1. Gauss’ Law Chapter 23

2. 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. Can you tell what the enclosed charge is now? Electric Flux Answer: -0.5Q

3. Can we find a simplified way to perform electric-field calculations? So far we considered problems like this: Given a charge distribution, what is the resulting E-field? (Adding vectors) Yes, we take advantage of a fundamental relationship between electric charge and electric field: Gauss’ law Now, we will consider the reverse: given an E-field, what is the underlying charge distribution? Let’s assume we have a closed container, e.g., a sphere of an imaginary material that doesn’t interact with an electric field We detect an electric field outside the Gauss surface and can conclude There must be a charge inside Gauss’ Law

4. We can even be quantitative about the charge inside the Gauss sphere The same number of field lines on the surface point into the Gauss sphere as there are field lines pointing out We say in this case the net electric flux is zero

5. (from Latin fluxus meaning flow): Flux is the scalar quantity, , which results from a surface integration over a vector field. In the case of electric flux,  E, the vector field is the electric -field and the surface is the surface of the Gauss volume Let’s consider the velocity vector field of a fluid flowing in a pipe The flux  v of the velocity vector field through the surface of area A is What is Flux?

6. Let’s interpret v A Volume element of fluid which flows through surface A in time dt volume flow rate through A What if we tilt A? Extreme case: Flux through A is zero

7. If we tilt area by an angle ϕ is a vector with the properties: Pointing to the surface If the surface is curved, the orientation of changes on the surface and we divide the surface into an infinite number of patches, each with area and sum the flux through each patch over the entire area:

8. Electric flux is defined in complete analogy The electric flux of a point charge inside a Gauss sphere Spherical Gauss surface (Note that the surface is closed) The symmetry of the problem makes the integration simple: We are going to calculate: indicates that we integrate over a closed surface is perpendicular to the surface and its magnitude is the same at each point of the surface

9. Considering the result obtained from calculating the electric flux of a point charge through a Gauss sphere. Do you expect that the flux depends on the radius of the sphere? 1.Yes, the larger the radius the more flux lines will penetrate through the surface 2.No, the flux is independent of the radius 3.I have to calculate again for a different radius Concept Check

10. More generally the result of the integral does not even depend on the specific form of the surface, ONLY on the amount of charge enclosed by the surface. result does NOT depend on the radius of the Gauss sphere Our considerations suggest: Flux through any surface enclosing the charge Q is given by Q/ε 0

11. For example, we can calculate the flux of the enclosed charge Q through the surface of a cube The box geometry suggests the use of Cartesian Coordinates: x y z a

12. With

13. And with

14. Let’s summarize findings into the general form of Gauss’s law The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface divided by ε 0

15. The differential 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. Electric Flux

16. 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. Electric Flux

17. Sample Problem 23.01

18. Sample Problem 23.02

20. Flux through a closed cylinder, uniform field Electric Flux

21. 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. Gauss’ Law

22. 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. Gauss’ Law

23. Surface S3.This surface encloses no charge, and thus q enc = 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. Gauss’ Law

24. Gauss’ Law and Coulomb’s Law

25. Sample Problem 23.03

27. A Charged Isolated Conductor

28. (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. External Electric Field

29. Sample Problem 23.05

30. 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 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. Applying Gauss’ Law: Cylindrical Symmetry

31. Figure (a-b) shows a portion of a thin, infinite, non- conducting 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 Non-conducting Sheet Applying Gauss’ Law: Planar Symmetry

32. 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. Two conducting Plates Applying Gauss’ Law: Planar Symmetry

33. 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 Two conducting Plates Applying Gauss’ Law: Planar Symmetry

34. 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. In the figure, applying Gauss’ law to surface S 2, for which r ≥ R, we would find that And, applying Gauss’ law to surface S 1, for which r < R, Applying Gauss’ Law: Spherical Symmetry

35. 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). Applying Gauss’ Law: Spherical Symmetry

36. Gauss’ Law Gauss’ law is the net flux of the electric field through the surface: Infinite non-conducting sheet Outside a spherical shell of charge Inside a uniform spherical shell Inside a uniform sphere of charge Eq. 23-15 Eq. 23-20 Applications of Gauss’ Law surface of a charged conductor Within the surface E=0. line of charge Eq. 23-6 Eq. 23-11 Eq. 23-6 Eq. 23-12 Eq. 23-13 Eq. 23-16 Summary