Question

A disk of radius $a/4$ having a uniformly distributed charge $6 C$ 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 $8 \,C$ is placed on the $x$-axis from $x=a / 4$ to $x=5 a / 4$. Two point charges $-7\, C$ and $3\, C$ are placed at $(a / 4,-a / 4,0)$ and $(-3 \,a / 4,3 \,a / 4,0)$, 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. $\frac{-2 C}{\epsilon_{0}}$
• B. $\frac{2 C }{\epsilon_{0}}$
• C. $\frac{10 C }{\epsilon_{0}}$
• D. $\frac{12 C}{\epsilon_{0}}$

Answer 1

Answer: (A)

As half part of the disk inside the cube so charge enclosed due to disk is
$Q_{d}=6 C / 2=3 C$.
If $\lambda$ is the line charge density,
$8 C =\int\limits_{ a / 4}^{5 a / 4} \lambda dx =\lambda(5 a / 4- a / 4)=\lambda a $
$\Rightarrow \lambda=\frac{8 C }{ a }$
Charge enclosed by cube due to rod is
$Q _{ r }=\int\limits_{ a / 4}^{ a / 2} \lambda dx =\frac{8 C }{ a }( a / 2- a / 4)=2 C$
The charge $-7 C$ inside the cube.
Thus the total charge enclosed by the cube is
$Q_{e n}=Q_{d}+Q_{r}-7 C=3 C+2 C-7 C=-2 C$
Using Gauss's law electric flux through the cube is
$\phi=\frac{Q_{e n}}{\epsilon_{0}}=\frac{-2 C}{\epsilon_{0}}$