A $2.0cm\times 3.0cm$rectangle lies in the $xz$-plane. What is the magnitude of the electric flux through the rectangle if

a. $\overrightarrow{E}=(100\hat{i}-200\hat{k})N/C$?

b. $\overrightarrow{E}=(100\hat{i}-200\hat{j})N/C$?

Given :

Dimensions of rectangle : $2.0cm\times 3.0cm$

a.$\overrightarrow{E}=(100\hat{i}-200\hat{k})N/C$

b.$\overrightarrow{E}=(100\hat{i}-200\hat{j})N/C$

Theory used :

The quantity of electric field passing through a closed surface is known as the Electric flux. Gauss's law indicates that the electric field across a surface is proportional to the angle at which it passes, hence we can determine charge inside the surface using the equation below.

${\Phi }_e=E·A·\cos \theta$

Where $\theta$is the angle formed between the electric field and the normal.

The rectangle is in $xz$-plane, this means the normal of the sheet is in$y$direction which means, the area has component $\hat{j}$. We can get $A$, the area of the flat sheet by

$A=(2cm\times 3cm)\hat{k}=6cm²=\underline{6x{10}^{-4}\hat{k}m²}$

(a) When the electric field would be $\overrightarrow{E}=(100\hat{i}-200\hat{k})N/C$, the electric flux will be :

$$\begin{gathered} {\Phi }_e=\overrightarrow{E}\overrightarrow{·A} \\ =(100\hat{i}-200\hat{k})N/C·(6x{10}^{-4}\hat{j})m² \\ =600\times {10}^{-4}(\hat{i}·\hat{j})-1200\times {10}^{-4}\left(\right(\hat{k}·\hat{j})) \\ =0-0Nm²/C \\ =\boxed{\mathbf{0}\mathbf{}\mathit{N}\mathbf{·}\mathit{m}\mathbf{²}\mathbf{/}\mathit{C}} \end{gathered}$$

(b)When the electric field would be $\overrightarrow{E}=(100\hat{i}-200\hat{j})N/C$, the electric flux will be :

$$\begin{gathered} {\Phi }_e=\overrightarrow{E}\overrightarrow{·A} \\ =(100\hat{i}-200\hat{k})N/C·(6x{10}^{-4}\hat{j})m² \\ =600\times {10}^{-4}(\hat{i}·\hat{j})-1200\times {10}^{-4}\left(\right(\hat{j}·\hat{j})) \\ =0-1200\times {10}^{-4}Nm²/C \\ =\boxed{\mathbf{-}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}\mathit{N}\mathbf{·}\mathit{m}\mathbf{²}\mathbf{/}\mathit{C}} \end{gathered}$$