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Chapter 23 Gauss’ Law Copyright © 2014 John Wiley & Sons, Inc. All rights reserved. 23-1 Electric Flux Learning Objectives 23.03 Identify that an area vector for a flat surface is a vector that 23.01 Identify that Gauss’ law is perpendicular to the surface relates the electric field at and that has a magnitude equal points on a closed surface to the area of the surface. (real or imaginary, said to be a Gaussian surface) to the net 23.04 Identify that any surface charge enclosed by that can be divided into area surface. elements (patch elements) that are each small enough and flat 23.02 Identify that the amount of enough for an area vector dA electric field piercing a surface to be assigned to it, with the (not skimming along parallel to vector perpendicular to the the surface) is the electric flux element and having a Φ through the surface. magnitude equal to the area of the element. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-1 Electric Flux Learning Objectives (Contd.) 23.05 Calculate the flux Φ 23.07 Calculate the net flux ϕ through a surface by through a closed surface, integrating the dot product of algebraic sign included, by the electric field vector E and integrating the dot product of the area vector dA (for patch the electric field vector E and elements) over the surface, in the area vector dA (for patch magnitude- angle notation and elements) over the full surface. unit-vector notation. 23.08 Determine whether a 23.06 For a closed surface, closed surface can be broken explain the algebraic signs up into parts (such as the sides associated with inward flux of a cube) to simplify the and outward flux. integration that yields the net flux through the surface. © 2014 John Wiley & Sons, Inc. All rights reserved. © 2014 John Wiley & Sons, Inc. All rights reserved. Flux of a Vector. Consider an airstream of velocity v that is aimed at a loop of area A . The velocity vector v is at angle with respect to the loop normal nˆ. The product vA cos is known as the flux. In this example the flux is equal to the volume flow rate through the loop (thus the name flux). Note 1 : depends on . It is maximum and equal to vA for 0 (v perpendicular to the loop plane). It is minimum and equal to zero for 90 (v parallel to the loop plane). Note 2 : vA cos v A. The vector A is parallel to the loop normal and has magnitude equal to A. n̂ n̂ (23-2) © 2014 John Wiley & Sons, Inc. All rights reserved. 23-1 Electric Flux The area vector dA for an area element (patch element) on a surface is a vector that is perpendicular to the element and has a magnitude equal to the area dA of the element. The electric flux dϕ through a patch element with area vector dA is given by a dot product: (a) An electric field vector pierces a small square patch on a flat surface. (b) Only the x component actually pierces the patch; the y component skims across it. (c) The area vector of the patch is perpendicular to the patch, with a magnitude equal to the patch’s area. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-1 Electric Flux Electric field vectors and field lines pierce an imaginary, spherical Gaussian surface that encloses a particle with charge +Q. Now the enclosed particle has charge +2Q. © 2014 John Wiley & Sons, Inc. All rights reserved. Can you tell what the enclosed charge is now? Answer: -0.5Q Flux of the Electric Field. Consider the closed surface shown in the figure. In the vicinity of the surface assume that we have a known electric field E. The flux of the electric field through the surface is defined as follows: 1. Divide the surface into small "elements" of area A. 2. For each element calculate the term E A EA cos . 3. Form the sum E A. n̂ n̂ 4. Take the limit of the sum as the area A 0. The limit of the sum becomes the integral: n̂ E dA E dA Flux SI unit: N m 2 / C Note 1 : The circle on the integral sign indicates that the surface is closed. When we apply Gauss' law the surface is known as "Gaussian." Note 2 : is proportional to the net number of electric field lines that pass through the surface. (23-3) © 2014 John Wiley & Sons, Inc. All rights reserved. Gauss' Law Gauss' law can be formulated as follows: The flux of E through any closed surface 0 net charge qenc enclosed by the surface. In equation form: 0 qenc Equivalently: ε0 E dA q enc Note 1 : Gauss' law holds for any closed surface. Usually one particular surface makes the problem of determining the electric field very simple. Note 2 : When calculating the net charge inside a closed n̂ surface we take into account the algebraic sign of each charge. Note 3 : When applying Gauss' law for a closed surface we ignore the charges outside the surface no matter how n̂ n̂ 0 qenc ε0 E dA q enc large they are. Example : Surface S1 : 0 1 q, Surface S2 : 0 2 q Surface S3 : 0 3 0, Surface S4 : 0 4 q q 0 Note : We refer to S1 , S 2 , S3 , S 4 as "Gaussian surfaces." (23-4) © 2014 John Wiley & Sons, Inc. All rights reserved. 23-1 Electric Flux Now we can find the total flux by integrating the dot product over the full surface. The total flux through a surface is given by The net flux through a closed surface (which is used in Gauss’ law) is given by where the integration is carried out over the entire surface. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-1 Electric Flux Flux through a closed cylinder, uniform field © 2014 John Wiley & Sons, Inc. All rights reserved. 23-2 Gauss’ Law Learning Objectives 23.09 Apply Gauss’ law to relate net flux through the closed the net flux ϕ through a closed surface. surface to the net enclosed 23.12 Derive the expression for charge qenc. the magnitude of the electric 23.10 Identify how the algebraic field of a charged particle by sign of the net enclosed using Gauss’ law. charge corresponds to the 23.13 Identify that for a charged direction (inward or outward) particle or uniformly charged of the net flux through a sphere, Gauss’ law is applied Gaussian surface. with a Gaussian surface that is 23.11 Identify that charge a concentric sphere. outside a Gaussian surface makes no contribution to the © 2014 John Wiley & Sons, Inc. All rights reserved. 23-2 Gauss’ Law Gauss’ law relates the net flux ϕ of an electric field through a closed surface (a Gaussian surface) to the net charge qenc that is enclosed by that surface. It tells us that we can also write Gauss’ law as Two charges, equal in magnitude but opposite in sign, and the field lines that represent their net electric field. Four Gaussian surfaces are shown in cross section. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-2 Gauss’ Law Surface S1.The electric field is outward for all points on this surface. Thus, the flux of the electric field through this surface is positive, and so is the net charge within the surface, as Gauss’ law requires Surface S2.The electric field is inward for all points on this surface. Thus, the flux of the electric field through this surface is negative and so is the enclosed charge, as Gauss’ law requires. Two charges, equal in magnitude but opposite in sign, and the field lines that represent their net electric field. Four Gaussian surfaces are shown in cross section. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-2 Gauss’ Law Surface S3.This surface encloses no charge, and thus qenc = 0. Gauss’ law requires that the net flux of the electric field through this surface be zero. That is reasonable because all the field lines pass entirely through the surface, entering it at the top and leaving at the bottom. Surface S4.This surface encloses no net charge, because the enclosed positive and negative charges have equal magnitudes. Gauss’ law requires that the net flux of the electric field through this surface be zero. That is reasonable because there are as many field lines leaving surface S4 as entering it. Two charges, equal in magnitude but opposite in sign, and the field lines that represent their net electric field. Four Gaussian surfaces are shown in cross section. © 2014 John Wiley & Sons, Inc. All rights reserved. Coulomb's law Gauss' law Gauss' Law and Coulomb's Law Gauss' law and Coulomb's law are different ways of describing the relation between electric charge and electric field in static cases. One can derive Coulomb's law from Gauss' law and vice versa. dA n̂ Here we will derive Coulomb's law from Gauss' law. Consider a point charge q. We will use Gauss' law to determine the electric field E generated at a point P at a distance r from q. We choose a Gaussian surface that is a sphere of radius r and has its center at q. We divide the Gaussian surface into elements of area dA. The flux for each element is: d EdA cos 0 EdA Total flux 2 EdA E dA E 4 r From Gauss' law we have: 0 qenc q 4 r 2 0 E q E q 4 r 2 0 This is the same answer we got in Chapter 22 using Coulomb's law. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-3 A Charged Isolated Conductor Learning Objectives 23.14 Apply the relationship between surface charge density σ and the area over which the charge is uniformly spread. 23.17 For a conductor with a cavity that contains a charged object, determine the charge on the cavity wall and on the external surface. 23.18 Explain how Gauss’ law is used to find the electric field magnitude E near an isolated conducting surface 23.15 Identify that if excess with a uniform surface charge charge (positive or negative) is density σ. placed on an isolated conductor, that charge moves 23.19 For a uniformly charged conducting surface, apply the to the surface and none is in relationship between the charge the interior. density σ and the electric field 23.16 Identify the value of the magnitude E at points near the electric field inside an isolated conductor, and identify the direction conductor. of the field vectors. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-3 A Charged Isolated Conductor (a) Perspective view (b) Side view of a tiny portion of a large, isolated conductor with excess positive charge on its surface. A (closed) cylindrical Gaussian surface, embedded perpendicularly in the conductor, encloses some of the charge. Electric field lines pierce the external end cap of the cylinder, but not the internal end cap. The external end cap has area A and area vector A. © 2014 John Wiley & Sons, Inc. All rights reserved. The Electric Field Inside a Conductor We shall prove that the electric field inside a conductor vanishes. e F v E Consider the conductor shown in the figure to the left. It is an experimental fact that such an object contains negatively charged electrons, which are free to move inside the conductor. Let's assume for a moment that the electric field is not equal to zero. In such a case a nonvanishing force F eE is exerted by the field on each electron. This force would result in a nonzero velocity v , and the moving electrons would constitute an electric current. We will see in subsequent chapters that electric currents manifest themselves in a variety of ways: (a) They heat the conductor. (b) They generate magnetic fields around the conductor. No such effects have ever been observed, thus the original assumption that there exists a nonzero electric field inside the conductor. We conclude that : The electrostatic electric field E inside a conductor is equal to zero. © 2014 John Wiley & Sons, Inc. All rights reserved. A Charged Isolated Conductor Consider the conductor shown in the figure that has a total charge q. In this section we will ask the question: Where is this charge located? To answer the question we will apply Gauss' law to the Gaussian surface shown in the figure, which is located just below the conductor surface. Inside the conductor the electric field E 0. Thus E A 0 (eq. 1). From Gauss's law we have: qenc 0 (eq. 2). If we compare eq. 1 with eq. 2 we get qenc = 0 . Thus no charge exists inside the conductor. Yet we know that the conductor has a nonzero charge q. Where is this charge located? There is only one place for it to be: On the surface of the conductor. No electrostatic charges can exist inside a conductor. All charges reside on the conductor surface. © 2014 John Wiley & Sons, Inc. All rights reserved. (23-7) An Isolated Charged Conductor with a Cavity Consider the conductor shown in the figure that has a total charge q. This conductor differs from the one shown on the previous page in one aspect: It has a cavity. We ask the question: Can charges reside on the walls of the cavity? As before, Gauss's law provides the answer. We will apply Gauss' law to the Gaussian surface shown in the figure, which is located just below the conductor surface. Inside the conductor the electric field E 0. Thus E A 0 (eq. 1). From Gauss's law we have: qenc 0 (eq. 2). If we compare eq. 1 with eq. 2 we get qenc = 0. Conclusion : There is no charge on the cavity walls. All the excess charge q remains on the outer surface of the conductor. © 2014 John Wiley & Sons, Inc. All rights reserved. (23- The Electric Field Outside a Charged Conductor n̂2 n̂3 S3 The electric field inside a conductor is zero. This is S2 n̂1 S1 not the case for the electric field outside. The electric field vector E is perpendicular to the conductor surface. If it were not, then E would have a component parallel E to the conductor surface. Since charges are free to move in the conductor, E would cause the free electrons to move, which is a contradiction to the assumption that we have stationary charges. We will apply Gauss' law using the cylindrical closed surface shown in the figure. The surface is further divided into three sections S1 , S2 , and S3 as shown in the figure. The net flux 1 2 3 . 1 EA cos 0 EA 2 EA cos 90 0 3 0 (because the electric field inside the conductor is zero). EA E qenc 1 . A 0 The ratio qenc 0 qenc is known as surface charge density E . A 0 © 2014 John Wiley & Sons, Inc. All rights reserved. Symmetry. We say that an object is symmetric under a particular mathematical operation (e.g., rotation, translation, …) if to an observer the object looks the same before and after the operation. Note: Symmetry is a primitive notion and as such is very powerful. Rotational symmetry Featureless sphere Observer Example of Spherical Symmetry Consider a featureless beach ball that can be rotated about a vertical axis that passes through its center. The observer closes his eyes and we rotate the sphere. When the observer opens his eyes, he cannot tell whether the sphere has been rotated or not. We conclude that the sphere has rotational symmetry about the rotation axis. Rotation axis (23-10) © 2014 John Wiley & Sons, Inc. All rights reserved. A Second Example of Rotational Symmetry Rotational symmetry Consider a featureless cylinder that can rotate about its central axis as shown in the figure. The observer closes his eyes and we rotate the cylinder. When he opens his eyes, he cannot tell whether the cylinder has been rotated or not. We conclude that the cylinder has rotational symmetry about the rotation axis. Featureless cylinder Rotation axis Observer (23-11) © 2014 John Wiley & Sons, Inc. All rights reserved. Example of Translational Symmetry: Consider an infinite featureless plane. An observer takes a trip on a magic carpet that flies above the plane. The observer closes his eyes and we move the carpet around. When he opens his eyes the observer cannot tell whether he has moved or not. We conclude that the plane has translational symmetry. Translational symmetry Observer Magic carpet Infinite featureless plane (23-12) © 2014 John Wiley & Sons, Inc. All rights reserved. Recipe for Applying Gauss’ Law 1. Make a sketch of the charge distribution. 2. Identify the symmetry of the distribution and its effect on the electric field. 3. Gauss’ law is true for any closed surface S. Choose one that makes the calculation of the flux as easy as possible. 4. Use Gauss’ law to determine the electric field vector: qenc 0 (23-13) © 2014 John Wiley & Sons, Inc. All rights reserved. 23-4 Applying Gauss’ Law: Cylindrical Symmetry Learning Objectives 23.20 Explain how Gauss’ law is used to derive the electric field magnitude outside a line of charge or a cylindrical surface (such as a plastic rod) with a uniform linear charge density λ. 23.22 Explain how Gauss’ law can be used to find the electric field magnitude inside a cylindrical non-conducting surface (such as a plastic rod) with a uniform volume charge density ρ. 23.21 Apply the relationship between linear charge density λ on a cylindrical surface and the electric field magnitude E at radial distance r from the central axis. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-4 Applying Gauss’ Law: Cylindrical Symmetry Figure shows a section of an infinitely long cylindrical plastic rod with a uniform charge density λ. The charge distribution and the field have cylindrical symmetry. To find the field at radius r, we enclose a section of the rod with a concentric Gaussian cylinder of radius r and height h. The net flux through the cylinder from Gauss’ Law reduces to yielding A Gaussian surface in the form of a closed cylinder surrounds a section of a very long, uniformly charged, cylindrical plastic rod. © 2014 John Wiley & Sons, Inc. All rights reserved. E 2 0 r n̂1 Electric Field Generated by a Long, Uniformly Charged Rod Consider the long rod shown in the figure. It is uniformly S1 charged with linear charge density . Using symmetry arguments we can show that the electric field vector points S2 n̂2 radially outward and has the same magnitude for points at the same distance r from the rod. We use a Gaussian surface S that has the same symmetry. It is a cylinder of radius r S3 and height h whose axis coincides with the charged rod. n̂3 We divide S into three sections: Top flat section S1 , middle curved section S2 , and bottom flat section S3 . The net flux through S is 1 2 3 . Fluxes 1 and 2 vanish because the electric field is at right angles with the normal to the surface: 3 2 rhE cos 0 2 rhE 2 rhE. From Gauss's law we have: If we compare these two equations we get: 2 rhE h E . 0 2 0 r © 2014 John Wiley & Sons, Inc. All rights reserved. qenc 0 h . 0 (23-14) 23-5 Applying Gauss’ Law: Planar Symmetry Learning Objectives 23.23 Apply Gauss’ law to derive the electric field magnitude E near a large, flat, nonconducting surface with a uniform surface charge density σ. 23.24 For points near a large, flat non-conducting surface with a uniform charge density σ, apply the relationship between the charge density and the electric field magnitude E and also specify the direction of the field. 23.25 For points near two large, flat, parallel, conducting surfaces with a uniform charge density σ, apply the relationship between the charge density and the electric field magnitude E and also specify the direction of the field. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-5 Applying Gauss’ Law: Planar Symmetry Non-conducting Sheet Figure (a-b) shows a portion of a thin, infinite, nonconducting sheet with a uniform (positive) surface charge density σ. A sheet of thin plastic wrap, uniformly charged on one side, can serve as a simple model. Here, Is simply EdA and thus Gauss’ Law, becomes where σA is the charge enclosed by the Gaussian surface. This gives © 2014 John Wiley & Sons, Inc. All rights reserved. E 2 0 Electric Field Generated by a Thin, Infinite, Nonconducting Uniformly Charged Sheet We assume that the sheet has a positive charge of surface density . From symmetry, the electric field vector E is perpendicular to the sheet and has a constant magnitude. Furthermore, it points away from the sheet. We choose a cylindrical Gaussian surface S with the caps of area A on either side of the sheet as shown in the figure. We divide S into three sections: S1 is the cap to the right of the sheet, S2 is the curved surface of the cylinder, and S3 is the cap to the left of the sheet. The net flux through S is 1 2 3 . n̂2 S3 n̂3 1 2 EA cos 0 EA. 3 0 S2 n̂1 ( = 90) 2 EA. From Gauss's law we have: S1 2 EA A 0 E . 2 0 © 2014 John Wiley & Sons, Inc. All rights reserved. qenc 0 (23-15) A 0 Ei S A S' The electric field generated by two parallel conducting infinite planes is charged with surface densities 1 and - 1. In figs. a and b we show the two plates isolated so that one does not influence the charge distribution of the other. The charge spreads out equally on both faces of each sheet. When the two plates are moved close to each other as shown in fig. c, then the charges on one plate attract those on the other. As a result the charges move on the inner faces of each plate. To find the field Ei between the plates we apply Gauss' law for the cylindrical surface S , which has caps of area A. q 2 A 2 The net flux Ei A enc 1 Ei 1 . To find the field E0 outside 0 0 the plates we apply Gauss' law for the cylindrical surface S', which has caps of area A. q 1 The net flux E0 A enc 1 0 E0 0. (23-16) 0 © 20140 John Wiley & Sons, Inc. All rights reserved. 0 E0 0 A' 0 2 1 23-5 Applying Gauss’ Law: Planar Symmetry Two conducting Plates Figure (a) shows a cross section of a thin, infinite conducting plate with excess positive charge. Figure (b) shows an identical plate with excess negative charge having the same magnitude of surface charge density σ1. Suppose we arrange for the plates of Figs. a and b to be close to each other and parallel (c). Since the plates are conductors, when we bring them into this arrangement, the excess charge on one plate attracts the excess charge on the other plate, and all the excess charge moves onto the inner faces of the plates as in Fig.c. With twice as much charge now on each inner face, the electric field at any point between the plates has the magnitude © 2014 John Wiley & Sons, Inc. All rights reserved. 23-6 Applying Gauss’ Law: Spherical Symmetry Learning Objectives 23.26 Identify that a shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge is concentrated at the center of the shell. charge, apply the relationship between the electric field magnitude E, the charge q on the shell, and the distance r from the shell’s center. 23.29 Identify the magnitude of the 23.27 Identify that if a charged electric field for points enclosed by particle is enclosed by a shell a spherical shell with uniform of uniform charge, there is no charge. electrostatic force on the 23.30 For a uniform spherical charge particle from the shell. distribution (a uniform ball of 23.28 For a point outside a charge), determine the magnitude spherical shell with uniform and direction of the electric field at interior and exterior points. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-6 Applying Gauss’ Law: Spherical Symmetry In the figure, applying Gauss’ law to surface S2, for which r ≥ R, we would find that And, applying Gauss’ law to surface S1, for which r < R, A thin, uniformly charged, spherical shell with total charge q, in cross section. Two Gaussian surfaces S1 and S2 are also shown in cross section. Surface S2 encloses the shell, and S1 encloses only the empty interior of the shell. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-6 Applying Gauss’ Law: Spherical Symmetry Inside a sphere with a uniform volume charge density, the field is radial and has the magnitude where q is the total charge, R is the sphere’s radius, and r is the radial distance from the center of the sphere to the point of measurement as shown in figure. A concentric spherical Gaussian surface with r > R is shown in (a). A similar Gaussian surface with r < R is shown in (b). © 2014 John Wiley & Sons, Inc. All rights reserved. n̂2 E0 The Electric Field Generated by a Spherical Shell of Charge q and Radius R Inside the shell : Consider a Gaussian surface S1 that is a sphere with radius r R and whose center coincides with that of the charged shell. q The electric field flux 4 r 2 Ei enc 0. 0 Thus Ei 0. Outside the shell : Consider a Gaussian surface S 2 n̂1 Ei Ei 0 E0 0 Thus E0 q 4 0 r (23-17) that is a sphere with radius r R and whose center coincides with that of the charged shell. q q The electric field flux 4 r 2 E0 enc . 2 q 4 0 r 2 0 . Note : Outside the shell the electric field is the same as if all the charge of the shell were concentrated at the shell center. © 2014 John Wiley & Sons, Inc. All rights reserved. n̂1 Eo Electric Field Generated by a Uniformly Charged Sphere of Radius R and Charge q Outside the sphere : Consider a Gaussian surface S1 that is a sphere with radius r R and whose center coincides with that of the charged shell. S1 The electric field flux 4 r 2 E0 qenc / 0 q / 0 Thus E0 q 4 0 r 2 . Inside the sphere : Consider a Gaussian surface S 2 n̂2 Ei that is a sphere with radius r R and whose center coincides with that of the charged shell. qenc 2 The electric field flux 4 r Ei . 0 S2 qenc 4 / r 3 R3 R3 q 2 q 3 q 4 r Ei 3 4 / R 3 r r 0 q Thus Ei r. 3 4 0 R © 2014 John Wiley & Sons, Inc. All rights reserved. (23-18) n̂1 E0 Electric Field Generated by a Uniformly Charged Sphere of Radius R and Charge q Summary : q Ei r 3 4 R 0 S1 E q 4 0 R 3 n̂2 Ei O E0 S2 R r q 4 0 r 2 (23-19) © 2014 John Wiley & Sons, Inc. All rights reserved. 23 Summary Gauss’ Law • Infinite non-conducting sheet • Gauss’ law is Eq. 23-13 Eq. 23-6 • the net flux of the electric field through the surface: • Outside a spherical shell of charge Eq. 23-15 Eq. 23-6 Applications of Gauss’ Law • Inside a uniform spherical shell • surface of a charged conductor Eq. 23-11 Eq. 23-16 • Inside a uniform sphere of charge • Within the surface E=0. • line of charge Eq. 23-20 Eq. 23-12 © 2014 John Wiley & Sons, Inc. All rights reserved.