- Home
- Standard 12
- Physics
1. Electric Charges and Fields
medium
The electric field in a region is given by $\vec E = \frac{3}{5}{E_0}\hat i + \frac{4}{5}{E_0}\hat j$ and $E_0 = 2\times10^3\, N/C$. Then, the flux of this field through a rectangular surface of area $0.2\, m^2$ parallel to the $y-z$ plane is......$\frac{{N - {m^2}}}{C}$
A
$240$
B
$320$
C
$0$
D
$560$
Solution
$\phi = \overrightarrow {\rm{E}} \cdot \overrightarrow {\rm{A}} $
$=\frac{3}{5} \mathrm{E}_{0}(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \cdot \mathrm{A} \hat{\mathrm{i}} $
$ =\frac{3}{5} \mathrm{E}_{0} \mathrm{A} $
$=\frac{3}{5} \mathrm{E}_{0} \times 0.2 $
$=\frac{3}{5} \times 2 \times 10^{3} \times 0.2 $
$ \phi =240 \frac{\mathrm{N}-\mathrm{m}^{2}}{\mathrm{C}} $
Standard 12
Physics
