The electric field in a region is given by $\vec{E}=\frac{3}{5} E_0 \vec{i}+\frac{4}{5} E_0 \vec{j}$

Question

The electric field in a region is given by $\vec{E}=\frac{3}{5} E_0 \vec{i}+\frac{4}{5} E_0 \vec{j}$ with $E_0=2.0 	imes 10^3 \mathrm{~N} \mathrm{C}^{-1}$. Find the flux of this field through a rectangular surface of area $0.2 \mathrm{~m}^2$ parallel to the $y-z$ plane.

Professor PC Thomas Classes · Answer

The electric flux $\Phi_E$ is given by the dot product of the electric field vector and the area vector:
$\Phi_E=\vec{E} \cdot \vec{A}$\ $\Phi_E=\left(\left(1.2 	imes 10^3ight) \vec{i}+\left(1.6 	imes 10^3ight) \vec{j}ight) \cdot(0.2 \vec{i})$\ $\Phi_E=\left(1.2 	imes 10^3ight)(0.2)(\vec{i} \cdot \vec{i})+\left(1.6 	imes 10^3ight)(0.2)(\vec{j} \cdot \vec{i})$\ 
Since $\vec{i} \cdot \vec{i}=1$ and $\vec{j} \cdot \vec{i}=0$, the equation simplifies to:
$
\begin{gathered}
\Phi_E=\left(1.2 	imes 10^3ight)(0.2) \
\Phi_E=2.4 	imes 10^2 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{gathered}
$