Question

A charge \[(5\sqrt 2 + 2\sqrt 5 )\] coulomb is placed on the axis of an infinite disc at a distance \[a\]from the center of the disc. The flux of this charge on the part of the disc having an inner and outer radius of $a$ and \[2a\] will be:

Answer 1

Hint: In this solution, we will first calculate the solid angle subtended by the inner radius and the outer radius of the disc. The flux through the disc will be the product of the charge and the difference of the solid angles subtended by the inner and outer radius.
Formula used: In this solution, we will use the following formula:
1. Solid angle subtended due to an angle: $W = 2\pi (1 - \cos \theta )$
2. Flux due to a point charge: $\phi = \dfrac{q}{{{\varepsilon _0}}}$ where $q$ is the charge near the surface and ${\varepsilon _0}$ is the permittivity
Complete step by step answer:

We’ve been given that a charge is placed on the axis of an infinite disc at a certain distance from it. Assuming that the charge is a point charge, we calculate the solid angle subtended by the charge on the inner radius as
$\tan {\theta _1} = \dfrac{a}{a} = 1$
$ \Rightarrow {\theta _1} = 45^\circ $
Hence we get,
$ \Rightarrow \cos {\theta _1} = \dfrac{1}{{\sqrt 2 }}$
Similarly, the angle subtended by the charge on the outer radius will be
$\tan {\theta _2} = \dfrac{{2a}}{a} = 2$
$ \Rightarrow {\theta _2} = 63.43^\circ $
Which gives us,
$ \Rightarrow \cos {\theta _2} = \dfrac{1}{{\sqrt 5 }}$
Hence the solid angle subtended by this hollow disc by the charge will be
$w = 2\pi (1 - \cos {\theta _1}) - 2\pi (1 - \cos {\theta _2})$
$ \Rightarrow w = 2\pi \left[ {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 5 }}} \right]$
Now if wanted to find the total flux emitted by the point charge, it would be distributed over $4\pi $ steradians and would have a value $\phi = \dfrac{q}{{{\varepsilon _0}}}$.
But in our case, the flux is only passing through $w = 2\pi \left[ {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 5 }}} \right]$ steradians.
Hence the flux will be
$\phi = \dfrac{q}{{{\varepsilon _0}}}\dfrac{w}{{4\pi }}$
So, we can write
\[\phi = \dfrac{{(5\sqrt 2 + 2\sqrt 5 )}}{{{\varepsilon _0}}} \times \dfrac{{2\pi \left[ {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 5 }}} \right]}}{{4\pi }}\]
Which can be simplified to
$\therefore \phi = \dfrac{3}{{2{\varepsilon _0}}}$
Note: To answer such questions, we must be familiar with the concepts of solid angles and their applications in determining the electric fluxes due to different charge configurations. This will impact the flux as we will only count the flux that passes between the inner and the outer radius region of the ring.