Q.The electric field in a region is given by $\vec{E}=5.0\hat{k}$ N/C. Calculate the electric flux through a square of side 10.0 cm in the following cases

[!TLDR] With $\vec E=5.0\hat k$ N/C and area $A=0.01\,\text{m}^2$: (a) flux $=EA=5.0\times10^{-2}$ Vm (normal along $\hat k$); (b) flux $=0$ (normal along $\hat j$, perpendicular to E); (c) flux $=EA\cos45^\circ\appro …

The field is $\vec{E}=5.0\hat{k}$ N/C (i.e. of magnitude 5.0 N/C, directed along the z-axis), and the square has side 10.0 cm $=0.1$ m, so its area is $A=(0.1)^2=0.01\,\text{m}^2$. Flux is $\phi=\vec{E}\cdot\vec{A}=EA\cos\theta$, where $\theta$ is the angle between $\vec E$ and the square's own area-normal vector.

1. The square lies in the XY plane, so its normal points along the z-axis, EXACTLY parallel to $\vec E$ ($\theta=0$): $\phi=EA\cos0^\circ=(5.0)(0.01)(1)=5.0\times10^{-2}\,\text{Vm}.$
2. The square lies in the XZ plane, so its normal points along the y-axis, exactly PERPENDICULAR to $\vec E$ (which is along z) ($\theta=90^\circ$): $\phi=EA\cos90^\circ=0.$
3. The square's normal makes an angle of $45^\circ$ with the Z axis (i.e. with $\vec E$ itself): $\phi=EA\cos45^\circ=(5.0)(0.01)(0.7071)\approx3.5\times10^{-2}\,\text{Vm}.$ …

Apply $\phi=\vec{E}\cdot\vec{A}=EA\cos\theta$ for each of the three orientations, identifying the angle between the field di …

Confusing the plane a surface LIES IN with the direction of its NORMAL vector (e.g. treating the XY-plane case as perpendicular to E instead of parallel, since the …