. Since $$1 \text{ m} = 100 \text{ cm}$$,
$$6.4 \text{ V m} = 6.4 \times 100 \text{ V cm} = 640 \text{ V cm}.$$

Electric flux through the cube = 640 V cm.