Electric Flux

Starting point: the flux through a surface in a uniform field is $${\Phi }_{E}=\vec{E}\cdot \vec{A}=EAcos\theta $$ Expressing the area of the circular disk $$A=\pi {R}^{2}$$ Substituting into the flux expression $${\Phi }_{E}=E\pi {R}^{2}cos\theta $$ Result: $${\Phi }_{E}=\pi {R}^{2}Ecos\theta $$ The flux depends on the square of the radius. Doubling $R$ quadruples the flux.

The flux decreases. The flux is proportional to $cos\theta $. Since $cos60^{\circ}=0.50$ is less than $cos30^{\circ}\approx 0.866$, increasing $\theta $ from 30° to 60° reduces the cosine factor. Fewer field lines pierce the surface at the steeper angle.

At the face $x=L$, the outward area vector points in the +x-direction. The field at that face is $\vec{E}=b{L}^{2} \hat{i}$, which is constant over that face because every point on it has $x=L$. The flux through that face is $$\Phi =\int \vec{E}\cdot d\vec{A}=\int b{L}^{2} dA=b{L}^{2}\int dA$$ The integral $\int dA$ over the square face equals ${L}^{2}$. So the flux through the face at $x=L$ is $b{L}^{4}$.

The total flux is $${\Phi }_{\text{total}}=b{L}^{4}+0+0+0+0+0=b{L}^{4}$$ The net flux is positive and nonzero. By Gauss’s law, a nonzero net flux through a closed surface means there is a nonzero enclosed charge. The positive sign means the net enclosed charge is positive. This is consistent because the field increases with $x$, so more field lines exit the right face than enter the left face. A source of field lines (positive charge) must exist inside.

The total flux is zero. The field lines enter through one flat end and exit through the other flat end. These two flat faces produce equal and opposite flux contributions. The curved surface has its area vector perpendicular to E-vector everywhere, so it contributes zero flux. Because no charge is enclosed, all entering field lines exit, and the net flux is zero.