Question

A cube has sides of length $l=0.300m$. It is placed with one corner at the origin as shown in the
figure. The electric field is not uniform, but it is given by $\overrightarrow{E}=(-5.00N{C}^{-1})x\hat{i}+\left(3.00N{C}^{-1}\right)z\hat{k}$.
The total flux passing through the cube is

• A. $-0.135N{m}^2{C}^{-1}$
• B. $-0.054N{m}^2{C}^{-1}$
• C. $-0.081N{m}^2{C}^{-1}$
• D. Zero

Answer 1

The correct option is B $-0.054N{m}^2{C}^{-1}$

For surface $z=0$ on $xy$ plane,
Normal $\overrightarrow{S_4}=l^2(-\hat{k})$

For surface $z=l$ on $xy$ plane,
Normal $\overrightarrow{S_2}=l^2\left(\hat{k}\right)$

For surface $x=0$ on $yz$ plane,
Normal $\overrightarrow{S_6}=l^2(-\hat{i})$

For surface $x=l$ on $yz$ plane,
Normal $\overrightarrow{S_5}=l^2\left(\hat{i}\right)$

For surface $y=0$ on $xz$ plane,
Normal $\overrightarrow{S_1}=l^2(-\hat{j})$

For surface $y=l$ on $xy$ plane,
Normal $\overrightarrow{S_3}=l^2(-\hat{j})$

Electric field $\overrightarrow{E}=-5x\hat{i}+3z\hat{k}$
Flux $\varphi =\overrightarrow{E}.\overrightarrow{A}$

${\varphi }_4=-3l^{2}z=0$
${\varphi }_2=3z{l}^2=3l^3$
${\varphi }_6=5l^{2}x=0$
${\varphi }_5=-5l^{2}x=-5l^3$
${\varphi }_1=0$
${\varphi }_3=0$

So the surfaces which have zero flux are $S_1,S_3,S_4$ and $S_6$.
Total flux through the cube = $-2l^3$ = $-2(0.3{)}^3$ = $-0.054$ $N{m}^2{C}^{-1}$