Question

A disk of radius $a/4$ having a uniformly distributed charge $6C$ is placed in the $x-y$ plane with its centre at $(-a/2,0,0)$. A rod of length 'a' carrying a uniformly distributed charge $8C$ is placed on the $x$-axis from $x=a/4$ to $x=5a/4$. Two point charges $-7C$ and $3C$ are placed at $(a/4,-a/4,0)$ and $(-3a/4,3a/4,0)$, respectively. Consider a cubical surface formed by six surfaces $x=\pm a/2,y=\pm a/2,z=\pm a/2$. The electric flux through this cubical surface is:

• A. $\frac{-2C}{{ϵ}_0}$
• B. $\frac{2C}{{ϵ}_0}$
• C. $\frac{10C}{{ϵ}_0}$
• D. $\frac{12C}{{ϵ}_0}$

Answer 1

Answer: (A)

As half part of the disk inside the cube so charge enclosed due to disk is
$Q_d=6C/2=3C$.
If $\lambda$ is the line charge density,
$8C=\underset{a/4}{\overset{5a/4}{\int }}\lambda dx=\lambda (5a/4-a/4)=\lambda a$
$\Rightarrow \lambda =\frac{8C}{a}$
Charge enclosed by cube due to rod is
$Q_r=\underset{a/4}{\overset{a/2}{\int }}\lambda dx=\frac{8C}{a}(a/2-a/4)=2C$
The charge $-7C$ inside the cube.
Thus the total charge enclosed by the cube is
$Q_{en}=Q_d+Q_r-7C=3C+2C-7C=-2C$
Using Gauss's law electric flux through the cube is
$\varphi =\frac{Q_{en}}{{ϵ}_0}=\frac{-2C}{{ϵ}_0}$