91Ó°ÊÓ

Given :

Diameter of the circle : $3.0-cm$

The electric field is : $\stackrel{→}{E}=(1500\hat{i}+1500\hat{j}-1500\hat{k})N/C$

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

$\stackrel{→}{A}=\mathrm{π}{\left(\frac{d}{2}\right)}^2=\mathrm{π}{\left(\frac{0.03m}{2}\right)}^2=\underline{7.1×{10}^{-4}\hat{j}m²}$

When the electric field would be $\stackrel{→}{E}=(1500\hat{i}+1500\hat{j}-1500\hat{k})N/C$, the electric flux will be :

$$\begin{gathered} {\mathrm{Φ}}_e=\stackrel{→}{E}\stackrel{→}{·A} \\ =(1500\hat{i}+1500\hat{j}-1500\hat{k})N/C·(7.1x{10}^{-4}\hat{j})m² \\ =10650×{10}^{-4}(\hat{i}·\hat{j})+10650×{10}^{-4}(\hat{j}·\hat{j})-10650×{10}^{-4}\left(\right(\hat{k}·\hat{j})) \\ =0+10650×{10}^{-4}-0Nm²/C \\ =\boxed{\mathbf{1}\mathbf{.}\mathbf{07}\mathbf{}\mathbf{}\mathit{N}\mathbf{·}\mathit{m}\mathbf{²}\mathbf{/}\mathit{C}} \end{gathered}$$