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Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge. The applications of Gauss Law are mainly to find the electric field due to infinite symmetries such as:
- Uniformly charged Straight wire
- Uniformly charged Infinite plate sheet
- Uniformly charged thin spherical shell
Gauss Law, also known as Gauss’s flux theorem or Gauss’s theorem, can be referred to as the law which explains the relation between electric charge distribution to the resulting electric field. The surface wherein Gauss law is applied is called the Gaussian surface.
The electrical field of the surface is calculated by applying Coulomb’s law, but to calculate the distribution of the electrical field on a closed surface, the Gauss law is required. It explains the electrical charge enclosed in the closed or the electrical charge present in the enclosed closed surface.
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Key Terms: Gauss law, Gauss theorem, Electric charge, Electric field, Electric Flux, Coulomb’s law, Maxwell’s law, Electrostatics
What is Gauss’ Law?
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Gauss Law refers to the total flux present within a closed surface which is 1/ε0 times the total electric charge enclosed by the closed surface.
- For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is \(q \over 6\pi\epsilon_0\).
- Gauss law was first formulated by Carl Friedrich Gauss, a German mathematician in 1835, and is one of the four equations of Maxwell’s laws.
- According to the definition of Gauss law, the following can be expressed:
| \(\oint{\vec{E}.\vec{d}s=\frac{1}{{{\in }_{0}}}q}\) |
Gauss law, in electrostatics, is known to include electric fields at the points on a closed surface and the net charge enclosed by it.
Gauss Law Explanation
Gauss Law Video
- An electric field’s total flux enclosed in a closed surface is in direct proportion to the electric charge enclosed in the particular surface.
- The net flux of the electric field via the given electric surface, divided by the enclosed charge must be a constant.
- Gauss’s law is considered true for any closed surface, despite the shape or size.
Gauss Law
The video below explains this:
Gauss's Law in Magnetism Detailed Video Explanation:
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Applications of Gauss Law
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Gauss's Law can be used to solve complex electrostatic problems involving unique symmetries such as cylindrical, spherical, or planar symmetry. It also helps in the calculation of the electrical field which is quite complex and involves tough integration. The
application of Gauss Theorem can be used to simplify the evaluation of the electrical field in a simple way.Major Gauss law applications are the following:
- Electric field due to a uniformly charged infinite straight wire.
- Electric field due to a uniformly charged infinite plate sheet.
- Electric field due to a uniformly charged thin spherical shell.
Thus, some of the important Gauss Law and its Application are:
Electric Field due to Infinitely Charged Wire
Consider an infinitely long wire with a linear load density of λ and a length of L. To calculate the electrical field, we assume that the Gaussian cylindrical surface is due to wire symmetry.
- As the electrical field E is radial in the direction, the flow through the end of the cylindrical surface will be zero, as the electrical field and the area vector are perpendicular to each other.
- The only flowing electric flux will be through the curved Gaussian surface.
- As the electric field is perpendicular to every point of the curved surface, its magnitude will be constant.
A cylindrical Gaussian Surface of radius r and length l
The surface area of the curved cylindrical surface will be 2πrl. The electric flux through the curve will be: E × 2πrl.
According to Gauss’s Law,
\(\phi = \frac{q}{\epsilon_0}\)
⇒ \(E \times 2 \pi rl = \frac{\lambda l}{\epsilon_0}\)
⇒ \(E = \frac{\lambda}{2 \pi \epsilon_0 r} \)
Vectorically, the above relation is:
\(\overrightarrow{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{n}\)
Where, \(\hat{n}\) is a radial unit vector pointing the direction of the electric field \(\overrightarrow{E}\)
The direction of electric field is radially outward in case of positive linear charge density
- Note 1: The direction of the electric field will be radially outward if linear charge density is positive and it will be radially inward if linear charge density is negative.
- Note 2: We considered only the enclosed charge inside the Gaussian surface
- Note 3: The assumption that the wire is infinitely long is important because, without this assumption, the electric field will not be perpendicular to the curved cylindrical Gaussian surface and will be at some angle with the surface.
Electric Field due to Infinite Plate Sheet
Imagine an infinite plane sheet, with surface charge density σ and cross-sectional area A. The position of the infinite plane sheet is given in the figure below:
Infinite Charge Sheet
The direction of the electric field due to the infinite charge sheet will be perpendicular to the plane of the sheet.
- Let’s consider a cylindrical Gaussian surface, whose axis is normal to the plane of the sheet. The electric field \(\overrightarrow{E}\) can be evaluated from Gauss’s Law.
- According to Gauss’s Law:
| \(\phi = \frac{q}{\epsilon_0}\) |
- From continuous charge distribution charge q will be σ A.
- Talking about net electric flux, we will consider electric flux only from the two ends of the assumed Gaussian surface.
- This is because the curved surface area and an electric field are normal to each other, thereby producing zero electric flux. So the net electric flux will be
⇒ Φ = EA – (– EA)
⇒ Φ = 2EA
Then we can write:
⇒ \(2EA = \frac{\sigma A}{\epsilon_0}\)
The term A cancels out which means the electric field due to the infinite plane sheet is independent of cross-section area A and equal to:
⇒ \(E = \frac{\sigma}{2 \epsilon_0}\)
In vector form, the above equation can be written as
\(\overrightarrow{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{n}\)
Here, \(\hat{n}\) is a unit vector depicting the direction of the electric field perpendicular and away from the infinite sheet.
- Note 1: The direction of the electric field is away from the infinite sheet if the surface charge density is positive and towards the infinite sheet if the surface charge density is negative.
- Note 2: Electric field due to the infinite sheet is independent of its position.
Electric Field due to thin Spherical Shell
Consider the thin spherical shell of the density of the surface charge and the radius "R." By observation, it is obvious that the shell has spherical symmetry. The electric field due to the spherical shell can be evaluated in two different positions:
- Electric Field Outside the Spherical Shell
- Electric Field Inside the Spherical Shell
Thus,
Electric Field Outside the Spherical Shell
To find an electric field outside the spherical shell, we take a point P outside the shell at a distance r from the center of the spherical shell.
- By symmetry, we take a Gaussian spherical surface with radius r and centre O.
- The Gaussian surface will pass through P, and experience a constant electric field \(\overrightarrow{E} \) all around as all points are equally distanced “r’’ from the centre of the sphere.
Diagram of a spherical shell with point P outside
Then, according to Gauss’s Law,
⇒ \(\phi = \frac{q}{\epsilon_0}\)
The enclosed charge inside the Gaussian surface q will be σ × 4 πR2. The total electric flux through the Gaussian surface will be:
⇒ Φ = E × 4 πr2
Then by Gauss’s Law, we can claim:
\(E \times 4 \pi r^2 = \sigma \times \frac{4 \pi R^3}{\epsilon_0}\)
\(E = \frac{\sigma R^2}{\epsilon_0 r^2}\)
Putting the value of surface charge density σ as q/4 πR2, we can rewrite the electric field as
\(E = \frac{kq}{r^2}\)
In vector form, the electric field is
\(\overrightarrow{E} = \frac{kq}{r^2} \hat{r}\)
Where, \(\hat{r}\) is a radius vector, depicting the direction of the electric field.
Note: If the surface charge density σ is negative, the direction of the electric field will be radially inward.
Electric Field Inside the Spherical Shell
Let's take point P inside the spherical shell to evaluate the electric field inside the spherical shell.
- By symmetry, we take again the spherical Gaussian surface passing through P, centred at O and radius r.
- Now, according to Gauss law,
| \(\phi = \frac{q}{\epsilon_0}\) |
The net electric flux will be E × 4 π r2.
But the enclosed charge q will be zero, as we know that surface charge density is dispersed outside the surface, therefore there is no charge inside the spherical shell.
Thus, by Gauss’s Law:
E = 4πr2 = 0
E = 0
Note: There is no electric field inside the spherical shell because of the absence of an enclosed charge
Some Other Applications of Gauss LawSome of the other Gauss Law and its applications are:
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Gauss Law Formula
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As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if Φ is total flux and ε0 is electric constant, the total electric charge Q enclosed by the surface is; Q = Φε0
The Gauss law formula is expressed by:
| \(\Phi = {Q \over \epsilon_o}\) |
- Q = total charge within the given surface,
- ε0 = the electric constant
Gauss Theorem
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Gauss theorem states that the net flux through a closed surface is directly proportional to the net charge in the volume enclosed by the closed surface.
| Φ = → E.d → A = qnet/ε0 |
In simpler words, the Gauss theorem in electrostatics relates the 'flow' of electrical field lines (flux) to charges within the enclosed surface. If there are no charges enclosed to the surface, the net electric flux remains zero. This means that the number of electric field lines entering the surface is equal to the field lines leaving the surface.
Important Result proposed by Gauss Theorem Statement
As per Gauss Theorem Statement:
- The electrical flux from any enclosed surface is due to the sources (positive charges) and sinks (negative charges) of the electrical fields enclosed by the surface.
- Any charges outside the surface do not contribute to the flow of electricity.
- Also, only electrical charges can act as sources or sinks for electrical fields. Changing magnetic fields, for example, cannot act as sources or sinks of electrical fields.
- The net flux for the surface on the left is non-zero as it encloses a net charge. The net flux for the surface on the right is zero since it does not enclose any charge.
Gauss Law in Magnetism
Note: The Gauss law is only a restatement of Coulomb's law. If you apply the Gauss theorem to a point charge enclosed by a sphere, you will get back Coulomb's law easily.
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Things to Remember
- The main application of Gauss law is to know the electric field produced due to: An infinitely charged uniform straight wire, A uniformly charged infinite plate sheet and, A uniformly charged thin spherical shell.
- Gauss law definition explains the electrical charge enclosed in the closed or electrical charge present in the enclosed closed surface.
- The total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface is known as Gauss Law.
- The Gauss law formula is expressed by; Φ = Q/ε0. Where, Q = total charge within the given surface, ε0 = the electric constant.
- Gauss's Law can be used to solve complex electrostatic problems involving unique symmetries such as cylindrical, spherical or planar symmetry.
Previous Year Questions
- A capillary tube of radius R is immersed in water and water rises in it to a height H … (BITSAT 2017)
- An electron falls from rest through a vertical distance h in a uniform and vertically upward-directed electric field E … (NEET 2018)
- A spherical conductor of radius 10 cm has a charge of 3.2 × 10−7 C … (NEET 2020)
- A cone of base radius R and height h is located in a uniform electric field … (JEE Mains 2014)
- A coil of 100 turns and area 2×10−2m2 is pivoted about a vertical diameter in a uniform magnetic field … (MET 1980)
- A point charge +q is placed at the centre of a cube of side L … (BITSAT 2010)
- Charge is a….. [KEAM]
- Two pith balls, each of mass 1.8 g are suspended from….
- A charge +q is placed at origin. There are two concentric spherical surfaces; S1 of radius R and S2 of radius….
- Two charges, each equal to q , are kept at x=−a…. [JEE Main 2013]
- If the electric dipole is slightly rotated from its equilibri
- The resistance of a wire is ′R′ ′ ohm. If it is melted and stretched to ′n′ times its original length, its new resistance will be :…..[NEET 2017]
- If the ammeter has a coil of resistance 480ohm and a shunt of 20ohm , the reading in the ammeter will be….[NEET 2015]
- If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be….[NEET 2003]
- um orientation, then its angular frequency ω is…. [JEE Main 2019]
- the kinetic energy of the particle becomes zero when the distance of the particle from the origin is…. [JEE Main 2020]
- The surface charge density (in Cm−2 ) of the earth is.. [ DUET 2009 ]
- One kilowatt hour is equal to….[NEET 1997]
- How to adjust a system of three identical capacitors to get high electrostatic energy with the given battery? [UPSEE 2016]
- The number of excess electrons on the drop is… [VITEEE 2011]
Sample Questions
Ques. Why is there no electric field inside a spherical shell? (1 Mark)
Ans: The enclosed charge q will be zero, as we know that surface charge density is dispersed outside the surface, therefore there is no charge inside the spherical shell. Therefore, E = 0.
Ques. If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change? (All India 2009, 2008) (1 Mark)
There will not be any change in electric flux through the Gaussian surface.
Ques. What is a Gaussian surface and why do we use it? (2 Marks)
Ans: A Gaussian surface is an imaginary closed surface present in a three-dimensional plane to calculate the flux of a vector field. Gaussian surfaces are used to carry out calculations of an electrically charged physical surface by taking advantage of its symmetries.
Ques. A charge Q is placed at the centre of cube of side L. What is the electric flux passing through each face of the cube? (2 Marks)
Ans: By Gauss law,
\(\phi = \frac{q}{\epsilon_0}\)
Where q is the total charge enclosed by the surface.
Total electric flux of cube = \(\phi = \frac{q}{\epsilon_0}\)
Since the charge q is present at the centre of the cube, electric flux is distributed symmetrically through all 6 faces.
Flux of each face = \(\frac{1}{6} \phi\)
Flux of each face = \(\frac{1}{6} \times \frac {q}{\epsilon_0} = \frac{q}{6 \epsilon_0}\)
Ques. A uniform electric field of magnitude E = 100 N/C exists in the space in X-direction. Using the Gauss theorem calculate the flux of this field through a plane square area of edge 10 cm placed in the Y-Z plane. Take the normal along the positive X-axis to be positive. (2 Marks)
Ans: The flux Φ = ∫ E.cosθ ds.
As the normal to the area points along the electric field, θ = 0.
Also, E is uniform, so Φ = E.ΔS = (100 N/C) (0.10m)2 = 1 N-m2.
Ques. What is the differential form of the Gauss theorem? (2 marks)
Ans. The differential form of the Gauss theorem represents the relationship between the electric field and the distribution of the charge at a specific point in space. According to the law the difference in the electric field (E) will be equal to the volume charge density (p) at a specific point. It can be represented as:
\(\triangle E = \rho/\epsilon_o\)
Here,
\(\epsilon_o\)= permittivity of free space
Ques. i) Define electric flux. Write its SI unit.
ii) Using Gauss’ law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of distance from it.
iii) How is the field directed if
(a) the sheet is positively charged
(b) negatively charged? (Delhi 2012) (3 marks)
ii) Gauss’s theorem states that the total electric flux through a closed surface is equal to the net charge enclosed by the surface.
Consider a uniformly charged infinite plane sheet with charge density \(\sigma\)
Consider a Gaussian surface inside
On applying Gauss's law, we get
\(\varphi = \oint \overrightarrow{E}.\overrightarrow{ds} = {\sigma ds \over \epsilon _o}\)
→ E.ds + E.ds + 0
→ \({\sigma ds \over \epsilon _o} = 2E.ds = {\sigma ds \over \epsilon _o}\)
→ E = \({\sigma \over 2\epsilon _o}\)
Thus, electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed usually away from the charge.
If the surface charge density s is negative, the electric field is directed towards the surface charge.
c) i) Away from the charged sheet.
ii) Towards the plane sheet.
Ques. A charge q is placed at the centre of a cube of side L. What is the electric flux passing through each face of the cube? (All India 2010; Foreign 2010) (3 Marks)
Ans. By Gauss’ theorem, total electric flux linked with a closed surface is given by
Φ = \({Q \over \epsilon_o}\)
where q is the total charge enclosed by the surface
Therefore total electric flux linked with cube:
Φ = \({q \over \epsilon_o}\)
As the charge is towards the centre, hence electric flux is symmetrically distributed through all the 6 faces:
Flux linked with each face = 1/6 \(\epsilon _o\)
= 1/6 x \({q \over \epsilon_o}\)
= \({q \over 6\epsilon_o}\)
Ques. Explain the process of choosing appropriate Gaussian Surfaces for different states. (3 Marks)
Ans: To choose an appropriate Gaussian surface, the ratio of charge needs to be considered. The dielectric constant is presented as a two-dimensional surface integral over the electric field symmetry of charge distribution.
There are three different cases to note:
- Spherical: The distribution of charge has spherical symmetry.
- Cylindrical: The distribution of charge has cylindrical symmetry.
- Planar: Distribution of charge has translational symmetry along a plane.
The surface size can be chosen on the basis of where the field needs to be calculated. Gauss’ Law can be used in locating fields when the symmetry is mentioned as it denotes the field direction.
Ques. How is electric flux related to Gauss law? (3 Marks)
Ans. Gauss law states that the total electric flux will be zero within an enclosed surface if the volume of the surface has some charge. To define the relation we first take Gauss law: ΦE = Q/ε0
Where ΦE refers to the electric flux through closed surface S enclosing any amount of volume V.
Q refers to the total charge closed within volume V
ε0 refers to the electric constant.
Electric flux can be defined as an integral of the electric field surface. It can be stated as: 
Here E is the electric field
dA is the vector that forms the infinitesimal element of the area of the surface.
Therefore flux is referred to as an important part of the electric field. This relation of Gauss law is called the integral form.
Ques. The figure shows three concentric thin spherical shells A, B and C of radii a, b, and c respectively. The shells A and C are given charges q and -q respectively and the shell B is earthed. Find the charges appearing on the surfaces of B and C. (5 Marks)
Ans: As shown in the previous worked-out example, the inner surface of B must have a charge -q from the Gauss law. Suppose, the outer surface of B has a charge q’.
The inner surface of C must have a charge -q’ from Gauss law. As the net charge on C must be -q, its outer surface should have a charge q’ – q. The charge distribution is shown in the figure.
The potential at B,
- Due to the charge q on A = q/4πε0b,
- Due to the charge -q on the inner surface of B = -q/4πε0b,
- Due to the charge q’ on the outer surface of B = q’/4πε0b,
- Due to the charge -q’, on the inner surface of C = -q’/4πε0c,
- Due to the charge q’ – q on the outer surface of C = (q’ – q)/4πε0c.
The net potential is, VB = q’/4πε0b – q/4πε0c
This should be zero as shell B is earthed. Thus, q’ = q × b/c
The charges on various surfaces are as shown in the figure:
Ques. Can the Gauss’ Law be used to find an electric field? (5 Marks)
Ans: Gauss Law determines the electric field of charge distributions symmetrically. The steps to solve the problems of an electric field in regard to Gauss Law is as follows:
- First, the spatial symmetry of the charge distribution needs to be identified.
- Choose an appropriate Gaussian surface with equal symmetry to the charge distribution. Identify the consequences as well.
- Evaluate integral ‘ΦsE’ over Gaussian surface. Then, calculate flux through the surface.
- Then, the amount of charge enclosed by the Gaussian surface needs to be calculated.
- Finally, the electric field of the charge distribution can be calculated.
There are three types of symmetry required to determine the electric field:
- Spherical Symmetry
- Cylindrical Symmetry
- Planar Symmetry
Inappropriate Coordinate Systems are to be calculated with Gaussian surface to achieve any particular symmetry.
Ques. Two large charged plane sheets of charge densities a and -2σ C/m2 are arranged vertically with a separation of d between them. Deduce expressions for the electric field at points
(i) to the left of the first sheet,
(ii) to the right of the second sheet, and
(iii) between the two sheets. (5 marks)
OR
An aspherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
(a) A charge q is placed at the center of the shell. Find out the surface charge density on the inner and outer surfaces of the shell.
(b) Is the electric field inside a cavity (with no charge) zero; independent of the fact whether the shell is spherical or not? Explain. (CBSE Outside Delhi, 2019) (5 Marks)
Ans.
Ques. State and prove the Gauss law theorem. (5 Marks)
Ans. In 1835, Friedrich Gauss stated the relationship between the electric field in an enclosed surface and the total charge which was present in that closed surface. Gauss law theorem defines the distribution of the electric charge as a result of the electric field.
When the surface area is calculated in a plane, perpendicular to the field and is multiplied by the electric field, it is known as the electric flux. It shows that the net electric flux of a particular surface is equal to \(\frac{1}{E\theta}\) times the net charge present within that enclosed surface.
Let’s take q as the point and r as the radius of the observed electric field.
Now let’s take ds as the area and E as the surface electric field.
de = E (vector) ds (vector) \(cos\theta\)
\(\theta = \int \overrightarrow {E.dA}.cos\theta\\ \theta = 0\\ cos 0 = 1\\ \theta= \int \overrightarrow {E.dA} = \overrightarrow{E}.4\pi r^2\)
The electric field at any point ‘P’ is
\(\overrightarrow{E} = {1 \over 4 \pi e \theta} . {q \over r^2}\)
Pointing the field radially outward at point “P”
\(\theta = {q \over e\theta}\)
Hence proved.
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