Electric Flux through a Cone or Disc - JEE Important Topic

Electric Flux is mainly defined as the value of the flow of the electric field among a given area, and it changes its characteristics with the number of lines present in the electric field passing through a virtual surface. For example, when we switch on the mosquito repellent, we can easily smell the fragrance after some interval. So, this process shows that the lines, also known as the electric field lines of fragrance travelling through the space of the room, are known as the electric flux. 

Electromagnetism and Electric Flux

If we consider electromagnetism, electric flux is termed to be the measure of the lines of the electric field that is crossing the particular surface. Although the electric field can’t travel on its own, it is the notion of describing the strength of the electric field at any valuable distance from the charge getting created in the field. The electric field E is able to generate the force on an electric charge at a given point in the space. 

The above information gave us a brief idea about the electric flux. Now let us focus on the concept more prominently. 

Let us imagine a hypothetical planar element which is of the area S, and an electric field exists on the surface of the place uniformly. We got to know that the number of field lines crossing a unit area normally placed to the electric field E at the given point is termed to be the measure of the strength of the electric field at that particular point. Now, draw a line which is normal to the surface and term it as positive normal to the surface. If the placement of the smaller planar element of an area S is normal to E at this particular point, then electric lines of the field passing this area are directly proportional to the dot product of E and S. The S.I. unit of the electric flux is denoted by V-m, known as the Volt metres and the dimension of the electric flux is classified as

\[N{{C}^{-1}}{{m}^{2}}\]  or \[Kg{{m}^{3}}{{s}^{-3}}{{A}^{-1}}\].

Electric Flux

The Formula of the Electric Flux

The formula of the electric flux is

\[{{\Phi }_{E}}=E.S=ES\cos \theta \]

Here, E stands for the electric field, S stands for the area of the surface, E stands for the magnitude of the field, and the other symbols denote the angle present between the electric lines of the field and the normal to S. 

Electric Flux Through a Cone and Disc

For a Cone, there are two surfaces to consider. One is the curved surface and the other is the base which can be considered a circular disc. First consider the curved surface only for flux calculation. Now, consider a charge placed at the middle of the base of the cone. In this case, half of the flux due to the charge passes through the cone while the other half will pass through in the other direction outside of the cone.

So, from Gauss’s Law, we know

\[\phi =\frac{q}{{{\varepsilon }_{0}}}\]

Now, the flux passing through the cone is halved. So,

\[\phi =\frac{q}{2{{\varepsilon }_{0}}}\]

This is the flux passing through the curved surface of the cone. If the net charge enclosed in the volume of a cone is zero, then automatically the flux through the cone will be zero.

For a disc of radius R, let us draw a perpendicular line from the centre of the disc to the point charge in space at a distance a. Now if lines are drawn from all the points on the circumference of the disc, then a solid angle will be formed at the charge position as shown in the figure. 

\[\phi ={{\phi }_{total}}\times \frac{\Omega }{4\pi }\][U1] 

=\[\frac{q}{{{\varepsilon }_{0}}}\times \frac{2\pi }{4\pi }\left( 1-\frac{R}{\sqrt{{{a}^{2}}+{{R}^{2}}}} \right)\]

=\[\frac{q}{2{{\varepsilon }_{0}}}\left( 1-\frac{R}{\sqrt{{{a}^{2}}+{{R}^{2}}}} \right)\]

Relation Between the Flux Through a Cone or Disc 

Here, we are finding the flux through the uncharged disc of the radius R because of the point charge +q, which is at a distance of x from the centre of the disc on the axis.

This problem can be solved by the Gauss law. The points on the periphery of the disc were connected to the point charge to obtain a cone. Now applying the law of Gauss, we need to consider the Gaussian sphere, including +q charge in its centre. Now, the charge q works for the total sphere, but the partition is needed where the disk is present only. Valuing the Gaussian surface, the sphere’s curved part is labelled as S1, linked to the flat disc attached at the end of the cone S2. 

Now, considering the integral surface of flux, this can be classified into the following, 

$\int\limits_{{{S}_{1}}}{\overrightarrow{E}.\overrightarrow{dS}}+\int\limits_{{{S}_{2}}}{\overrightarrow{E}.\overrightarrow{dS}}=0$ 

As mentioned above, the surface, the flat, consists of a normal pointing towards the charge. To flip this equation, add a negative sign, respectively. The result shows that the electric field due to the disc has equal flux passing within them, and this is how we find the flux through the disc. 

Conclusion

Here we learned about the basic concepts and the capacity of the electric flux, which determines the flux as the number of electric field lines passing through a given space at a particular time. The formula of the electric flux will make your concepts clear about the electric field due to the disc and how to derive the mathematical solutions.