Question

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

• A. ${2\epsilon }_0$
• B. $zero$
• C. ${\epsilon }_0$
• D. ${6\epsilon }_0$

Answer 1

The correct option is A ${2\epsilon }_0$

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

Thus, $\overrightarrow{E}{|}_{EFGH}=\left(6y\right)\hat{j}$

$\overrightarrow{E}{|}_{ABCD}=(-4\hat{i}+6y\hat{j})$

$\overrightarrow{E}{|}_{CDGH}=(-4x\hat{i}+6\hat{j})$

$\overrightarrow{E}{|}_{AEFB}=-\left(4x\hat{i}\right)$

$\overrightarrow{E}{|}_{AEHD}=(-4x\hat{i}+6y\hat{j})$

$\overrightarrow{E}{|}_{FGCB}=(-4x\hat{i}+6y\hat{j})$

Area vector $\to$

${\overrightarrow{A}}_{ABCD}=\left(Area\right)\widehat{Area}$

$=(1\times 1)\hat{i}=\hat{i}$

${\overrightarrow{A}}_{EFGH}=\left(Area\right)\widehat{Area}$

$=(1\times 1)(-\hat{i})$

$=-1\hat{i}$

${\overrightarrow{A}}_{CDGH}=(1\times 1)\left(\hat{j}\right)$

$=\hat{j}$

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

$=-\hat{j}$

${\overrightarrow{A}}_{AEHD}=(1\times 1)\hat{k}$

$=\hat{k}$

${\overrightarrow{A}}_{FGCB}=(1\times 1)(-\hat{k})$

$=-\hat{k}$

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

${\varphi }_{ABCD}=E{|}_{ABCD}.Area{|}_{ABCD}=(-4\hat{i}+6y\hat{j}).\left(1\hat{i}\right)$

as at $ABCD(x=1)$

$=-4$

${\varphi }_{EFGH}=E{|}_{EFGH}.A{|}_{EFGH}=\left(6y\hat{j}\right).(-1\hat{i})$

$=0(asx=0atEFGH)$

${\varphi }_{CDGH}=E{|}_{CDGH}.A_{CDGH}=(-4x\hat{i}+6\hat{j}).\left(1j\right)$

$=6(asy=1atCDGH)$

${\varphi }_{AEFB}=E{|}_{AEFB}.A{|}_{AEFB}=(-4x\hat{i}).(-1\hat{j})$

$=0(asy=0atAEFB)$

${\varphi }_{AEHD}=E{|}_{AEHD}.A_{AEHD}=(-4x\hat{i}+6y\hat{j}).\left(1\hat{k}\right)$

$=0$

${\varphi }_{FGCB}=E{|}_{FGCB}.A{|}_{FGCB}=(-4x\hat{i}+6y\hat{j})(-1\hat{k})$

$=0$

${\varphi }_{net}{|}_{ABCDEFGH}={\varphi }_{ABCD}+{\varphi }_{EFGH}+{\varphi }_{CDGH}+{\varphi }_{AEFB}+{\varphi }_{AEHD}+{\varphi }_{FGCB}$

$=6-4$

$=2$

${\varphi }_{net}=\frac{q_{enc}}{{\epsilon }_0}$ (By Gauss's law)

$\Rightarrow q_{ene}=2{\epsilon }_0$