Question

The electric field in a region is given by $\overrightarrow{E}=\alpha x\hat{i}$. Here, $\alpha$ is a positive constant of proper dimension. Find the total flux passing through a cube bounded by the coordinates, $x=l,x=2l,y=0,y=l,z=0,z=l.$

• A. $\alpha l^3$
• B. $2\alpha l^3$
• C. $3\alpha l^3$
• D. $4\alpha l^3$

Answer 1

The correct option is A $\alpha l^3$

Given,
Electric field, $\overrightarrow{E}=\alpha x\hat{i}$


Since, electric field is directed along positive $x-$ direction. Therefore, field lines are perpendicular to faces $\text{ABCD}$ and $\text{EFGH}$.
At all other four faces field line are tangential. So, net flux passing through these four faces will be zero.

For face $\text{ABCD}$,
$\because x=l$
So, $\overrightarrow{E}=\alpha l\hat{i}$

Also, area, $\overrightarrow{A}=-l^2\hat{i}$

Now, electric flux,
${\varphi }_1=\overrightarrow{E}.\overrightarrow{A}=\left(\alpha l\hat{i}\right).(-l^2\hat{i})$

$\Rightarrow {\varphi }_1=-\alpha l^3.....\left(1\right)$

Similarly, for face $\text{EFGH}$,

$\because x=2l$

So, $\overrightarrow{E}=2\alpha l\hat{i}$

Also, area, $\overrightarrow{A}=l^2\hat{i}$

$\therefore {\varphi }_2=2\alpha l^3.....\left(2\right)$

Now, net electric flux,

${\varphi }_{\text{net}}={\varphi }_1+{\varphi }_2$

$\Rightarrow {\varphi }_{\text{net}}=-\alpha l^3+2\alpha l^3=\alpha l^3$

So, $\left(a\right)$ is correct choice.