Maxwell's equations let us calculate the forces charged particles exert on each other. While Newton's laws only hold for particles moving with speeds much less than the speed of light, Maxwell's equations hold for particles moving with any speed. We say that Maxwell's equations are relativistically correct.
Maxwell's equations are a set of four equations. The first of these equations is Gauss's Law. It states that the electric flux through any closed surface is equal to the total charge inside divided by e0.
In physics, Gauss's law, also known as Gauss's flux theorem, is a law relating the distribution of electric charge]]to the resulting electric field. Gauss's law states that: electric flux through any closed surface is proportional to the enclosed electric charge.
In integral form the equation is:
It is one of the four Maxwell's equations, which form the basis of classical electrodynamics, and is also closely related to Coulomb's law. The law was formulated by Carl Friedrich Gauss in 1835, but was not published until 1867.
Gauss's law has two forms, an integral form and a differential form. They are related by the divergence theorem, also called "Gauss's theorem". Each of these forms can also be expressed two ways: In terms of a relation between the electric field E and the total electric charge, or in terms of the electric displacement field D and the free electric charge.
Gauss's law has a close mathematical similarity with a number of laws in other areas of physics. In fact, any "inverse-square law" can be formulated in a way similar to Gauss's law. For example, Gauss's law itself is essentially equivalent to the inverse-square Coulomb's law, and Gauss's law for gravity is essentially equivalent to the inverse-square Newton's law of gravity. Gauss's law can be used to demonstrate that there is no electric field inside a Faraday cage with no electric charges. Gauss's law is something of an electrical analogue of Ampère's law, which deals with magnetism. Both equations were later incorporated into Maxwell's equations.
Physics textbooks often derive Gauss's law using Coulomb's law for the force between electric charges. However if we consider Gauss's law as a more fundamental law than Coulomb's law, then the fact that electrical forces are inverse square forces can be considered as a consequence of Gauss's law.
The fact that the electric field is zero inside a conducting shell also derives from Gauss's law and is the principle behind the Faraday cage.

Flux There are two common uses of the word flux in Physics. We will use the first, concerned with transport phenomena, to gain an understanding of the second, concerned with electric flux. Imagine a uniform flow of liquid flowing at velocity through a surface of area A - it might look something like Figure 1. The volumetric flow rate, Φ for now, would then depend on the angle between the surface and the fluid flow. From Figure 1 one can see that flow rate Φ can be written in terms of this angle by the following expression:

(Eq 1)
Here the vertical component of the flow (the sin term) was left out, as it is parallel to the surface and thus cannot contribute to the flow through the surface.

Figure 1. A uniform stream of velocity v is flowing through a plane of area A. The component of v perpendicular to the plane is v*cos(Θ), where Θ is the angle between v and a normal (blue) to the plane.

Equation 1 can be rewritten in vector notation. In order to do this we must first define the area vector whose magnitude is the area of our imaginary surface and whose direction is normal to that surface. Then the flux can be written as the scalar (or dot) product of the velocity vector and the area vector as

(Eq 2)
Now with the vector notation of flux and an understanding that the flux represents the flow of something (in our example - the flow of a liquid) under our belts, let us begin to examine the flux of an electric field. Consider the surface in Figure 2. Here an arbitrary Gaussian surface has been immersed in a nonuniform electric field. The surface has been divided into small squares of area ΔA; each small enough to consider it flat and thereby neglect any curvature. Each of the squares has an area vector that lies perpendicular to the Gaussian surface and each square also contains a constant electric field vectorthat points in the direction of the electric field at that point. This statement might seem odd because the Gaussian surface is immersed in a nonuniform electric filed, but since each square is small the electric field can be considered uniform across the square.

Figure 2. A Gaussian surface of arbitrary shape immersed in an electric field. The surface is divided into small squares of area ΔA. The electric field vectors E and the area vectors ΔA for the three representative squares, marked 1, 2 and 3, are shown.

Video 1 demonstrates flux throug an arbitrarily shaped surface, and video 2 demonstrates flux through a closed box.

Now the electric flux Φ can be computed by algebraically summing each of the scalar dot products of the area vectors and the electric field vectors. This is done with Equation 3:

(Eq 3)
This however is not the finial definition of the electric flux. Image now, that we take smaller and smaller areas, or, in other words, divide the Gaussian surface into more pieces. Doing this turns the finite sum of Eq 3 into a cyclic integral because ΔA now approaches a differential limit dA. From this Eq 3 becomes Eq 4:

(Eq 4)

Gauss’ Law

Gauss’ law provides a relationship between the net electrical flux Φ through a closed surface and the net charge qenc enclosed by that surface. Gauss’ law states that the net electric flux out of a closed surface is equal to the charge enclosed by the surface divided by the permittivity:

(Eq 5)
Plugging Eq 5 into Eq 6 and rearranging yields
(Eq 6)
In Eq 5 and 6 the charge enclosed by the surface qenc is simply the algebraic sum of all the charges (positive and negative) within the Gaussian surface. Charges outside the surface are not included in qenc. However, the electric field in Eq 6 is the electric field due to all charges inside or outside the Gaussian surface. All charges can be included in the determination of the electric field because chargers outside the surface contribute zero net flux through the surface – the number of field lines entering the surface due to charge outside the surface equal the number of field line exiting the surface due to the same charge.

Gauss’ and Coulomb’s Relation

In the Electric Charge section we learned about Coulomb’s law which gives the electrostatic force of attraction or repulsion. In this section we have learned about Gauss’ law which relates the net electrical flux through a Gaussian surface to the net charge includes by that surface. However, both Gauss’ law and Coulomb’s law are just different ways of describing the relationship between electric charge and electric fields. Thus reason would say that one should be able to derive one of the laws from the other. Let us examine the derivation of Coulomb’s law from Gauss’ law as this requires less intensive math.

Figure 3 shows a spherical Gaussian surface around a positive point charge. The surface has radius r and has been divided into small differential areas dA. As can be seen from the picture, the area vectors, as well as the electric field vectors, are perpendicular to the surface (and parallel to each other). Thus, the angle Θ between them equals zero and Eq 6 can be written as:

Then, if we recall that the electric field can be considered constant if the areas are small enough, the electric field E can be brought outside the integral. Furthermore, because the only charge enclosed by the surface is q, qenc = q. This simplifies the above expression to:

The integral is now just the sum of all the differential areas of the surface, or the surface area of the sphere (4πr^2): or (Eq 7) which is Coulomb’s law for a point charge.

Figure 3. A spherical Gaussian surface of radius r, centered on a positive point charge q.

Applying Gauss’ Law

Cylindrical Symmetry

There are several situations that can be examined using the cylindrical symmetry approach. They are: a line of charge, a non-conducting cylinder of charge and a conducting cylinder of charge. Let us first look at the infinite line of charge with a uniform linear charge density λ. Consider the cylindrical Gaussian surface in Figure 4. From the figure we can see that the electric fields lines point radially outward and are parallel to the two end caps. The electric flux can then be written as:

or (Eq 8)

Figure 4. A closed cylindrical Gaussian surface, surrounding a section of an infinitely long, uniformly charged, thin rod.

The same expression can be found for a conducting cylinder. The only difference between a line of charge and a conducting cylinder is that an additional assumption is needed. A line of charge is considered infinitely thin, while cylinders and other 3-dimensional objects have volume. Assume r is the radius of the Gaussian cylinder and R is the radius of the charged cylinder. For Gaussian cylinders with r > R (meaning the Gaussian surface is outside the radius of the conductor) the expression is exactly the same. But for r < R the electric field E = 0, since the Gaussian surface is inside the conductor (see Figure 5).

Figure 5. A closed cylindrical Gaussian surface, surrounding a section of an infinitely long, conducting cylinder.

For a charged cylinder and r > R the above equation for the electric field holds true. However, for charged cylinder with r < R the electric field is not equal to zero (see Figure 6). Starting from

and letting r < R, we obtain
(Eq 9)

Figure 6. A closed cylindrical Gaussian surface, surrounding a section of an infinitely long, uniformly charged, nonconducting cylinder.

The following link provides a three dimensional animation of cylindrical symmetry (found here)

Planar Symmetry

Just as there were several cylindrical symmetry approaches to finding the electric field with Gauss’ law, there are several planar symmetric systems. The three discussed here include a sheet of charge, a conducting surface, and parallel conducting plates.

Consider the thin, infinite, nonconducting sheet of charge in Figure 7. The sheet has a uniform surface charge density σ. A closed cylinder with end caps of area A is a useful Gaussian surface in this context. From the figure it can be seen that the electric field is perpendicular to the surface and thus the electric field lines are parallel to the cylinder wall. This means only the end caps contribute to the flux. The charge included by the cylinder is simply, and not surprisingly, the product of the charge density and the cross sectional area of the cylinder.

Figure 7. A closed cylindrical Gaussian surface, stuck through a uniformly charged, infinite, thin, nonconducting sheet.

The electric field due to a conducting surface is similar but is not identical. Consider the conductor or Figure 8. Here a conductor with surface density σ is at equilibrium, or in other words the free electrons in the conductor are not moving. Again, a cylindrical Gaussian surface oriented perpendicular to the surface is used to determine the electric field. From the picture only, the electric flux through the top of the Gaussian surface need be considered as the other end cap is inside the conductor. The flux is then given by:

or (Eq 10)

Figure 8. A closed cylindrical Gaussian surface, emerging from an infinite, conducting plate.

If two oppositely charged conductors are brought together, Gauss' law can be used to calculate the electric field between the two infinite plates. Presuming the plates to be at equilibrium with zero electric field inside the conductors, the result from a charged conducting surface can be used:

However, looking at Figure 9 one can see that the electric field between the plates has a magnitude twice that of one charged sheet. Thus the 2 in the denominator cancels out, leaving us with the equation: (Eq 11)

Figure 9. Electric field between two uniformly and oppositely charged, infinite conducting sheets.

The following link provides a three dimensional view of planar symmetry (found here)

Spherical Symmetry

Using a spherical Gaussian surface we can prove the two shell theorems that were presented in the Electric Charge section. The two theorems state: “a shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell” and “if a charged particle is located inside a shell of uniform charge, there is no electrostatic force on the particle from the shell.”

Looking at Figure 10, the electric field of the conducting sphere of charge Q can be obtained by a simple application of Gauss' law. Consider a spherical Gaussian surface with radius r > R, the electric field has the same magnitude at every point along the surface and is directed outward. The electric flux is then just the electric field times the area of the spherical surface:

or (Eq 12)
For r < R, the Gaussian surface lies within the charged shell. Since all the shell’s charge will reside on the conducting surface, a Gaussian surface at r < R will enclose no charge. Thus E = 0. This proves the above two theorems.

Figure 10. A conducting sphere of radius R and charge Q, surrounded by a spherical Gaussian surface of radius r.

Next, let us examine the electric field due to a sphere of uniform charge. If r < R than not all of the charge would be enclosed by the spherical Gaussian surface (see Figure 11). The charge inside the Gaussian surface with radius r would then be given by the ratio of the volumes:

Noting that the charge outside the Gaussian surface (Q - Q' ) does not contributed to the electric flux, and substituting Q' in place of Q, the electric flux can be written as:

or (Eq 13)

Figure 11. A nonconducting sphere of radius R and uniform charge Q, encapsulating a spherical Gaussian surface of radius r.

Example 1:
An electron is released from rest at a perpendicular distance of 9.0 cm from a line of charge on a very long non-conducting rod. That charge is uniformly distributed, with 6.0 µC per meter. What is the magnitude of the electron’s initial acceleration?
We combine Newton’s second law (F =ma) with the definition of electric field (F=qE) and

To get:

This gives acceleration to be 1.867 X 10^17 m/s^2.

Example 2:

The figure below shows cross sections through two large, parallel non-conducting sheets with identical distributions of positive charge with surface charge density σ = 1.77 X 10^-22 C/m^2. In unit-vector notation, what is E at points (a) above the sheets, (b) between them, and (c) below them?

Maxwell's equations let us calculate the forces charged particles exert on each other. While Newton's laws only hold for particles moving with speeds much less than the speed of light, Maxwell's equations hold for particles moving with any speed. We say that Maxwell's equations are relativistically correct.

Maxwell's equations are a set of four equations. The first of these equations is

Gauss's Law. It states that the electric flux through any closed surface is equal to the total charge inside divided by e0.In physics,

Gauss's law, also known asGauss's flux theorem, is a law relating the distribution of electric charge]]to the resulting electric field. Gauss's law states that: electric flux through any closed surface is proportional to the enclosed electric charge.In integral form the equation is:

It is one of the four Maxwell's equations, which form the basis of classical electrodynamics, and is also closely related to Coulomb's law. The law was formulated by Carl Friedrich Gauss in 1835, but was not published until 1867.

Gauss's law has two forms, an

integral formand adifferential form. They are related by the divergence theorem, also called "Gauss's theorem". Each of these forms can also be expressed two ways: In terms of a relation between the electric fieldEand the total electric charge, or in terms of the electric displacement fieldDand thefreeelectric charge.Gauss's law has a close mathematical similarity with a number of laws in other areas of physics. In fact, any "inverse-square law" can be formulated in a way similar to Gauss's law. For example, Gauss's law itself is essentially equivalent to the inverse-square Coulomb's law, and Gauss's law for gravity is essentially equivalent to the inverse-square Newton's law of gravity. Gauss's law can be used to demonstrate that there is no electric field inside a Faraday cage with no electric charges. Gauss's law is something of an electrical analogue of Ampère's law, which deals with magnetism. Both equations were later incorporated into Maxwell's equations.

Physics textbooks often derive Gauss's law using Coulomb's law for the force between electric charges. However if we consider Gauss's law as a more fundamental law than Coulomb's law, then the fact that electrical forces are inverse square forces can be considered as a consequence of Gauss's law.

The fact that the electric field is zero inside a conducting shell also derives from Gauss's law and is the principle behind the Faraday cage.

FluxThere are two common uses of the word

fluxin Physics. We will use the first, concerned with transport phenomena, to gain an understanding of the second, concerned with electric flux. Imagine a uniform flow of liquid flowing at velocity through a surface of areaA- it might look something like Figure 1. Thevolumetric flow rate, Φ for now, would then depend on the angle between the surface and the fluid flow. From Figure 1 one can see that flow rate Φ can be written in terms of this angle by the following expression:(Eq 1)

Here the vertical component of the flow (the

sinterm) was left out, as it is parallel to the surface and thus cannot contribute to the flowthroughthe surface.Equation 1 can be rewritten in vector notation. In order to do this we must first define the

area vectorwhose magnitude is the area of our imaginary surface and whose direction is normal to that surface. Then the flux can be written as the scalar (or dot) product of the velocity vector and the area vector as(Eq 2)

Now with the vector notation of flux and an understanding that the flux represents the flow of something (in our example - the flow of a liquid) under our belts, let us begin to examine the flux of an electric field. Consider the surface in Figure 2. Here an arbitrary

Gaussian surfacehas been immersed in a nonuniform electric field. The surface has been divided into small squares of area ΔA;each small enough to consider it flat and thereby neglect any curvature.Each of the squares has an area vector that lies perpendicular to the Gaussian surface and each square also contains a constant electric field vector that points in the direction of the electric field at that point. This statement might seem odd because the Gaussian surface is immersed in a nonuniform electric filed, but since each square is small the electric field can be considered uniform across the square.Video 1 demonstrates flux throug an arbitrarily shaped surface, and video 2 demonstrates flux through a closed box.

Now the electric flux Φ can be computed by algebraically summing each of the scalar dot products of the area vectors and the electric field vectors. This is done with Equation 3:

(Eq 3)

This however is not the finial definition of the electric flux. Image now, that we take smaller and smaller areas, or, in other words, divide the Gaussian surface into more pieces. Doing this turns the finite sum of Eq 3 into a cyclic integral because Δ

Anow approaches a differential limitdA.From this Eq 3 becomes Eq 4:(Eq 4)

## Gauss’ Law

Gauss’ law provides a relationship between the net electrical flux Φ through a closed surface and the net chargeqencenclosed by that surface. Gauss’ law states that the net electric flux out of a closed surface is equal to the charge enclosed by the surface divided by the permittivity:(Eq 5)

Plugging Eq 5 into Eq 6 and rearranging yields

(Eq 6)

In Eq 5 and 6 the charge enclosed by the surface

qencis simply the algebraic sum of all the charges (positive and negative)withinthe Gaussian surface. Charges outside the surface are not included inqenc. However, the electric field in Eq 6 is the electric field due to all charges inside or outside the Gaussian surface. All charges can be included in the determination of the electric field because chargers outside the surface contribute zero net flux through the surface – the number of field lines entering the surface due to charge outside the surface equal the number of field line exiting the surface due to the same charge.## Gauss’ and Coulomb’s Relation

In the Electric Charge section we learned about Coulomb’s law which gives the electrostatic force of attraction or repulsion. In this section we have learned about Gauss’ law which relates the net electrical flux through a Gaussian surface to the net charge includes by that surface. However, both Gauss’ law and Coulomb’s law are just different ways of describing the relationship between electric charge and electric fields. Thus reason would say that one should be able to derive one of the laws from the other. Let us examine the derivation of Coulomb’s law from Gauss’ law as this requires less intensive math.Figure 3 shows a spherical Gaussian surface around a positive point charge. The surface has radius

rand has been divided into small differential areasdA. As can be seen from the picture, the area vectors, as well as the electric field vectors, are perpendicular to the surface (and parallel to each other). Thus, the angle Θ between them equals zero and Eq 6 can be written as:Then, if we recall that the electric field can be considered constant if the areas are small enough, the electric field

Ecan be brought outside the integral. Furthermore, because the only charge enclosed by the surface is q,qenc= q. This simplifies the above expression to:The integral is now just the sum of all the differential areas of the surface, or the surface area of the sphere (4πr^2):

or (Eq 7)

which is Coulomb’s law for a point charge.

## Applying Gauss’ Law

## Cylindrical Symmetry

There are several situations that can be examined using the cylindrical symmetry approach. They are: a line of charge, a non-conducting cylinder of charge and a conducting cylinder of charge. Let us first look at the infinite line of charge with a uniform linear charge density λ. Consider the cylindrical Gaussian surface in Figure 4. From the figure we can see that the electric fields lines point radially outward and are parallel to the two end caps. The electric flux can then be written as:or (Eq 8)

The same expression can be found for a conducting cylinder. The only difference between a line of charge and a conducting cylinder is that an additional assumption is needed. A line of charge is considered infinitely thin, while cylinders and other 3-dimensional objects have volume. Assume

ris the radius of the Gaussian cylinder andRis the radius of the charged cylinder. For Gaussian cylinders withr>R(meaning the Gaussian surface is outside the radius of the conductor) the expression is exactly the same. But forr<Rthe electric field E = 0, since the Gaussian surface is inside the conductor (see Figure 5).For a charged cylinder and

r>Rthe above equation for the electric field holds true. However, for charged cylinder withr<Rthe electric field is not equal to zero (see Figure 6). Starting fromand letting

r<R, we obtain(Eq 9)

The following link provides a three dimensional animation of cylindrical symmetry (found here)

## Planar Symmetry

Just as there were several cylindrical symmetry approaches to finding the electric field with Gauss’ law, there are several planar symmetric systems. The three discussed here include a sheet of charge, a conducting surface, and parallel conducting plates.Consider the thin, infinite, nonconducting sheet of charge in Figure 7. The sheet has a uniform surface charge density σ. A closed cylinder with end caps of area

Ais a useful Gaussian surface in this context. From the figure it can be seen that the electric field is perpendicular to the surface and thus the electric field lines are parallel to the cylinder wall. This means only the end caps contribute to the flux. The charge included by the cylinder is simply, and not surprisingly, the product of the charge density and the cross sectional area of the cylinder.The electric field due to a conducting surface is similar but is not identical. Consider the conductor or Figure 8. Here a conductor with surface density σ is at equilibrium, or in other words the free electrons in the conductor are not moving. Again, a cylindrical Gaussian surface oriented perpendicular to the surface is used to determine the electric field. From the picture only, the electric flux through the top of the Gaussian surface need be considered as the other end cap is inside the conductor. The flux is then given by:

or (Eq 10)

If two oppositely charged conductors are brought together, Gauss' law can be used to calculate the electric field between the two infinite plates. Presuming the plates to be at equilibrium with zero electric field inside the conductors, the result from a charged conducting surface can be used:

However, looking at Figure 9 one can see that the electric field between the plates has a magnitude twice that of one charged sheet. Thus the 2 in the denominator cancels out, leaving us with the equation:

(Eq 11)

The following link provides a three dimensional view of planar symmetry (found here)

## Spherical Symmetry

Using a spherical Gaussian surface we can prove the two shell theorems that were presented in the Electric Charge section. The two theorems state:“a shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell”and“if a charged particle is located inside a shell of uniform charge, there is no electrostatic force on the particle from the shell.”Looking at Figure 10, the electric field of the conducting sphere of charge Q can be obtained by a simple application of Gauss' law. Consider a spherical Gaussian surface with radius

r > R, the electric field has the same magnitude at every point along the surface and is directed outward. The electric flux is then just the electric field times the area of the spherical surface:or (Eq 12)

For

r < R, the Gaussian surface lies within the charged shell. Since all the shell’s charge will reside on the conducting surface, a Gaussian surface atr < Rwill enclose no charge. Thus E = 0. This proves the above two theorems.Next, let us examine the electric field due to a sphere of uniform charge. If

r<Rthan not all of the charge would be enclosed by the spherical Gaussian surface (see Figure 11). The charge inside the Gaussian surface with radius r would then be given by the ratio of the volumes:Noting that the charge outside the Gaussian surface (

Q - Q') does not contributed to the electric flux, and substitutingQ'in place ofQ, the electric flux can be written as:or (Eq 13)

Example 1:

An electron is released from rest at a perpendicular distance of 9.0 cm from a line of charge on a very long non-conducting rod. That charge is uniformly distributed, with 6.0 µC per meter. What is the magnitude of the electron’s initial acceleration?

We combine Newton’s second law (F =ma) with the definition of electric field (F=qE) and

To get:

This gives acceleration to be 1.867 X 10^17 m/s^2.

Example 2:

The figure below shows cross sections through two large, parallel non-conducting sheets with identical distributions of positive charge with surface charge density σ = 1.77 X 10^-22 C/m^2. In unit-vector notation, what is E at points (a) above the sheets, (b) between them, and (c) below them?

Answer: