A disc 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. Conisder a cubical surface formed by isx surfaces `x=+-a//2`, `y=+-a//2`, `z=+-a//2`. The electric flux through this cubical surface is

Options

A. (a) `(-2C)/(epsilon_0)`
B. (b) `(2C)/(epsilon_0)`
C. (c) `(10C)/(epsilon_0)`
D. (d) `(12C)/(epsilon_0)`

Answer

A

Explanation

From the figure it is clear that the charge enclosed in the cubical surface is `3C+2C-7C=-2C`. Therefore the electric flux through the cube is

`phi=(q_(i n))/(epislon_0)=(-2C)/(epislon_0)`