The electric field in a region is given by $\overrightarrow{E}=\frac{E_{0}x}{l}\overrightarrow{i}$ 

Find the charge contained inside a cubical volume bounded by the surfaces $x=0,x=a,y=0,y=a,z=0andz=a.$ 

Take $E_0=5\times {10}^{3}N{C}^{-1},l=0.02m,anda=0.01m.$

 

Given

$\overrightarrow{E}=\frac{E_0}{l}\widehat{i},l=2cm,a=1cm,E_0=5\times {10}^{3}N{C}^{-1}$

We see that flux passes mainly through surface area ABDC and EFGH. As the AEFB and CHGD are parallel to the flux again in ABDC=0, the flux only passes through the surface area EFGHE = 0.

Flux $=E_0\frac{a}{l}\times area=5\times {10}^3\times \frac{a}{l}\times a^2$

$=5\times {10}^3\times \frac{a^3}{l}$

$=5\times {10}^3\times \frac{{\left(0.01\right)}^3}{2\times {10}^{-2}}=2.5\times {10}^{-1}$

So, $q={\epsilon }_0\times flux$

$=8.85\times {10}^{-12}\times 2.5\times {10}^{-1}$

$=22.125\times {10}^{-13}=2.2125\times {10}^{-12}C$