Question

Electric field in a given by $\overrightarrow{E}$ = $-4x\hat{i}+6y\hat{j}$. Then find the charge enclosed in the cube of side $1$m oriented as shown in the diagram.

• A. ${ϵ}_o$
• B. $2{ϵ}_o$
• C. $4{ϵ}_o$
• D. $6{ϵ}_o$

Answer 1

The correct option is C $4{ϵ}_o$

Given, electric field $E=-4x\hat{i}+6y\hat{j}$. A cube of side $1m$. We have to find the charge enclosed in a cube.

For surface $BCGF$, area $=\overrightarrow{A}=(1\times 1)\hat{i}=\hat{i}$

$\overrightarrow{E}=-4x\hat{i}+6y\hat{j}$

$x=1=$ constant $0\le y\le 1$, $0\le z\le 1$

${\varphi }_1=\overrightarrow{E}.\overrightarrow{A}=(-4x\hat{i}+6y\hat{j}).\hat{i}$

$=-4x=-4\left(1\right)=-4N{m}^2/C$

For surface $ADHE$, $\overrightarrow{A}=(1\times 1)(-\hat{i})=-\hat{i}$

$\overrightarrow{E}=-4x\hat{i}+6y\hat{j}$

$x=0=$ constant

${\varphi }_2=\overrightarrow{E}.\overrightarrow{A}=(-4x\hat{i}+6y\hat{j}).(-\hat{i})$

$=4x=4\left(0\right)=0$

For surface $ABFE$,

$\overrightarrow{A}=(1\times 1)(-\hat{j})=-\hat{j}$

$\overrightarrow{E}=-4x\hat{i}+6y\hat{j}$

$y=0=$ constant, $0\le x\le 1$, $0\le z\le 1$

So, $d{\varphi }_3=\overrightarrow{E}.\overrightarrow{dA}$

$=\overrightarrow{E}.\left(zdx\right)(-\hat{j})$

$d{\varphi }_3=(-4x\hat{i}+6y\hat{j})\left(zdx\right)(-\hat{j})$

$d{\varphi }_3=-6z\left(0\right)=0$

Similarly, for surface $CDGE$, ${\varphi }_4=0$

For surfacev $EFGH$, $\overrightarrow{A}=(1\times 1)\hat{k}=\hat{k}$

${\varphi }_5=\overrightarrow{E}.\overrightarrow{A}=(-4x\hat{i}+6y\hat{j}).\hat{k}=0$

For surface $ABCD$, $\overrightarrow{A}=(1\times 1)\widehat{-k}=\widehat{-k}$

${\varphi }_6=\overrightarrow{E}.\overrightarrow{A}=(-4x\hat{i}+6y\hat{j}).\left(\widehat{-k}\right)$

Total flux $={\varphi }_1+{\varphi }_2+{\varphi }_3+{\varphi }_4+{\varphi }_5+{\varphi }_6$

$=-4N{m}^2/C$

Now applying gauss' law, $\varphi =\frac{q}{{\epsilon }_0}$

$-4=\frac{q}{{\epsilon }_0}$

$q=4{\epsilon }_0$