Electric Flux - Definition, Formula, FAQs

Electric flux measures how an electric field passes through an area that was defined already. Electric flux is the rate of flow of electric lines of force across a particular area chosen as a surface . This point is essential in emsmagnetics because it enables us to see what happens to the electric fields in different situations. The total electric flux through any closed surface equals the charge enclosed within it.

Electric flux within electromagnetism is very important because it helps in proper comprehension of how electric fields come into contact with or affect surfaces. The value of electric flux passing through specific areas enables us to evaluate the magnitude as well as pattern of an electric field. Since this principle is fundamental for solving complicated problems that incorporate field propagation and fee distribution electric flux mastery is essential. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of nine questions have been asked on this concept.

What is Electric flux?

To understand the concept of electric flux let us take an example

Let us consider an electric field that is uniform in both magnitude and direction. Let these field lines penetrate a rectangular surface of area A whose plane is oriented perpendicular to the field lines. As the number of of lines per unit area is proportional to the magnitude of electric field. So the total number of lines penetrating is proportional to E.A. The magnitude of this product is known as Electric flux, which is denoted by

ϕE=EA

So,

Electric flux through an area is the number of electric field lines passing normally through the area.

Flux through an area dA is given by

dϕ=E→⋅d→A=EdAcos⁡θ

here, θ is the angle between the area vector and the electric field.

Total flux through area A is

ϕ=∫E→⋅d→A

Flux is a scalar quantity so they can be added algebracally.

The SI unit of flux is volt-meter.

Electric Flux Through a Closed Surface

To understand this concept, let us take an example

Consider a cylindrical surface of radius R, length , in a uniform electric field E. Compute the electric flux if the axis of the cylinder is parallel to the field direction. For this Sol. we can divide the entire surface into three parts, right and left plane faces and curved portion of its surface. Hence, the surface integral consists of the sum of the three terms:

ϕE=∮E⋅dA=∮left end E⋅dA+∮right end E⋅dA+∮curved E⋅dA

Here, left end and electric field is making 180^o and the right end and the electric field is making 0^o . Also one can notice that the curved surface is making 90^o with the direction of electric field. So,

(ϕE)left end =∮left end E→⋅dA→=∮left end E→dA→cos⁡180∘=−EπR2(ϕE)right end =∮right end E→⋅dA→=∮right end E→A→cos⁡0∘=EπR2(ϕE)curved =∮curved surfice E→⋅dA→=∮curved surface EdA(cos⁡90∘)=0 Total flux =(ϕE)right end +(ϕL)left end +(ϕE)curved surface =(+EπR2)+(−EπR2)+0=0

Similarly we can find the electric flux through any closed surface by an electric field.

Solved Examples Based on Electric Flux

Example 1: A cone of base radius R and height h is located in a uniform electric field E→ parallel to its base. The electric flux entering the cone is :
1) 12EhR
2) EhR
3) 2EhR
4) 4EhR

Solution:

Electric field \vec{E} through any area \vec{A} -

ϕ=E→⋅A→=EAcos⁡Θ S.I unit −( volt )m or N−m2c

wherein

Area of Δ facing =12×h×2R

∴ϕ=EhR

Example 2: The electric field in a region of space is given by, E→=E0i^+2E0j^ where E0=100 N/C. The flux of this field through a circular surface of radius 0.02 m parallel to the Y -Z plane is nearly :
1) 0.125Nm2IC
2) 0.02Nm2IC
3) 0.005Nm2IC
4) 3.14Nm2IC

Solution:

E→=E0i^+2E0J^E0=100 W/CE→=100i^+200J^A=πr2=227×0.02×0.02A=1.25×10−3i^ m2∴ New flux ∴ϕ=EAcos⁡θϕ=(100i^+200J^).1.25×10−3i^ϕ=1.25×10−1Nm2/c=0.125Nm2/C

Hence, the answer is the option (1).

Example 3: A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by
1) 2πR2E
2) πR2/E
3) (πR2−πR)/E
4) zero

Solution:

Electric field E through any area A

ϕ=E→⋅A→=EAcos⁡Θ S.I unit −( volt )m or N−m2c

wherein

Flux through surface AϕA=E×πR2 and ϕB=−E×πR2
Flux through curved surfaceC =∫Eds=∫Edscos⁡90∘=0

Total flux through cylinder =ϕA+ϕB+ϕC=0

Example 4: Electric field at a point varies as r0 for
1) An electric dipole
2) A point charge
3) A plane infinite sheet of charge
4) A line charge of infinite length

Solution:

Electric field E through any area A

ϕ=E→⋅A→=EAcos⁡Θ S.I unit −( volt )m or N−m2c

wherein

E=σ(2ε0)

Example 5: Total electric flux coming out of a unit positive charge put in air is:

1) ε0
2) ε0−1
3) (4pε0)−1
4) 4πε0

Solution:

Flux may be Positive Negative or zero

For closed body.

wherein

$\text { Total flux coming out from unit charge }=\vec{E} d \vec{s}=1 / \varepsilon_0=\varepsilon_0^{-1}$

Summary

The number of electric field lines crossing through a specific surface makes up the electric flux. It is related to the intensity of the electric field as well as to the magnitude of the surface area passing through the field lines. Algebraically, it is the scalar product of the surface’s area vector and its electric field. Once the surface has been closed off, the total electric flux entering it is connected to its charge enclosed within the surface by Gauss’ Law.