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Electric Flux is the rate of flow of an electric field through an area. Electric flux is proportional to the number of electric field lines passing through a virtual surface.
- Flux refers to the presence of a force field in a physical medium.
- In electronics, flux refers to an electrostatic field and any magnetic field.
- Flux is shown as "lines" in a plane that contains or intersects electric charge poles or magnetic poles.
- The total number of electric field lines passing a given area in a unit of time is defined as the electric flux.
Similar to the example above, if the plane is normal to the flow of the electric field, the total flux is given as:
| \(\begin{array}{l}\phi _{p}=EA\end{array}\) |
The S.I. unit of electric flux is Volt Metres (V m). The dimensions of the electric flux physics are NC−1 m2 or Kg m3 s−3 A−1.
Read Also: Electric Charges and Fields: Important Questions
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Key Terms: Electric Flux Formula, Electric Field, Net Electric Flux, Newton, Coulomb’s Law, Electrostatics, Gauss Law, Electric Charge
What is Electric Flux?
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In Electrostatics, Electric Flux is described as several electric field lines, passing per unit area. It is an additional physical quantity that is used to measure the strength of an electric field and build the fundamentals of electrostatics in the world of Physics. Electric Flux can also be described as the product of the electric field and surface area projected in a direction perpendicular to the electric field.
Explanation of Electric Flux
Assume a liquid flowing with velocity ‘v’, over a very small surface dS, in normal direction with respect to the surface. The rate of flow of water is calculated as volume passing through the given area per unit time which is v dS.
Volume = Area × Length
Differentiating the given relation with respect to time, we have: (dVolume/ dt) = dS × (dlength/ dt)
Flow of liquid = dS × v
Where,
- dS = surface area perpendicular to the velocity v of water
- v = velocity of liquid (calculated as the rate of change of length per unit time)
This flow of liquid, which is given by the quantity v dS, is called flux. If there is some angle θ between the velocity of liquid and surface area dS and flux of liquid, can be calculated from the given relation:
Flow of liquid = v dS cos θ
If we put in place an electric field with flowing water, we define it as electric flux.
Because the strength of the electric field is directly proportional to the number of lines passing per unit area, electric flux also indicates the strength of the electric field. The S.I unit of electric flux is given in Newton meters squared per coulomb. The electric field stands for the symbol Φ (phi) and is defined by: Φ = E Δ S
Terms Related to Electric Flux
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| Important Concepts Related to Electric Flux | ||
|---|---|---|
| Continuous Charge Distribution | Unit of Electric Field | Physical Significance of Electric Field |
| Charge density formula | Parallel Combination of Resistances | Electric Charges and Fields |
Electric Flux Formula
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The total number of electric field lines passing a given area in a unit of time is described as electric flux. Very much like the example above, if the given plane is normal to the flow of the electric field, the total flux is defined by the following electric flux equation:
| \(\begin{array}{l}\phi _{p}=EA\end{array}\) |
On tilting this plane at an angle of θ, the calculated area is given by Acosθ and the total flux over this surface is given as:
| \(\begin{array}{l}\phi=EAcos\theta\end{array}\) |
Where,
- E = magnitude of the electric field
- A = area of the surface through which the electric flux is to be determined
- Θ = angle made by the plane and the axis parallel to the direction of flow of the electric field
Electric Flux Detailed Infograph
Solved Example of Electric Flux FormulaExample: A uniform electric field with a magnitude of E = 400 N/C incident on a plane with a surface of area A = 10m2 and makes an angle of θ = 30∘ with it. Find the electric flux through the surface. Solution: Electric flux is defined as the amount of electric field passing through a surface of area A with the formula \(\Phi_e=\vec{E} \cdot \vec{A}=E\, A\,\cos\theta \) Where “\(\cdot\)” is the dot product between the electric field and area vector and \(\theta\) is the angle between \(\vec{E}\) and the normal vector (a vector of magnitude one and perpendicular to the surface) to the plane. In this problem, the angle between the electric field and normal to the plane is 30∘ so we get: \(\begin{align*} \Phi_e &=EA\,\cos \theta\\&=400\times 10\times \underbrace{\cos 30^\circ}_{\sqrt{3}/2}\\&=\boxed{2000\sqrt{3}\quad {\rm N \cdot m^2/C}}\end{align*}\) |
Observations in Electric Flux
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Electric Flux is a variable quantity as it relies on the direction of the electric field and the orientation of the planar object. Assume one is asked to pass out water through a ring, on which position does he/she expect the maximum flow of water? This happens as the area of the ring becomes perpendicular to the flow of water. The same thing is true for electric flux.
Electric Flux can also be defined as:
| “The rate of flow of electric field via a given area, which varies directly with the number of electric field lines passing through a virtual surface.” |
A few observations related to electric flux are as given below:
- Electric flux becomes zero, when the electric field is parallel to the surface area ΔS, as the angle becomes 90°and the value of cos 90° is zero
- The direction of ΔS is calculated by area vector, as it has both magnitude and direction. The area vector’s direction is constantly normal and out of the plane surface.
- Electric flux turns negative when the electric field and area vector is antiparallel
- The direction of the area vector is constantly out of the surface.
- If there is a curved surface, we calculate ΔS by dividing a large surface area into short segments and integrating it with proper limits.
Frequently Asked QuestionsQues. What is going to happen to the flux due to a charge present within the surface, in case the charge is placed outside of a closed surface? (1 mark) Ans. Gauss’s principle claims that the net electric flux which is coming out of the surface is always going to be 0 only in case there is no charge bound within that surface. The charge is going to generate no field lines due to the fact that it is kept outside of the surface. This means that there will be no flux inside the surface. |
Applications of Electric Flux
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Electric Flux constitutes the fundamentals of electrostatics in the world of Physics and has varied applications such as:
- Electric Flux helps in the determination of electric field
- Electric Flux facilitates the evaluation of the electric field in complex figures
- Gauss Theorem, one of the most useful theorems of electrostatics, depends primarily on electrostatics
Gauss Law Explanation
Gauss Law Video
Handwritten Notes on Electric Flux ( Electric charges and Fields)
Important Handwritten Notes on Electric Flux
SI Unit of Electric Flux
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Talking about the unit, the SI base unit of electric flux is voltmeters (V m) which is also equal to newton-meters squared per coulomb (N m2 C-1). Besides, the base units of electric flux are kg·m3·s-3·A-1.
| Electrical Flux SI Unit: Voltmeters (V m), or, N m2 C−1 |
Dimensional Formula of Electric Flux
In electromagnetism, electric flux is the measure of the distribution of the electric field through a given surface. The dimensional formula of electric flux is: [M L3 T−3 A−1]
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Things to Remember
- Electric field lines help in describing the basic nature of the electric field in two-dimensional pictorial forms. To demonstrate the electric field in a 3-D formation the concept of electric flux is used.
- In Electrostatics, Electric Flux is described as several electric field lines, passing per unit area. It is an additional physical quantity that is used to measure the strength of an electric field and build the fundamentals of electrostatics in the world of Physics.
- It can also be described as the product of the electric field and surface area projected in a direction perpendicular to the electric field.
- Electric Flux is a variable quantity as it relies on the direction of the electric field and the orientation of the planar object.
- Observations related to electric flux include that Electric flux becomes zero, when the electric field is parallel to the surface area ΔS, as the angle becomes 90°and the value of cos 90° is zero.
- Electric flux turns negative when the electric field and area vector is antiparallel. The direction of the area vector is constantly out of the surface.
- Electric Flux helps in the determination of electric fields. Electric Flux facilitates the evaluation of the electric field in complex figures.
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Sample Questions
Ques: What is the basic nature of an electric field line? (1 mark)
Ans: An electric field line comes into origination on a positive electric charge, ending on a negative charge further.
Ques: What is the relationship between electric flux and gauss’s law? (1 mark)
Ans: The mathematical relationship which is established between an enclosed charge and electric flux is known as Gauss’s law (it is in the case of an electric field).
Ques: When its radius is increased, how does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected? (CBSE 2016) (2 marks)
Ans. According to the Gauss’ law, the electric flux through a closed surface is given by,
\(\begin{array}{l}\oint{\vec{E}.\vec{d}s=\frac{1}{{{\in }_{0}}}q}\end{array}\)
Here, q is the charge enclosed by the gaussian surface
Meaning, on increasing the radius of the gaussian surface, charge q remains unchanged.
Thus, flux through the gaussian surface will not be affected when its radius is increased.
Ques: (i) What is electric flux? What is its SI unit?
(ii) A small metal sphere carrying charge +Q is situated at the centre of a spherical cavity within a large uncharged metallic spherical shell as shown in the figure given below. Use Gauss’ law to find the expressions for the electric field at points P1 and P2. (2 marks)
Ans. (i) Electric flux over an area in an electric field represents the total number of electric lines of force that cross the area in a direction normal to the plane of the area. The S.I. unit of electric flux is N-m2/C.
(ii) Using Gauss’ theorem,
The field at point P2 = 0 as the electric field inside the conductor is zero.
Ques: A charge q is placed at the centre of a cube of side L. What will be the electric flux passing through each face of the cube? (All India, Foreign 2010) (3 marks)
Ans. By Gauss’ theorem, the total electric flux attached with a closed surface is given by,
Where, q = the total charge enclosed by the closed surface.
Therefore, total electric flux linked with cube,
As charge is at centre, thus the electric flux is symmetrically distributed through all 6 faces.
Ques: The following figure shows three charges, +2q, -q and +3q. The two charges +2q and -q are enclosed within the surface S. What is the electric flux through the surface S due to this configuration? (Delhi 2010) (3 marks)
Ans. Electric flux through the closed surface S is,
Charge +3q is outside the closed surface S due to which it would not be taken into consideration in applying Gauss’ theorem.
Ques: A thin straight infinitely long conducting wire that has charge density λ is enclosed by a cylindrical surface of radius r and length l and its axis coinciding with the length of the wire. Obtain the expression for the electric flux through the surface of the cylinder. (All India 2011) (3 marks)
Ans. A thin straight infinitely long conducting wire is a uniform linear charge distribution. Let us consider that q charge be enclosed by the cylindrical surface.
Charge enclosed by the cylindrical surface,
…. (i)
By Gauss’ theorem,
Therefore, total electric flux through the surface of cylinder,
Ques: Find the flux of a uniform electric field held E = 5 x 103 i N/C through a square of 10 cm on a side whose plane is parallel to the YZ plane. What would be the flux through the same square if the plane makes an angle of 30 degree with the X-axis? (Delhi 2014) (5 marks)
Ans. Given, the electric field intensity
E = 5 x 103 i N/C
The magnitude of electric field intensity
E = 5 x 103 i N/C
Sides of square, S = 10cm = 0.1m
Area of square, A = (0.1)2 = 0.01m2
The plane of the square is parallel to the YZ-plane … (1)
Therefore, the angle between the unit vector normal to the plane and the electric field is zero.
That is, θ = 0°
Therefore, flux through the plane is,
Ques: A hollow cylindrical box of length 1m and area of cross section 20 cm2 is placed in a three dimensional coordinate system as shown below. The electric field in the region is given by ? E = 50 xi, where E is in NC-1 and x is in metre. Find
(i) net flux through the cylinder
(ii) charge enclosed by the cylinder. (Delhi 2013) (5 marks)
Ans.
As the electric field is only along the X axis, therefore the flux will pass only through the cross section of the cylinder.
Magnitude of electric field at cross section A,
Magnitude of electric field at cross section B,
The corresponding electric fluxes are,
(ii) Using Gauss’ law,
Thus, the charge enclosed by the cylinder is 1.1 x 10-12 C.
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