Question

The electric field in a region is given by $\overrightarrow{E}=\frac{E_{0}x}{l}\hat{i}$. Find the charge contained inside a cubical volume bounded by the surfaces $x=0,x=a,y=0,y=a,$$z=0$ and $z=a$.
Take
$E_0=5\times {10}^3\text{N/C},l=2\text{cm,}$
$a=1\text{cm},{\epsilon }_0=8.86\times {10}^{-12}{\text{C}}^2/{\text{Nm}}^2$

• A. $1.1\times {10}^{-12}\text{C}$
• B. $2.2\times {10}^{-12}\text{C}$
• C. $0.6\times {10}^{-14}\text{C}$
• D. None of the above

Answer 1

The correct option is B $2.2\times {10}^{-12}\text{C}$

Given,
electric field, $\overrightarrow{E}=\frac{E_{0}x}{l}\hat{i}....\left(1\right)$
Electric flux enters from the surface parallel to $y-z$ plane at $x=0$.

But from $\left(1\right)$, $E=0$ at $x=0$.
Hence, flux entering the cube ${\varphi }_{\text{entering}}=0$

Flux leaves the cube from the surface parallel to $y-z$ plane at $x=a$.

Flux leaving the cube
${\varphi }_{\text{leaving}}=\left(\frac{E_{0}x}{l}\right)\left(a^2\right)=\left(\frac{E_{0}a}{l}\right)\left(a^2\right)$ (at $x=a$)

Substituting the values, we get
${\varphi }_{\text{leaving}}=\frac{(5\times {10}^3)({10}^{-2}{)}^3}{(2\times {10}^{-2})}$

$\Rightarrow {\varphi }_{\text{leaving}}=2.5\times {10}^{-1}$

$\therefore {\varphi }_{\text{leaving}}=0.25{\text{Nm}}^2/\text{C}$

At all other four surfaces, electric lines are tangential. Hence, flux is zero.

$\therefore {\varphi }_{\text{net}}={\varphi }_{\text{leaving}}=\left(0.25\right)=\frac{q_{\text{in}}}{{\epsilon }_0}$

$\Rightarrow q_{\text{in}}=\left(0.25\right)\left({\epsilon }_0\right)$

$\Rightarrow q_{\text{in}}=\left(0.25\right)(8.86\times {10}^{-12})$

$\Rightarrow q_{\text{in}}=2.2\times {10}^{-12}\text{C}$

Hence, option (b) is correct option.