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 $({\rho }_o=8.85\times {10}^{–12}\frac{C}{m^3})$ is :

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $1$

Answer 1

The correct option is C $\frac{3}{4}$

Given $A=1{\text{m}}^2$
We know that,
$dq=\rho dV$
$\int dq=\int \rho Adx$
$q=A\int \rho dx$
$\int \rho dx=$Area under $\rho -x$ curve
$\therefore q_{enclosed}=\frac{1}{2}(\frac{1}{2}+1){\rho }_o$
$\Rightarrow q_{enclosed}=\frac{3}{4}{\rho }_o$
So, total flux through the cube, $\varphi =\frac{q_{enclosed}}{{ϵ}_o}=\frac{3{\rho }_o}{4{ϵ}_o}=\frac{3}{4}$
$(\because {ϵ}_o=8.85\times {10}^{-12})$