An electric field $$\vec{E} = 4x\hat{i} - (y^2 + 1)\hat{j}$$ N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as $$\phi_I$$ and $$\phi_{II}$$ respectively. The absolute difference between $$(\phi_I - \phi_{II})$$ is (in Nm$$^2$$/C) ___________.

For surface ABCD (Top face at $$z = 2$$):

$$\text{The outward normal area vector points along the z-direction: } d\vec{A} = dx\,dy\,\hat{k}$$

$$\vec{E} = 4x\hat{i} - (y^2 + 1)\hat{j}$$

$$\phi_I = \iint \vec{E} \cdot d\vec{A} = \iint [4x\hat{i} - (y^2 + 1)\hat{j}] \cdot [dx\,dy\,\hat{k}] = 0$$

For surface BCGF (Right face at $$x = 3$$):

$$\text{The outward normal area vector points along the x-direction: } d\vec{A} = dy\,dz\,\hat{i}$$

$$\text{At } x = 3\text{, the x-component of the electric field is: } E_x = 4(3) = 12\text{ N/C}$$

$$\phi_{II} = \iint E_x \,dy\,dz = 12 \int_{0}^{2} dy \int_{0}^{2} dz = 12 \times (2 \times 2) = 48$$

$$|\phi_I - \phi_{II}| = |0 - 48| = 48$$