Question

As shown in the figure, \[q\] charge is placed at the open end of the cylinder with one end open. What is the total flux emerging from the surface of the cylinder?
 

Answer 1

Hint: First of all, we will find the expression which gives the electric flux through the cylinder which is the summation of the flux through the curved surface and the two flat surfaces. We will find out the total flux linked to the cylinder and the flux linked to one flat surface. Lastly, we will subtract to get the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
A charge \[q\] is placed at the open end of the cylinder with one end open. We are asked to find the total flux emerging from the surface of the cylinder.

To begin with, we know that the charge \[q\] is placed at the centre of the open end of a cylindrical vessel then the amount of charge which will contribute to the flux is half the amount. This is due to the reason that one half will lie inside the surface while the other will be lying outside the surface.
Let us proceed to solve the problem,
We know, the total electric flux passing through the cylinder is given by the summation of the flux through the curved surface and the two flat surfaces.Let the electric flux passing through the curved surface is \[{\phi _1}\] and the electric flux passing through the flat surface is \[\phi \] . Again, the flux passing through the cylinder is \[{\phi _{\text{t}}}\] .
We know,
\[{\phi _{\text{t}}} = \dfrac{q}{{{\varepsilon _0}}}\]
We have, according to Gauss law:
${\phi _{\text{t}}} = {\phi _1} + 2 \times \phi \\
\Rightarrow {\phi _1} = {\phi _{\text{t}}} - 2\phi \\$
Again, we can write flux through one flat surface is:
$\phi = \dfrac{{\left( {\dfrac{q}{{{\varepsilon _0}}}} \right)}}{{4\pi }} \times 2\pi \\
\Rightarrow \phi = \dfrac{q}{{2{\varepsilon _0}}} \\$
To find the electric flux emerging from the surface of the cylinder, we will have to subtract the flux through one flat surface from the total flux linked to the cylinder, which is shown below (since one end is open):
${\phi _1} = {\phi _{\text{t}}} - \phi \\
\Rightarrow {\phi _1} = \dfrac{q}{{{\varepsilon _0}}} - \dfrac{q}{{2{\varepsilon _0}}} \\
\therefore {\phi _1} = \dfrac{q}{{2{\varepsilon _0}}} \\$

Hence, the total flux emerging from the surface of the cylinder is \[\dfrac{q}{{2{\varepsilon _0}}}\].

Note: While solving this problem, most of the students seem to have confusion regarding the flux linked to the flat surface. Since there is one end open, we must consider the one closed end. If you consider both the flat ends, then the result obtained will be zero. This is due to the fact that both the fluxes will cancel out each other leaving no net flux. The total flux depends on the strength of the field, the size of the surface area through which it passes, and how the field is oriented.