Question

Electric field in the region is given by $\overrightarrow{E}=(2\hat{i}+3\hat{j}+4\hat{k}){\text{NC}}^{-1}$ . The flux associated with a square plate of side $5\text{cm}$ as shown in the figure is

• A. $10{\text{Nm}}^2{\text{C}}^{-1}$
• B. $-10{\text{Nm}}^2{\text{C}}^{-1}$
• C. $20{\text{Nm}}^2{\text{C}}^{-1}$
• D. $-20{\text{Nm}}^2{\text{C}}^{-1}$

Answer 1

The correct option is C $20{\text{Nm}}^2{\text{C}}^{-1}$

Coordinates of
$\text{B}=(5\cos 53{}^{\circ },0,5\sin 53{}^{\circ })$

$\therefore \text{B}=(3,0,4)$

Coordinates of $\text{D}=(0,5,0)$


Now, from figure
$\overrightarrow{AB}=\left[\right(3-0)\hat{i}+(0-0)\hat{j}+(4-0\left)\hat{k}\right]=(3\hat{i}+4\hat{k})$
$\overrightarrow{AD}=\left[\right(0-0)\hat{i}+(5-0)\hat{j}+(0-0\left)\hat{k}\right]=\left(5\hat{j}\right)$

Area vector=$\overrightarrow{A}=\overrightarrow{AB}\times \overrightarrow{AD}=(3\hat{i}+4\hat{k})\times \left(5\hat{j}\right)$
$\therefore \overrightarrow{A}=15\hat{k}-20\hat{i}...\left(1\right)$

Given,
$\overrightarrow{E}=2\hat{i}+3\hat{j}+4\hat{k}...\left(2\right)$

We know thar flux $\varphi$ associated with the area vector $\overrightarrow{A}$ and electric field vector $\overrightarrow{E}$ is $\varphi =\overrightarrow{E}.\overrightarrow{A}$.

From $\left(1\right)$ and $\left(2\right)$ we have,
$\varphi =(2\hat{i}+3\hat{j}+4\hat{k}).(15\hat{k}-20\hat{i})$

$\therefore \varphi =60-40=20{\text{Nm}}^2{\text{C}}^{-1}$

Hence, option (c) is the correct answer.