Question

Electric field in a region is given by $\overline{E}=\frac{3}{5}{E}_0\hat{i}+\frac{4}{5}{E}_0\hat{j}$ where $E_0=2\times {10}^{3}N{C}^{-1}$. The flux through a rectangular surface of area $0.2m^2$ parallel to $YZ$ plane is (in volt meter)

• A. $240$
• B. $60$
• C. $40$
• D. $80$

Answer 1

The correct option is C $40$

$\overline{E}=\frac{3}{5}{E}_0\hat{i}+\frac{4}{5}{E}_0\hat{j}$

$E_0=2\times {10}^{2}N/C$

The flux through a rectangular surface of area $=0.2m^2$ parallel to $Y-Z$ plane is $-$

$\varphi =E.A$

$=(\frac{3}{5}\times 2\times {10}^2\hat{i}+\frac{4}{5}\times 2\times {10}^2\hat{j})\times \left(0.2\right)$

$=2\times {10}^2\left[\sqrt{{\left(\frac{1}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2\times 0.2}\right]$

$=2\times {10}^2\times \frac{2}{10}\sqrt{\frac{9}{25}+\frac{16}{25}}$

$=4\times 10$

$=40$