Physics
Gauss Law
Gauss's law, a fundamental principle in physics, states that the total electric flux through a closed surface is proportional to the total charge enclosed by the surface. It provides a convenient way to calculate electric fields due to symmetric charge distributions. The law is a key tool in solving problems related to electric fields and charges.
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12 Key excerpts on "Gauss Law"
- eBook - PDF
- Daniel Fleisch(Author)
- 2008(Publication Date)
- Cambridge University Press(Publisher)
1 Gauss’s law for electric fields In Maxwell’s Equations, you’ll encounter two kinds of electric field: the electrostatic field produced by electric charge and the induced electric field produced by a changing magnetic field. Gauss’s law for electric fields deals with the electrostatic field, and you’ll find this law to be a powerful tool because it relates the spatial behavior of the electrostatic field to the charge distribution that produces it. 1.1 The integral form of Gauss’s law There are many ways to express Gauss’s law, and although notation differs among textbooks, the integral form is generally written like this: I S ~ E ^ n da ¼ q enc e 0 Gauss’s law for electric fields (integral form). The left side of this equation is no more than a mathematical description of the electric flux – the number of electric field lines – passing through a closed surface S , whereas the right side is the total amount of charge contained within that surface divided by a constant called the permittivity of free space. If you’re not sure of the exact meaning of ‘‘field line’’ or ‘‘electric flux,’’ don’t worry – you can read about these concepts in detail later in this chapter. You’ll also find several examples showing you how to use Gauss’s law to solve problems involving the electrostatic field. For starters, make sure you grasp the main idea of Gauss’s law: Electric charge produces an electric field, and the flux of that field passing through any closed surface is proportional to the total charge contained within that surface. 1 In other words, if you have a real or imaginary closed surface of any size and shape and there is no charge inside the surface, the electric flux through the surface must be zero. If you were to place some positive charge anywhere inside the surface, the electric flux through the surface would be positive. If you then added an equal amount of negative charge inside the surface (making the total enclosed charge zero), the flux would again be zero. - eBook - PDF
- William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
6 | GAUSS'S LAW Figure 6.1 This chapter introduces the concept of flux, which relates a physical quantity and the area through which it is flowing. Although we introduce this concept with the electric field, the concept may be used for many other quantities, such as fluid flow. (credit: modification of work by “Alessandro”/Flickr) Chapter Outline 6.1 Electric Flux 6.2 Explaining Gauss’s Law 6.3 Applying Gauss’s Law 6.4 Conductors in Electrostatic Equilibrium Introduction Flux is a general and broadly applicable concept in physics. However, in this chapter, we concentrate on the flux of the electric field. This allows us to introduce Gauss’s law, which is particularly useful for finding the electric fields of charge distributions exhibiting spatial symmetry. The main topics discussed here are 1. Electric flux. We define electric flux for both open and closed surfaces. 2. Gauss’s law. We derive Gauss’s law for an arbitrary charge distribution and examine the role of electric flux in Gauss’s law. 3. Calculating electric fields with Gauss’s law. The main focus of this chapter is to explain how to use Gauss’s law to find the electric fields of spatially symmetrical charge distributions. We discuss the importance of choosing a Gaussian surface and provide examples involving the applications of Gauss’s law. 4. Electric fields in conductors. Gauss’s law provides useful insight into the absence of electric fields in conducting materials. So far, we have found that the electrostatic field begins and ends at point charges and that the field of a point charge varies inversely with the square of the distance from that charge. These characteristics of the electrostatic field lead to an important mathematical relationship known as Gauss’s law. This law is named in honor of the extraordinary German mathematician and scientist Karl Friedrich Gauss (Figure 6.2). - No longer available |Learn more
- James Babington(Author)
- 2016(Publication Date)
- Mercury Learning and Information(Publisher)
We know already that it is due to the body having a net charge of the same type as the first. That they should be proportional is also evident. From Coulomb’s observation of an inverse square law, one can see that if we were to surround the source body with a sphere that was centered on where the source body was, then the product of the area of the sphere and the size of the electric field should be a numerical constant. In fact it should also be proportional to how the charge is distributed on the source body. This is all encapsulated in Gauss’s law. Carl Friedrich Gauss (1777–1855), one of the strongest mathematicians of the 19th century (he was named the “Prince of Mathematicians”). He worked on a huge number of different areas in mathematics and physics. Gauss published this particular result in 1838; however it had already been discovered earlier by George Green in 1828 and was later rediscovered by Lord Kelvin in 1842. We define the electric flux through a closed two dimensional surface which is the boundary of the volume to be:- (3.13) In integral form, Gauss’s law states that the electric flux for a closed surface area that encloses a certain amount of charge is proportional to the enclosed charge:- (3.14) The constant is the permittivity of free space mentioned earlier so that when we consider a single point charge and its electric field at some distance we get back Coulomb’s law Equation (3.1). (Exercise: verify this) Since it is an integral equation we are making a global statement about the system. However, we can also make a local statement (that is, recast the equation so that it holds at each point in space) by using one of our useful integral relations. The two-dimensional surface integral can be converted to the volume integral by the divergence theorem Equation (2.33) whence, (3.15) We have now replaced here the discrete charge summation with a continuous distribution of charge density integrated, as in earlier examples - Ozgur Ergul(Author)
- 2021(Publication Date)
- Wiley(Publisher)
1 Gauss’ Law We start with Gauss’ law, which describes how electric fields are created by electric charges. S ¯ D (¯ r, t ) · ds = Q enc ( t ) ¯ ∇ · ¯ D (¯ r, t ) = q v (¯ r, t ) There are two different forms of Gauss’ law. First, in the integral form, it states that the flux of the electric field through a surface is equal to the enclosed electric charge. This is a very strong theorem, especially considering that it does not matter how the electric charge is distributed inside the volume. At the same time, it is useful only in symmetric cases to find electric fields from electric charges. Second, in the differential form, Gauss’ law provides pointwise information: i.e. the divergence of the electric flux density is the electric charge density. While the divergence operation (as a kind of derivative) is a loss of information, this is also an important tool to relate electric fields to electric charges. Application of Gauss’ law at boundaries further leads to important boundary conditions that must be satisfied by electric fields across boundaries. In general, both forms of Gauss’ law are valid for all cases, including static and dynamic scenarios. On the other hand, Gauss’ law alone is not sufficient to analyze most dynamic cases, and other of Maxwell’s equations are required to completely investigate electromagnetic radiation. However, in static cases, Gauss’ law is an excellent tool, hand-in-hand with Coulomb’s law in a vacuum. 1.1 Integral Form of Gauss’ Law Gauss’ law provides the relationship between electric charges and electric fields: i.e. it shows that electric fields are created by electric charges. Consider a surface S bounding an electric Introduction to Electromagnetic Waves with Maxwell’s Equations , First Edition. ¨ Ozg¨ ur Erg¨ ul. c 2022 John Wiley & Sons Ltd. Published 2022 by John Wiley & Sons Ltd. Companion website: www.wiley.com/go/ergulmax- eBook - PDF
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
The relationship is called Gauss’ law, which was developed by German mathematician and physicist Carl Friedrich Gauss (1777–1855). Let’s first take a quick look at some simple examples that give the spirit of Gauss’ law. Figure 23.1.1 shows a particle with charge +Q that is surrounded by an imaginary concentric sphere. At points on the sphere (said to be a Gaussian sur- face), the electric field vectors have a moderate magnitude (given by E = kQ/r 2 ) and point radially away from the particle (because it is positively charged). The electric field lines are also outward and have a moderate density (which, recall, is related to the field magnitude). We say that the field vectors and the field lines pierce the surface. Figure 23.1.2 is similar except that the enclosed particle has charge +2Q. Because the enclosed charge is now twice as much, the magnitude of the field vec- tors piercing outward through the (same) Gaussian surface is twice as much as in Fig. 23.1.1, and the density of the field lines is also twice as much. That sentence, in a nutshell, is Gauss’ law. Gaussian surface Field line E Figure 23.1.1 Electric field vectors and field lines pierce an imaginary, spherical Gaussian surface that encloses a particle with charge +Q. Figure 23.1.4 (a) An electric field vector pierces a small square patch on a flat surface. (b) Only the x com- ponent actually pierces the patch; the y component skims across it. (c) The area vector of the patch is perpendic- ular to the patch, with a magnitude equal to the patch’s area. ( a) ( b ) ( c ) y x y E A ΔA ΔA y x E x E y ΔA x θ Figure 23.1.2 Now the enclosed particle has charge +2Q. Figure 23.1.3 Can you tell what the enclosed charge is now? Guass’ law relates the electric field at points on a (closed) Gaussian surface to the net charge enclosed by that surface. - David Halliday, Jearl Walker, Patrick Keleher, Paul Lasky, John Long, Judith Dawes, Julius Orwa, Ajay Mahato, Peter Huf, Warren Stannard, Amanda Edgar, Liam Lyons, Dipesh Bhattarai(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
Keep in mind that we want to determine the net flux through a surface because that is what Gauss’ law relates to the charge enclosed by the surface. (The law is coming up in the next module.) Note that flux is a scalar (yes, we talk about field vectors but flux is the amount of piercing field, not a vector itself). 23.2 Gauss’ law LEARNING OBJECTIVES After reading this module, you should be able to: 23.2.1 apply Gauss’ law to relate the net fux Φ through a closed surface to the net enclosed charge q enc 23.2.2 identify how the algebraic sign of the net enclosed charge corresponds to the direction (inward or outward) of the net fux through a Gaussian surface 23.2.3 identify that charge outside a Gaussian surface makes no contribution to the net fux through the closed surface 23.2.4 derive the expression for the magnitude of the electric feld of a charged particle by using Gauss’ law 23.2.5 identify that for a charged particle or uniformly charged sphere, Gauss’ law is applied with a Gaussian surface that is a concentric sphere. KEY IDEAS • Gauss’ law relates the net fux Φ penetrating a closed surface to the net charge q enc enclosed by the surface: 0 Φ = q enc . • Gauss’ law can also be written in terms of the electric feld piercing the enclosing Gaussian surface: 0 ∮ E ⋅ d A = q enc . Gauss’ law relates the net flux Φ of an electric field through a closed surface (a Gaussian surface) to the net charge q enc that is enclosed by that surface. It tells us that 0 Φ = q enc . (23.7) By substituting the definition of flux, Φ = ∮ E ⋅ d A, we can also write Gauss’ law as 0 ∮ E ⋅ d A = q enc . (23.8) The two equations for Gauss’ law hold only when the net charge is located in a vacuum or (what is the same for most practical purposes) in air. In chapter 25, we modify Gauss’ law to include situations in which a material such as mica, oil, or glass is present.- eBook - PDF
- David Halliday, Robert Resnick, Kenneth S. Krane(Authors)
- 2019(Publication Date)
- Wiley(Publisher)
A third reason is that Gauss’ law is valid in the case of rapidly moving charges, but Coulomb’s law can be used only for charges that are at rest or moving very slowly. Finally, as we show later in this chapter, Coulomb’s law can be de- rived as a special case of Gauss’ law, and thus Gauss’ law is more general than Coulomb’s law. For these reasons, Gauss’ law is considered to be more fundamental than Coulomb’s law and is included as one of the four basic GAUSS’ LAW C oulomb’s law can always be used to calculate the electric field for any discrete or continuous distribution of charges at rest. The sums or integrals might be complicated (and a computer might be needed to evaluate them numerically), but the resulting electric field can always be found. Some cases discussed in the previous chapter used simplifying arguments based on the symmetry of the physical situation. For example, in calculating the electric field at points on the axis of a charged circular loop, we used a symmetry argument to deduce that components of perpendicular to the axis must vanish. In this chapter we discuss an alternative to Coulomb’s law, called Gauss’ law, that provides a more useful and instructive approach to calculating the electric field in situations having certain symmetries. The number of situations that can directly be analyzed using Gauss’ law is small, but those cases can be done with extraordinary ease. Although Gauss’ law and Coulomb’s law give identical results in the cases in which both can be used, Gauss’ law is considered a more fundamental equation than Coulomb’s law. It is fair to say that whereas Coulomb’s law provides the workhorse of electrostatics, Gauss’ law provides the insight. E B E B CHAPTER 27 CHAPTER 27 equations of electromagnetism (Maxwell’s equations, which we discuss in Chapter 38). Before we introduce Gauss’ law, we first need to define and discuss a new quantity, the flux of the electric field. - eBook - PDF
- Pierluigi Zotto, Sergio Lo Russo, Paolo Sartori(Authors)
- 2022(Publication Date)
- Società Editrice Esculapio(Publisher)
Gauss’s law has been demonstrated starting from Coulomb’s law for electrostatic fields, so in this case the two laws are perfectly equivalent. The equivalence keeps its validity for slowly varying time dependent electric fields, but it does not hold anymore for rapidly vary- ing electric fields (e.g. electromagnetic waves): while Coulomb’s law is not true anymore, Gauss’s law retains its validity. Then Gauss’s law is more general than Coulomb’s law and it is actually one of the fundamental laws of electromagnetism. The choice k = 1 4πε 0 ( ) for the constant in Coulomb’s law was adopted in order to simplify Gauss’s law. 3.5 Gauss’s Law Applications Gauss’s law simplifies the calculation of electric fields in the cases which show a high degree of symmetry. In particular it is relevant for cases in which cylindrical, planar or spherical symmetries exist, since in these cases it is easy to identify a Gauss surface featur- ing a constant magnitude of the normal component Ei u n of the electric field. If this occurs, the flux of the electric field through that surface is simply EidA u n = A ∫ E cos θ dA = A ∫ E cos θ dA A ∫ = EA cos θ . Gauss surfaces of regular solids featuring such a property are easily identified. a) Indefinite Wire (or Indefinite Cylinder) Consider an indefinite wire charged with a uniform charged density λ. The symmetry argument which has been introduced in order to per- form the direct calculation of paragraph 2.11a, allows to un- derstand that the field must be normal to the wire and it must have the same intensity at the same distance from it, i.e. - eBook - ePub
- Melvin Schwartz(Author)
- 2012(Publication Date)
- Dover Publications(Publisher)
Eq. (1-4-3) ] we convert this to a surface integral.The integral on the right is just equal to 4π. Thus we haveThis result is one of the four basic equations of Maxwell and lies at the foundation of electrostatics. It says, in effect, that electric field can only have a net flux into or out of a region if there is charge within the region. We use this to derive Gauss’ law by integrating over a volume V.The left side of Eq. (2-2-4) can be transformed into a surface integral by means of Gauss’ theorem.The right side of Eq. (2-2-4) is just 4π times the total charge Q within the volume. We have thenIn other words, the total flux of electric field out of any given volume is just equal to 4π times the charge within the volume.Gauss’ law greatly facilitates the determination of electric field in situations characterized by high symmetry. We illustrate its application by calculating the field everywhere due to a spherically symmetrical, uniform charge distribution of radius R and total charge Q. For convenience we set the origin of our coordinate system at the center of the sphere, as shown in Fig. 2-2 , and let r be the distance from the origin to the point at which we wish to determine the field. Obviously, by symmetry, the field must be radial in direction with magnitude only dependent upon r. We set an imaginary spherical surface at radius r and observe that the total flux of E out of the volume enclosed by this surface is just 4πr2 E(r). If r < R then the charge within the surface is just Qr3 /R3 . We have then2-3A FEW WORDS ABOUT MATERIALS; CONDUCTORSAt this point we must say a few words about the electrical nature of materials. Later, in Chap. 10 , we will try to give a more detailed and extensive picture of what actually goes on at the microscopic level. For now we will be rather brief and somewhat incomplete.As we all know, matter is constituted of positively charged heavy nuclei surrounded by negatively charged light electrons. The scale of physical dimensions of macroscopic bodies is determined by the natural radius of an electron cloud about the nucleus, about 10–8 cm. Hence about 1024 - eBook - ePub
- A. L. Stanford, J. M. Tanner(Authors)
- 2014(Publication Date)
- Academic Press(Publisher)
Our first consideration is the calculation of electric fields at points in space caused by collections of fixed point charges. This procedure will then be extended to include the calculation of electric fields caused by continuous charge distributions.Certain symmetrical arrangements of charge produce electric fields that may be determined by a simple, powerful, problem-solving procedure. This process is based on a relationship called Gauss’s law, which, in turn, is based on the concept of electric flux. Therefore, we will illustrate electric flux and see how it is used in Gauss’s law to calculate the electric fields of symmetrical charge distributions.Electric fields and the way that electric fields are related to Gauss’s law permit us to deduce how static charges position themselves in and on conducting materials. Therefore, our final consideration is how static charges and the electric fields associated with those charges distribute themselves in and around electrical conductors.12.1 Electric Fields of Point Charges
In the preceding chapter we saw that a point charge q located a distance r from a fixed point charge Q experiences a force F given by Coulomb’s law,F = k(12-1)q Qr 2r ˆ(12-1)where is a unit vector in the direction from Q toward q . We saw also that the electric field E at the position of q is defined to beE =(12-2)F q(12-2)Combining Equations (12–1) and (12-2) , we obtain the electric field of the point charge Q at a distance r from Q :E = k(12-3)Qr 2r ˆ(12-3)A point in space at which we calculate a field is referred to as a field point . Thus, the electric field E at a field point P located at a distance r from a point charge Q is in the same direction as would be the force F on a positive test charge q if it were placed at P , and the magnitude of E is equal to F/q = kQ/r 2 . It follows, then, that the principle of superposition applies not only to electrostatic forces but also to electric fields. In other words, the total electric field (at a field point) caused by n fixed charges Q 1 , Q 2 , …Qnmay be calculated by finding the electric field at P produced by each of those n charges as if each were the only charge present and summing those fields vectorially - Lawrence N. Dworsky(Author)
- 2014(Publication Date)
- Wiley-IEEE Press(Publisher)
The rationale for using this term will become clear shortly. Consider a point charge q surrounded by a virtual spherical shell of radius r 0. The surface area of this shell is. Since is a function only of r, it is a constant everywhere on this shell; also, since it is pointed radially outward everywhere, it is normal to the shell at intersection. The integral of (the magnitude of) over the surface of the shell is (1.5) By examining the situation for an arbitrary collection of charges and an arbitrary surface surrounding them, we can generalize this result to Gauss’s law 3 (1.6) where the integral is over the entire surface. is a differential area with vector direction normal to the plane of the area and q is the total charge enclosed. Returning to equation (1.5), we have (1.7) For a spherical shell centered at q, we obtain only, pointing radially outward, and therefore (1.8) which is essentially identical to equation (1.2). In other words, Gauss’s and Coulomb’s laws are equivalent. Suppose that we have a sphere of charge of radius a, centered at the origin, of uniform charge density ρ [expressed in coulombs per cubic meter (C/m 3)] (see Figure 1.4). FIGURE 1.4 Sphere of uniform charge density ρ. From the symmetry of the situation, we know again that only, pointing radially outward. For any r ≤ a, the charge enclosed is (1.9) Putting this result into Gauss’s law, we have (1.10) or (1.11) The field goes to 0 at r = 0, because there is no charge enclosed. It increases with increasing r. At r = a all of the charge is enclosed and again using Gauss’ law, for r ≥ a, we obtain (1.12) and then (1.13) If ρ is not a constant but is instead a function of r (and only r), then it must be brought inside the integral of equation (1.9) and the integral properly evaluated. The electric field outside the sphere of charge (r ≥ a) depends only on the total charge in the sphere, irrespective of the details of ρ(r)- eBook - PDF
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
682 CHAPTER 23 GAUSS’ LAW E (10 7 N/C) E s 0 1 2 r (cm) 3 4 5 Figure 23-53 Problem 48. ••38 In Fig. 23-48a, an electron is shot directly away from a uniformly charged plastic sheet, at speed v s = 2.0 × 10 5 m/s. The sheet is nonconducting, flat, and very large. Figure 23-48b gives the electron’s vertical velocity component v versus time t until the return to the launch point. What is the sheet’s surface charge density? Module 23-6 Applying Gauss’ Law: Spherical Symmetry •44 Figure 23-52 gives the magni- tude of the electric field inside and outside a sphere with a positive charge distributed uniformly throughout its volume. The scale of the vertical axis is set by E s = 5.0 × 10 7 N/C. What is the charge on the sphere? •45 Two charged concentric spherical shells have radii 10.0 cm and 15.0 cm. The charge on the inner shell is 4.00 × 10 −8 C, and that on the outer shell is 2.00 × 10 −8 C. Find the electric field (a) at r = 12.0 cm and (b) at r = 20.0 cm. •46 Assume that a ball of charged particles has a uniformly distributed negative charge density except for a narrow radial tunnel through its center, from the surface on one side to the surface on the opposite side. Also assume that we can position a proton anywhere along the tunnel or outside the ball. Let F R be the magnitude of the electrostatic force on the proton when it is located at the ball’s surface, at radius R. As a multiple of R, how far from the surface is there a point where the force magnitude is 0.50F R if we move the proton (a) away from the ball and (b) into the tunnel? •47 SSM An unknown charge sits on a conducting solid sphere of radius 10 cm. If the electric field 15 cm from the center of the sphere has the magnitude 3.0 × 10 3 N/C and is directed radially inward, what is the net charge on the sphere? ••48 A positively charged particle is held at the center of a spherical shell. Figure 23-53 gives the magnitude E of the electric field versus radial distance r.
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