Learn About Electric Flux
We define the Electric Flux as the rate of flow of the electric field through a given area and it varies directly with the number of electric field lines going via a virtual surface.
Switch on the mosquito repellent, you smell the fragrance after some time. So, here, lines (electric field lines) of fragrance passing through the area of the room are the electric flux.
On this page, you will learn what is electric flux, electric flux definition in detail.
Important Terms Used in the Concept of Electric Flux
Some of the important terms that are used in the concept of electric flux and one must understand these terms in order to understand the concept of electric flux. These terms are as follows
Electric Field: electric field is a field or space around a stable or moving charge in the form of a charged particle or between the two voltages. The other charged objects or particles in this space also experience some force exerted by this field, the intensity and type of force exerted will be dependent on the charge a particle carries.
Electric Charge: It is a physical property of matter that causes it to experience a force while in an electromagnetic field. It also provides the particle to have an electric field of itself. An electron has a charge of -1 while the proton has a charge of +1. Neutrons are neutral with charge equal to the 0 and also do not interfere with any other electric charges.
Electric displacement field: The term electric displacement field or electric induction is a vector field which is used in the Maxwell's equation, and In Physics, it is denoted by ‘D’.
What Is Electric Flux?
In terms of electromagnetism, electric flux is the measure of the electric field lines crossing the surface. Although an electric field cannot flow by itself, it is a way of describing the electric field strength at any distance from the charge creating the field. The electric field E can generate a force on an electric charge at any point in space.
Concept of Electric Flux
In the above text, we understood the electric flux definition. Now, let’s understand the concept of electric flux.
Let’s consider a hypothetical (imaginary) planar element of area ΔS and a uniform electric field exists on the plane surface.
We know that the number of field lines crossing a unit area placed normal to the electric
field E at a point is a measure of the strength of the electric field at that point.
Now, draw a line normal to the surface and call one side of it the positive normal to the surface.
If we place a smaller planar element of an area ΔS normal to \[\vec{E}\] at this point electric field lines crossing this area are proportional to EΔS.
Points to Remember
Please note that it doesn’t mean that the number of field lines crossing this area is equal to E ΔS.
As the orientation of the area element also decides the amount of actual eclectic flux experienced, the area in the electric flux is a vector quantity.
S.I. unit of electric flux is volt metres (V m) and the dimensions of the electric flux are -
\[N C^{-1} m^{2} or Kg m^{3} s^{-3} A^{-1}.\]
Flux of Electric Field
If we tilt the area element by an angle Ө (or we tilt/rotate \[\vec{E}\] with respect to the area element by an angle Ө, the number of electric field lines crossing the area will be smaller.
As the projection of area element normal to \[\vec{E}\] is ΔS CosӨ (or the component of \[\vec{E}\] normal to area element is E CosӨ).
Therefore, the number of electric field lines crossing area ΔS is proportional to the following components in the equation:
Unit of Electric Flux
Dimensional Formula of Electric Flux
What is Flux in Electrical?
Let’s suppose that we draw a vector of the magnitude ΔS along with the positive normal. ΔS is called the area-vector. The equation for the electric flux through a given area is:
Where,
Φ = Electric flux which is proportional to the number of field lines cutting the area element.
Ө is the smaller angle between \[\vec{E}\] and ΔS.
Now, let’s look at the following cases to determine the electric flux at certain angles:
When \[\vec{E}\] is normal to the area element times the magnitude of area element,
Ө = 0°, Φ is maximum. (∵ Cos 0° =1)
When \[\vec{E}\] is along with the area element, Ө = 90°, Φ is zero.
When Ө > 90°, CosӨ is negative,i .e., Φ is negative.
Now, Let’s Solve Some Problems to Understand the Flux of Electric Field:
Question1: Let’s Say that Charge q is Placed at the Centre of a Sphere. Taking Outward Normal as Positive. Determine the Flux of the Electric Field via the Surface of the Sphere Due to the Enclosed Charge.
Solution: Let us take a small element ΔS on the surface of the sphere
(Figure.a). The electric field here is radially outward and has the following magnitude:
= \[\frac{q} {( 4 \times π \times 𝛆o \times r^{2})}\]
Here,
q is the charge inside the sphere
r is the radius of the sphere
𝛆o is the permittivity of free space
As the positive normal is also outward, Ө = 0° and flux via this element are given by:
Δ Φ = \[\vec{E}\].ΔS = E ΔS Cos 0° = E ΔS
=> Δ Φ = \[\frac{q} {(4 \times π \times 𝛆o \times r^{2} \times ΔS)}\]
Summing over all the elements of the spherical surface
Φ = ∑ Δ Φ = q / (4 x π x 𝛆o x r² ∑ ΔS)…(1)
The surface area of the sphere ΔS = 4 x π x r²….. putting this value in eq (1), we get:
= \[\frac{q} { (4 \times π \times 𝛆o \times r^{2} \times 4 \times π \times r^{2})}\]
This is the required equation for the electric flux enclosed in the sphere. It means that the electric flux equation remains the same.
Question2: A rectangular surface with the measurement of sides 10 cm and 12 cm is placed inside a uniform electric field of 22 V m -1, in such a manner that the normal to the surface of the rectangle and the direction of the electric field, makes an angle of 60o. Simply find the flux of the electric field through the rectangular surface.
Solution: Given
The uniform electric field = E = 22 V m -1 and the angle formed between the area vector and the electric field vector is 60o.
The flux of electric field passing through such a rectangular surface can be given by -
Φ = \[\vec{E}\].S = E ΔS cos Ө
ΔS = 10 cm12 cm = 120 cm2 = 1.210 -2 m2
Φ = 221.210 -2 cos 60o.
Φ = 26.410 -2 (½). = 0.132 N m 2 C -1
FAQs on Electric Flux
1. A uniform electric field exists in space. find the flux of this field via a cylindrical surface with the axis parallel to the field.
From Fig.2, look at the small area ΔS on the cylindrical surface.
The normal to the cylindrical area is perpendicular to the axis of the cylinder but the electric field is parallel to the axis of the cylinder and hence the equation becomes the following:
Δ Φ = \[\vec{E}\] . \[\vec{\Delta S}\]
Since the electric field passes perpendicular to the area element of the cylinder, so the angle between E and ΔS becomes 90°. In this way, the equation f the electric flux turns out to be the following:
Δ Φ = \[\vec{E}\] . \[\vec{\Delta S}\] = E ΔS Cos 90°= 0 (∵ Cos 90° = 0)
This is true for each small element of the cylindrical surface. The total flux of the surface is zero.
2. What do you mean by the electric flux in electrical?
Let’s consider two types of electric fields for determining the electric flux in each:
1. Uniform Electric Field
If the electric field is uniform, the electric flux passing through the vector surface area S is:
Where E is the magnitude of the electric field has units of V/m, S is the surface area, and Ө Is the angle between E and the normal to S.
Δ Φ = E.S = E S Cos Ө
2. Non-uniform Electric Field
If the electric field is non-uniform, the electric flux dΦE via a small surface area dS is given by
Δ Φ = E.ds Cos Ө
Here, we see that the electric field E is multiplied by the component of area perpendicular to the field.
3. What are the properties of the electric flux lines?
Electric field or flux lines, are the lines of electric field and apply force on any charge that interests them. Some of the properties of the flux lines are -
1. Flux lines originate from a positive charge and end at a negative charge.
2. The number of flux lines is like the measurement of the strength of the electric field; the more flux lines present in an electric field the more powerful that field will be.
3. All the flux lines are parallel to each other until they terminate at a negative charge.
4. What are the important points to remember in the topic of electric flux?
Listed below are some of the important notes and points to remember regarding the topic of electric flux.
1. The orientation of the area elements impacts the amount of electric flux, hence the area element is a vector quantity.
2. The direction of the vector of area elements, is perpendicular to the surface itself.
3. S.I. unit of electric flux is volt metres (V m) and the dimensions of the electric flux are - Kg m3 s-3 A-1 or NC -1m 2 .
4. In the formula of finding electric flux, Ө is the angle between the E and the area vector (ΔS).
5. What are some of the applications of gauss law?
Some of the major applications of the gauss law are as follows
1. Gauss law can be used to find the electric field of a point charge, infinite line, infinite sheet or infinite sphere of charge.
2. One can also use this law to find the electric flux passing through a closed surface.
3. Formulas to calculate the Electric Field for three different distributions of charges can be derived from the law.
