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A disk of radius $a / 4$ having a uniformly distributed charge $6 \mathrm{C}$ 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 $8 \mathrm{C}$ is placed on the $x$-axis from $x=a / 4$ to $x=5 a / 4$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ 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

$\frac{-2 \mathrm{C}}{\varepsilon_0}$
$\frac{2 \mathrm{C}}{\varepsilon_0}$
$\frac{10 \mathrm{C}}{\varepsilon_0}$
$\frac{12 \mathrm{C}}{\varepsilon_0}$
Solution
As half part of the disk inside the cube so charge enclosed due to disk is $Q _{ d }=6 C / 2=3 C$.