Gauss's Law

12 Key excerpts on "Gauss's Law"

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  A Student's Guide to Maxwell's Equations

  • 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.

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  University Physics Volume 2

  • 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).

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  Basic Electromagnetic Theory

  • 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

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  Halliday and Resnick's Principles of Physics

  • David Halliday, Robert Resnick, Jearl Walker(Authors)
  • 2020(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 shows a particle with charge +Q that is surrounded by an imaginary concentric sphere. At points on the sphere (said to be a Gaussian surface), the elec- tric 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 mag- nitude). We say that the field vectors and the field lines pierce the surface. Figure 23-2 is similar except that the enclosed particle has charge +2Q. Because the enclosed charge is now twice as much, the magnitude of the field vectors piercing outward through the (same) Gaussian surface is twice as much as in Fig. 23-1, and the density of the field lines is also twice as much. That sentence, in a nutshell, is Gauss’ law. Let’s check this with a third example with a particle that is also enclosed by the same spherical Gaussian surface (a Gaussian sphere, if you like, or even the catchy G-sphere) as shown in Fig. 23-3. What is the amount and sign of the enclosed charge? Well, from the inward piercing we see immediately that the charge must be negative. From the fact that the density of field lines is half that of Fig. 23-1, we also see that the charge must be 0.5Q. (Using Gauss’ law is like being able to tell what is inside a gift box by looking at the wrapping paper on the box.) The problems in this chapter are of two types. Sometimes we know the charge and we use Gauss’ law to find the field at some point. Sometimes we know the field on a Gaussian surface and we use Gauss’ law to find the charge enclosed by the surface. However, we cannot do all this by simply comparing the density of field lines in a drawing as we just did.

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  Halliday's Fundamentals of Physics, 1st Australian & New Zealand Edition

  • 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.

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  Fundamentals of Physics, Volume 2

  • 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.

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  General Physics Electromagnetism Optics

  • 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.

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  Physics for Students of Science and Engineering

  • A. L. Stanford, J. M. Tanner(Authors)
  • 2014(Publication Date)
  • Academic Press

    (Publisher)

  S , referred to as a gaussian surface. Notice that a gaussian surface need not correspond to a physical surface; it may be any convenient imaginary closed surface. As we will see, an important class of electrostatic problems may be treated using Gauss’s law, but before we explore the utility of Gauss’s law, let us examine one further simple situation that relates that law to Coulomb’s law.

  Consider a closed spherical surface S having a radius R and a positive point charge Q at its center, as shown in Figure 12.10 Every element of area dS is radially outward and has a magnitude given by

  Figure 12.10 A closed spherical surface S having radius R and enclosing a charge Q located at the center of the sphere.

  E = k

  Q

  R 2

  =

  1

  4 π

  ∈ o

  Q

  R 2

  (12-28)

  (12-28)

  Equation (12–28) was obtained, we should recall, from Coulomb’s law, F = (kQq/r 2 ) , and the definition of electric field, E = F /q . The electric flux for the closed spherical surface is

  Φ E

  =

  ∮ S

  E · d S =

  Q

  4 π

  ∈ o

  R 2

  ∮ S

  d S =

  Q

  4 π

  ∈ o

  R 2

  · 4 π

  R 2

  (12-29)

  (12-29)

  in which we have used the fact that the surface area of a sphere of radius R is 4 π R 2 . Then Equation (12–29) becomes

  ∮ s

  E · d S =

  Q net

  ∈ o

  which is precisely Gauss’s law, Equation (12–27) .

  E 12.9  

  Charge is uniformly distributed throughout a region of space with a density ρ = 1.5 × 10−8 C/m3 . Calculate the electric flux for a spherical surface (radius = 0.20 m) in this region.

  Answer: 57 N·m2 /C

  E 12.10  

  At each point on the surface of the cube shown in the accompanying figure, the electric field is in the positive x direction. On face A , E is constant and given by E A = 95 N/C; and on face B , EB = 40 N/C. Calculate the electric flux for the entire cubical surface.

  Answer: 14 N·m2 /C

  E 12.11  

  Calculate the net charge contained in the cube of Exercise 12.10 .

  Answer. 1.2 × 10−10 C

  Exercise 12.10

  E 12.12  

  Charge having a uniform density of 8.0 × 10−10 C/m3

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  Principles of Electrodynamics

  • 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 have

  This 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 then

  In 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 then

  2-3A FEW WORDS ABOUT MATERIALS; CONDUCTORS

  At 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

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  Laws, Concepts and Applications of Electron and Electromagnetism

  • (Author)
  • 2014(Publication Date)
  • Academic Studio

    (Publisher)

  ________________________ WORLD TECHNOLOGIES ________________________ Chapter 8 Gauss's Law for Magnetism In physics, Gauss's Law for magnetism is one of Maxwell's equations, the four equations that underlie classical electrodynamics. It states that the magnetic field B has divergence equal to zero, in other words, that it is a solenoidal vector field. It is equivalent to the statement that magnetic monopoles do not exist. Rather than magnetic charges, the basic entity for magnetism is the magnetic dipole. (Of course, if monopoles were ever found, the law would have to be modified, as elaborated below.) Gauss's Law for magnetism can be written in two forms, a differentia l form and an integral form . These forms are equivalent due to the divergence theorem. Note that the terminology Gauss's Law for magnetism is not universally used. The law is also called Absence of free magnetic poles. (or some variant); one reference even explicitly says the law has no name. It is also referred to as the transversality requirement because for plane waves it requires that the polarization be transverse to the direction of propagation. Differential form The differential form for Gauss's Law for magnetism is the following: where denotes divergence, B is the magnetic field. Integral form The integral form of Gauss's Law for magnetism states: ________________________ WORLD TECHNOLOGIES ________________________ where S is any closed surface (a closed surface is the boundary of some three-dimensional volume; the surface of a sphere or cube is a closed surface, but a disk is not), d A is a vector, whose magnitude is the area of an infinitesimal piece of the surface S , and whose direction is the outward-pointing surface normal. The left-hand side of this equation is called the net flux of the magnetic field out of the surface, and Gauss's Law for magnetism states that it is always zero.

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  Introduction to Numerical Electrostatics Using MATLAB

  • 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)

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  Fundamentals of Physics, Extended

  • 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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