Question

The electric field in a region is given by $\overrightarrow{E}=(1200\hat{i}+1600\hat{j})\text{N/C}$. Find the flux of this field through a rectangular surface of area $0.2{\text{m}}^2$ parallel to the $y-z$ plane.

• A. $340{\text{N-m}}^2/\text{C}$
• B. $140{\text{N-m}}^2/\text{C}$
• C. $240{\text{N-m}}^2/\text{C}$
• D. $440{\text{N-m}}^2/\text{C}$

Answer 1

The correct option is C $240{\text{N-m}}^2/\text{C}$

Given:
$\overrightarrow{E}=(1200\hat{i}+1600\hat{j})\text{N/C};$
Surface area, $A=0.2{\text{m}}^2$

From the diagram, we can see that positive normal to the rectangular surface area is along $+x-$ axis.

Thus, area vector, $\overrightarrow{A}=\left(0.2\hat{i}\right){\text{m}}^2$

Now electric flux $\varphi$ is written as

$\varphi =\overrightarrow{E}.\overrightarrow{A}$

$\Rightarrow \varphi =(1200\hat{i}+1600\hat{j}).\left(0.2\hat{i}\right)$

$\Rightarrow \varphi =1200\times 0.2+1600\times 0=240{\text{N-m}}^2/\text{C}$

So, option $\left(b\right)$ is correct.