Question

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 difference between $$(\phi_I -\phi_{II})$$ is (in $$Nm^2 /C)$$

Answer 1

Correct option is A. -48
flux through $$ABCD$$

$$\phi_1 = \displaystyle E.dA$$

Area vector of $$ABCD$$ is in z-axis $$(\hat{K})$$

And $$\vec{E}$$ is in x-direction & y-direction.

So, Area vector is perpendicular to $$\vec{E}$$. So, $$\phi = 0 \, (\because \cos 90 = 0)$$

Flux through $$BCEF$$

$$\phi_2 = \displaystyle \int E.dA = E.A$$

$$= [(4 \times \hat{i} - (y^2 + 1) \hat{j}] . 4\hat{i} = 16 x - 0 = 16 x$$

$$\phi_2 = 16X$$, where $$x = 3 , \phi_2 = 48 $$

$$\phi_1 - \phi_2 = -48 \dfrac{Nm^2}{c} $$