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. Then the total flux (in S.I. units) through the cube if $({\rho }_0=8.85\times {10}^{-12}{\text{Cm}}^{-3})$ is $\frac{x}{8}$. Find the value of $x$ (integer only)

Answer 1

Net flux through the cube, ${\varphi }_{net}=\frac{q_{en}}{{\epsilon }_0}$

To find $q_{en}$, let's divide the cube into small elements and consider a small element of width $dx$ as shown.

$\text{Charge on the small element}dq=\left(\rho \right)\left(Adx\right)$

$\text{Total charge}Q=\int \rho Adx[\because \int \rho dx=\text{Area under the curve}]$

$Q=A\int \rho dx$

$\text{Here A is the surface area}=1\times 1=1{\text{m}}^2$

Now the total charge of cube,

$Q=\left(1\right)\times \left[\frac{1}{2}(1+\frac{1}{2}){\rho }_0\right]=\frac{3}{4}{\rho }_0$



$\Rightarrow \text{Net flux}{\varphi }_{net}=\frac{q_{en}}{{\epsilon }_0}=\frac{\frac{3}{4}{\rho }_0}{{\epsilon }_0}=\frac{3}{4}[\because {\rho }_0=8.85\times {10}^{–12}]$

Given, the net flux ${\varphi }_{net}=\frac{x}{8}$

$\therefore \frac{x}{8}=\frac{3}{4}\Rightarrow x=6$