Question

The volume charge density as a function of distance $X$ from one face inside a unit cube is varying as shown in the figure. If the total flux (in Wb) through the cube if $P_0=8.85\times {10}^{-12}{\text{C/m}}^3$ is $\frac{x}{4}$ then find the value of $x$ (Give integer value)?

Answer 1

Let $P$ be the volume charge density.

Charge in volume $dV$ be $dq$.

For a unit cube, $dq=P\times dx\times A[A=1{\text{m}}^2]$

$\Rightarrow dq=Pdx$

So, total charge enclosed within unit cube
$q_{enclosed}=\int dq=\int Pdx$

$\int Pdx$ is Area under the given graph.

So total charge enclosed ,

$Q=\frac{1}{2}\times P_0\times (\frac{3}{4}-\frac{1}{4}+1)$

$Q=\frac{3P_0}{4}$
So, total flux ,$\varphi =\frac{Q}{{\epsilon }_0}=\frac{3P_0}{4{\epsilon }_0}$

Substituting the given data we get,

$\varphi =\frac{3\times 8.85\times {10}^{-12}}{4\times 8.85\times {10}^{-12}}=\frac{3}{4}$

Comparing with the data given in the question we get, $x=3$

Accepted answer : $3$