Question

The electric field in a region is given by $\overrightarrow{E}=\frac{3}{5}{E}_0\overrightarrow{i}+\frac{4}{5}{E}_0\hat{j}$ with $E_0=2.0\times {10}^{3}N{C}^{-1}$. Find the flux of this field through a rectangular surface of area $0.2m^2$ parallel to the $y-z$ plane.

Answer 1

$Mg-T=D$
$T=mg$
$f=mg$
$T=N$
$N=mg$
$\varphi =\overrightarrow{E}\cdot \hat{n}ds$
$\overrightarrow{E}=3/4E_{0}i+4/5E_0\hat{i}$
Rectangle in $yz$ plane
So Area vector along $\hat{i}$
$\varphi =(3/5E_0\hat{i}+4/5E_0\hat{j})\cdot \left(0.2\right)\hat{i}$
$\frac{3}{5}\times E_0\times 0.2$
$=\frac{3}{5}\times 2\times {10}^3\times 0.2$
$=0.6\times 0.4\times {10}^3$
$=240N{m}^2{c}^{-1}m$.