91Ó°ÊÓ

Given :

Dimensions of rectangle : $2.0cm×3.0cm$

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

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

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.

${\mathrm{Φ}}_e=E·A·\cos \mathrm{θ}$

Where $\mathrm{θ}$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

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

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

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

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

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