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Gauss law, in a closed surface, shows that the net flux of an electric field is directly proportional to the enclosed electric charge. Gauss Law claims that a closed surface's total electric flux is equivalent to the enclosed charge of that surface divided by permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field. Gauss law formula can be given by:
⇒ ϕ = Q/ϵ0
Here,
- Q = Total charge within the given surface
- ε0 = Electric constant
Gauss law, in a closed surface, indicates that the net flux of an electric field is directly proportional to the enclosed electric charge. Gauss Law for magnetism is considered one of the four equations of Maxwell’s laws of electromagnetism. It was first formulated by Carl Friedrich Gauss in 1835.
It connects the electric fields at the points on a closed surface and its enclosed net charge. Gauss Law is studied in relation to the electric charge along a surface and the electric flux. Gauss’ law can be derived from Coulomb's law and vice versa.
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Key Terms: Gauss Law, Electric Flux, Gaussian Surface, Maxwell’s Laws, Coulomb's Law, Electric Charge, Charge
What is Gauss Law?
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Gauss law, in a closed surface, indicates that the net flux of an electric field is directly proportional to the enclosed electric charge.
Gauss Law
- Gauss law was articulated by Carl Friedrich Gauss, who was a German mathematician, in the year 1835, and is one among the four equations of Maxwell’s laws.
- As per Gauss law, the total flux linked with a closed surface is 1/ε0 times the enclosed charge by it. Thus,
| \(\oint{\vec{E}.\vec{d}s=\frac{1}{{{\in }_{0}}}q}\) |
- Gauss law associates electric fields at the points on a closed surface and the net charge enclosed by that surface.
- The total flux of an electric field enclosed in a closed surface is directly proportional to the electric charge enclosed in the particular surface.
- The net flux of the electric field through the given electric surface, divided by the enclosed charge should be a constant. Gauss’s law is true for any closed surface, regardless of its shape or size.
- It includes the sum of all charges enclosed by the surface and these charges may be situated anywhere inside the surface.
The video below explains this:
Gauss's Law in Magnetism Detailed Video Explanation:
Gauss Law Explanation
Gauss Law Video
Gaussian Surface
In the case when there are some charges inside and some outside the enclosed surface, the electric field is calculated due to all the charges, both inside and outside. (The term Q, which is denoted on the right side of Gauss’s law, however, represents only the total charge inside the enclosed surface and not outside.)
- The surface to which Gauss’s law is applied is called the Gaussian surface.
- Gauss’s law can be applied to any surface, given that the Gaussian surface does not pass through any discrete charge.
- This is likely because the electric field present due to a system of discrete charges is not well defined at the location of any charge (moving near the charge, the field grows without any bounds).
- However, it can be said that the Gaussian surface can pass through a continuous charge distribution.
Gaussian Surface
Gauss law is easier to calculate the electrostatic field when the system has some symmetry. The choice of a suitable Gaussian surface can facilitate it. Gauss law has an inverse square relation based on the distance comprised in Coulomb's law.
Also Check: Verify the laws of parallel combination of resistances using a metre bridge experiment
Gauss Law Formula
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According to the Gauss theorem, the total charge enclosed in a closed surface is in proportion to the total flux of the surface. Thus, if ϕ is total flux and ϵ0 is electric constant, then the total electric charge Q which is enclosed by the surface can be represented as, Q = ϕ ϵ0
Gauss law formula can be denoted by:
| ϕ = Q/ϵ0 |
Here,
- Q = Total charge within the given surface
- ε0 = Electric constant
Read More:
| Related Concepts | ||
|---|---|---|
| Electrostatic Forces | Force multiple charges | Application of Gauss Law |
| Electrolytic Capacitor | Dielectric Constant | Charge transfer |
| Displacement Current | Gaussian Surface | Circuit Diagram |
Gauss Theorem
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As per Gauss theorem, the net flux passing via a closed surface is in direct proportion to the net charge in the volume enclosed by it.
Gauss Theorem
Thus, Φ = → E.d → A = qnet/ε0.
- Gauss theorem corresponds to the ‘flow’ of electric field lines (flux), within a closed surface, to the charges.
- Assuming that the charges are enclosed by a surface, the net electric flux will be zero.
- Thus, the number of electric field lines that enter the surface is equivalent to the field lines exiting the surface.
Gauss theorem statement claims an important corollary as well:
- In any closed surface, the electric flux is only due to the sources (positive charges) and sinks (negative charges) of the given electric fields that are enclosed by it.
- The charges outside the surface do not contribute to the electric flux.
- It is only the electric charges that can serve as sources or sinks of the electric fields.
- Hence, the changing magnetic fields cannot function as sources or sinks of electric fields.
Note: Gauss law is considered a form of restatement of Coulomb's law. In case Gauss theorem is applied to a point charge enclosed by a sphere, Coulomb’s law can be easily obtained.
Solved ExamplesQues. Can Gauss’ law be applied to all surfaces? Ans. Gauss law is considered valid for any closed surface and for any distribution of charges. Ques. An enclosed gaussian surface is placed in the 3D space where its electrical flux is going to be measured. It has been mentioned that the gaussian surface is spherical and is enclosed by 30 electrons. It also has a radius of 0.5 meters. Thus determine the electric flux that passes through the surface. Ans. As per the question, we can say that the net charge enclosed in the surface can be calculated using the formula of electric flux. It can be obtained using charge multiplication for the electron with electrons appearing on the surface. Thus, by means of it, free space permittivity and the electric flux can be shown. Hence, ⇒ Ф = Q/є0 = [30 (1.60 * 10-19)/8.85 * 10-12] = 5.42 * 10-12 Newton*meter/Coulomb |
What is Electric Flux?
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Electric Flux can be defined as:
- The electric flux for a given area is given by the multiplication of the electric field passing through the given area and the area of the surface in a plane perpendicular to the field.
- The electric field of the surface is calculated by applying Coulomb’s law, but to calculate the electric field distribution in a closed surface, Gauss law is used. It gives the electric charge enclosed in a closed surface.
Relation Between Electric flux and Gauss law
Assuming the relation between electric flux and Gauss law, the law expresses that the net electric flux within a closed surface must be zero, considering that the volume by the surface contains a net charge.
To establish the relation, we will first take a look at the Gauss law.
Gauss law can be denoted as:
ΦE = Q/εo
Where,
- ΦE = Electric Flux via a closed surface enclosing a volume.
- Q = Total Charge enclosed within volume
- εo = Electric Constant
Meanwhile, the electric flux ΦE can be denoted by:
ΦE = ∫∫ E . dA
Where,
- E = Electric Field
- dA = Vector denoting Infinitesimal Element of Area (surface)
Thus, flux is known to be an integral of the electric field and this form of Gauss law is termed as the integral form.
Read More: Unit of Electric Field
Gauss Law Equation
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Gauss Law can be represented using the following integral equation:
| ∫E⋅dA = Q/ε0 … (1) |
Here,
- E = Electric field vector
- Q = Total electric charge enclosed inside the surface
- ε0 = Electric permittivity of free space, and
- A = Outward pointing normal area vector
Flux can be defined as the measure of the field’s strength passing via a surface. Electric flux can be represented as:
| Φ = ∫E⋅dA … (2) |
Electric field can be considered as flux density. Gauss law indicates that the net electric flux via a given closed surface is zero, until and unless the volume enclosed by that surface comprises a net charge.
Gauss law for electric fields can be understood by neglecting electric displacement (d). In matters, the dielectric permittivity might not be equivalent to the permittivity of free space (which is, ε≠ε0). The density of electric charges can further be segregated into a “free” charge density (ρf) and a “bounded” charge density (ρb). Thus, Ρ = ρf + ρb
Applications of Gauss Law
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Some of the applications of Gauss law are:
- For an infinite line of charge which is at a distance ‘r’, E = (1/4 × πrε0) (2π/r) = λ/2πrε0, where λ denotes the linear charge density.
- The intensity of the electric field near a plane sheet of charge is given by E = σ/2ε0K, where σ denotes the surface charge density.
- The intensity of the electric field near a plane-charged conductor in a medium of dielectric constant K, is given by E = σ/Kε0. If the dielectric medium is air, then the equation is given by Eair = σ/ε0.
- The electric field between two parallel plates of a condenser is given by E = σ/ε0, where σ denotes the surface charge density.
Electric Field due to Infinite Wire (Gauss Law Application)Assume an infinitely long line of charge that has a charge per unit length being λ. The cylindrical symmetry of this situation can be considered. By means of symmetry, the electric fields of the points radially move away from the line of charge, with no component parallel to the line of charge. Thus, a cylinder can be used (with an arbitrary radius (r) and length (l)) on the centre of the line of charge of the Gaussian surface. The electric field is found to be perpendicular to the curved surface of the cylinder. Hence, the angle that forms between the electric field and area vector remains zero and cos θ = 1. The top and bottom surfaces are parallel to the electric field, thus forming an angle between the area vector and the electric field at 90 degrees, with cos θ = 0.
Electric Field due to Infinite Wire Hence, the electric flux is due to the curved surface. As per Gauss Law, Φ = → E.d → A ⇒ Φ = Φcurved + Φtop + Φbottom ⇒ Φ = → E . d → A = ∫E . dA cos 0 + ∫E . dA cos 90° + ∫E . dA cos 90° ⇒ Φ = ∫E . dA × 1 As per the radial symmetry, the curved surface can be found in an equidistant condition from the line of charge, with the electric field on the surface having a constant magnitude. Thus, ⇒ Φ = ∫E . dA = E ∫dA = E . 2πrl The enclosed net charge of the surface: ⇒ qnet = λ.l Now, after applying the Gauss theorem, ⇒ Φ = E × 2πrl = qnet/ε0 = λl/ε0 ⇒ E × 2πrl = λl/ε0 ⇒ E = λ/2πrε0 |
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Things to Remember
- Gauss Law refers to the total flux of an electric field surrounded in a closed surface directly proportional to the electric charge enclosed in the particular surface.
- The net flux of the electric field when it moves through the given electric surface and is divided by the enclosed charge must be a constant.
- Gauss law is true for any closed surface matter irrespective of its shape or size.
- Gauss law includes the sum of all charges enclosed by the surface and these charges may be situated anywhere inside the surface.
- The equation for Gauss Law is ∫E⋅dA=Q/ε0.
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Sample Questions
Ques: Who was the German mathematician credited with formulating Gauss’ law? (1 mark)
Ans: The Gauss law was articulated by Carl Friedrich Gauss in the year 1835.
Ques: What is electric flux? (1 mark)
Ans: The electric flux is described as the electric field passing through a given area multiplied by the area of the surface in a plane perpendicular to the field.
Ques: What is the relation of Gauss’ Law to Coulomb's law? (1 mark)
Ans: Gauss law has an inverse square relation based on the distance contained in Coulomb's law.
Ques: What is the integral equation given for Gauss’ Law? (1 mark)
Ans: The Gauss Law is given by the following integral equation:
∫E⋅dA=Q/ε0,
Where E is the electric field vector, Q is the total electric charge enclosed inside the surface, ε0 is the electric permittivity of free space, and A is the outward pointing normal area vector.
Ques: What is the main assertion of Gauss' law? (1 mark)
Ans: The total flux of the electric field through the given electric surface, divided by the enclosed charge should be a constant. It is directly proportional to the electric charge enclosed in the surface.
Ques: What is a Gaussian surface? (1 mark)
Ans: The Gaussian surface is the surface we choose for the application of the Gauss law.
Ques: What happens to the electric field in case the charges are inside as well as outside? (1 mark)
Ans: In case the surface is so chosen that there are some charges inside and some outside, the electric field is due to all the charges, both inside and outside S.
Ques: Does Gauss’ law depend on the shape or size of the surface? (1 mark)
Ans: No, Gauss' law is true for any closed surface, irrespective of its shape of size.
Ques: The surface charge density of a large plane charge sheet is σ = 2.0 × 10-6 C-m-2 on an X-Y plane. Determine the flux of the electric field via a circular area with a radius of 1 cm lying in the region where x, y, and z is found to be positive with its normal, forming an angle 600 with the Z-axis. (3 marks)
Ans: Electric field near the plane charge sheet = E = σ/2ε0 (away from the sheet)
Area = πr2
= 3.14 × 1 cm2
= 3.14 × 10-4 m2.
As is given, the angle formed between the normal to the area and the field = 600.
Hence, as per the Gauss theorem, the flux = \(\vec{E}.\Delta \vec{S}\)
Thus, E.ΔS cos θ
= σ/2ε0 × pr2 cos 60º
= \(\frac{2.0\times10^{-6}C/m^{2}}{2\times8.85\times10^{-12}C^{2}/N-m^{2}}\times(3.14\times10^{-4}m^{2})\frac{1}{2}\)
= 17.5 N-m2C-1
Ques: A particle of mass 5 × 10-6g has been placed over a horizontal sheet of charge with density 4.0 × 10-6 C/m2. Determine the charge that should be provided to the particle such that, if released, it doesn’t drop down. Also mention the number of electrons that are to be removed to obtain the charge. How much mass can be decreased after removing the electrons? (3 marks)
Ans: Electric field in the sheet’s front, E = σ/2ε0
= (4.0 × 10-6)/(2 × 8.85 × 10-12)
= 2.26 × 105 N/C
Considering a charge q is allotted to the particle, then the electric force qE functions in an upward direction, thus balancing the weight of the particle in case:
⇒ q × 2.26 × 105 N/C = 5 × 10-9 kg × 9.8 m/s2
or, ⇒ q = [4.9 × 10-8]/[2.26 × 105]C = 2.21 × 10-13 C
One electron charge is = 1.6 × 10-19C.
Hence, the total number of electrons that should be removed,
= [2.21 × 10-13]/[1.6 × 10-19] = 1.4 × 106
Therefore, the decreased mass after removing the electrons = 1.4 × 106 × 9.1 × 10-31 kg
= 1.3 × 10-24 kg.
Ques: How is an appropriate Gaussian Surface for different cases is chosen? (3 marks)
Ans: In order to choose an appropriate Gaussian Surface, the different cases to keep in mind are:
- Spherical: Charge distribution is spherically symmetric.
- Cylindrical: Charge distribution is cylindrically symmetric.
- Pillbox: Charge distribution having translational symmetry along a plane.
Ques: Define electric flux and write its SI unit. The electric field components in the figure shown are : Ex = x, Ey = 0, Ez = 0 where = 100 N Cm. Calculate the charge within the cube, assuming a = 0.1m. (2018) (3 marks)
Ans: Electric flux refers to the total number of lines in the electric field that intersects an area in the electric field.
Φ = E. ds
SI unit = Nm²/C
Provided, Ex = α X
α = 100
Ex = 100x
Total flux,
Φ = -100 * a * (a²) + 100 * 2a(a²)
= 100a³
Φ = q/ε0
q = Φ. ε0 = 8.85 x 10¹² x100(0.1)³
= 8.85 x 10-¹³ C
=0.885pC
Ques: a) Use Gauss’s theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density.
b) An infinitely large thin plane sheet has a uniform surface charge density +. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. (2017) (3 marks)
Ans: Let's consider A as the gaussian surface’s intersectional cylinder.
Φ = E * 2A
(The flux can pass through only 2 circular intersection points of the cylinder)
According to the Gauss law: Φ = q/ε0
E x 2A = 
b) E is considered as the electric field = σ/2ε0 is fixed
The work done: W = Fd
Displacement D = r
Therefore w = qEd = σqd/2ε0
Ques: What is the electric flux through a cube of side 1 cm which encloses an electric dipole? (2015) (1 mark)
Ans: Since the total charge present in the closed surfaces bound by a cube is equal to 0 (dipole has equal and opposite charges), according to the Gauss law we can say that the total flux through the cube is equal to 0.
Ques: Given a uniform electric field N/C, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What could be the flux through the same square if the plane makes a 300 angle with the x-axis? (2014) (2 marks)
Ans: Flux through the surface is represented by:
Φ = E.dS
Area S = EAcos(90) = 0.1 * 0.1 = 0.1m2
Φ = 5 * 103 * 0.01 = 50Nm2C-1
When an angle of 300 is formed with the square with an a-axis, the angle formed by the regular electric field is 90o - 30o = 60o
Hence Φ = EScos(60) = 25Nm2C-1
Ques: State Gauss’s law for magnetism. Explain its significance. Write the four important properties of the magnetic field lines due to a bar magnet. (2019) (5 marks)
Ans: Gauss’s law states that the magnetic flux in any magnetic field accounts for 0 and the number of lines in the magnetic field that enters any closed surface is equal to the number of lines in the magnetic field which is leaving the enclosed surface.
The significance of Gauss's law refers to the fact that there is an absence of magnetic monopoles because when it comes to monopoles the magnetic flux is not equal to 0. In the case of the dipole, any enclosed surface has the magnetic flux approaching the inward direction to the south pole and equal flux approaching the outward direction to the north pole.
Properties of the magnetic field lines due to a bar magnet refer to the following:
- The lines in the magnetic field are in the form of closed continuous curves.
- Two lines in the magnetic field cannot intersect.
- Tangent to the lines of the magnetic field at any given point provides the direction to the strength of the magnetic field at that point.
- The magnetic field lines that are large and in close proximity represent stronger magnetic fields and vice versa.
Ques: Write three points of differences between para-, dia- and ferromagnetic materials, giving one example for each. (2019) (5 marks)
Ans: Three differences between paramagnetic, diamagnetic, and ferromagnetic materials are:
Paramagnetic:
- The materials which are weak in getting attracted to a magnet are known as paramagnetic materials.
- The paramagnetic materials are small and positive when it comes to their magnetic response and the proximity between the lines in the field increases inside the material.
- When the temperature rises the paramagnetic nature of the material falls.
- Examples: Calcium, aluminium, sodium, etc.
Diamagnetic:
- The diamagnetic materials are weak in repulsion to a magnet.
- The diamagnetic materials are small and negative when it comes to their magnetic response and the proximity between the lines in the field decreases inside the material.
- Every material has the presence of diamagnetism and they act as superconductors at low temp.
- Examples: Copper, Bismuth, Silicon, etc.
Ferromagnetism:
- Ferromagnetic materials are the ones that are powerfully magnetized when kept in the external magnetic field.
- The ferromagnetic materials are large and positive when it comes to their magnetic response and the proximity between the lines in the field increases inside the material.
- With a drop in the temp, the ferromagnetism also falls. They turn paramagnetic when placed at Curie temp.
- Examples: Nickel, Iron, and Cobalt.
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