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Chapter 22 Gauss’s Law

Chapter 22 Gauss’s Law. Gauss’s Law is a relationship between the field at all the points on the surface and the total charge enclosed within the surface. Gauss’s Law is part of the key to using symmetry considerations to simplify electric-field calculations.

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Chapter 22 Gauss’s Law

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  1. Chapter 22 Gauss’s Law • Gauss’s Law is a relationship between the field at all the points on the surface and the total charge enclosed within the surface. • Gauss’s Law is part of the key to using symmetry considerations to simplify electric-field calculations. • In Chapter 21 we asked the question “Given a charge distribution, what is the electric field produced by that distribution at point P?” • The answer from Chapter 21 is: The total field at point P is the vector sum of the fields due to all the point charges. Chapter 22

  2. Introduction • The movement of electrons can be shocking (pardon the pun). • If you look at the girl’s hair (figure to the right), you’ll see the electrons coating each individual hair fiber and then repelling each other. • Gauss imagined a flow through a surface placed around a charge and then considered outcomes that we will study in Chapter 22.

  3. The charge within the box determines the magnitude and direction of the Electric field outside the box. Twice the charge, twice as many lines penetrating closed surface Chapter 22

  4. Electric Flux is the “Flow” of the Electric Field in or out of an enclosed surface The net electric flux through the surface of the box is directly proportional to the magnitude of the net charge enclosed in the box. Chapter 22

  5. Electric Flux = The net sum of the product of the perpendicular component of E and the surface area Gauss’s Law – Qualitative statement The net electric flux (number of lines) penetrating a closed surface is directly proportional to the total electric charge enclosed by the closed surface and is independent of the size of the enclosed surface. Chapter 22

  6. The Electric Flux is the scalar product of E and A En E Et If the surface (A) is not flat and/or the electric field (E) is not uniform, A is divided into small elements dA and integrated over the entire surface (A). Chapter 22

  7. Electric Flux Through a disk - Example 22-1 (p.756) E = 2 x 103 N/C; the disk of radius 0.1m is at an angle. (a) Disk angle = 30o; electric flux E through the disk? Answer: E= 54 N-m2/C (b) Disk angle = 90o ; (the disk parallel to E ) electric flux E through the disk? Answer: E= 0 N-m2/C (c) Disk angle = 0o; (the disk is normal to E) electric flux E through the disk? Answer: E= 63 N-m2/C Chapter 22

  8. Electric Flux through a sphere - Example 22-3 (p.757) q = 3 µC, r = 0.2 m Determine electric flux through the sphere Setup Execute Evaluate The is independent of the radius. Larger r produces a smaller E and a larger A Chapter 22

  9. Gauss’s Law for a point charge in a sphere (22-6) (where E is the total electric flux penetrating surface A) Chapter 22

  10. Mathematical Representation of Gauss’s Law - General Case Figure 22-13 Chapter 22

  11. Applications of Gauss’s Law Excess charge on a solid conductor resides entirely on the conductor’s surface Note: A Gaussian surface does not have to be a physical surface Total electric flux penetrating imaginary gaussian surface Qencl = 0 Therefore: E and E penetrating gaussian surface = 0 This means that the electric field (E) inside the charged solid conductor is zero. Chapter 22

  12. Problem 22.8 Chapter 22

  13. Electric Field (E) outside a Charged Solid Conductor Sphere (Ex 22-5, p.762) (imaginary) q (r  R) Figure 22-18 Chapter 22

  14. Electric field of a line charge (Example 22.6, p.763) Problem: Electric charge is distributed uniformly along an infinite long, thin wire. The charge per unit length is λ (assume positive). Find the electric field. Identify: The system has cylindrical symmetry. The field must point away from the + charge Solution: The field lines are radial and lie in a plane perpendicular to the wire. The field magnitude depends only on the radial distance from the wire. Execute: No flux out the ends. All flux out the walls of the cylinder, EA Total flux: Total enclosed charge: From Gauss’s law: Solve for E: Evaluation: Not valid for short wire. For a short wire use the method of Example 21.11(p.731). note: independent of length Chapter 22

  15. Field between oppositely charged parallel conducting plates (Ex 22-8) Two charged parallel plates . Charge per unit area is +σ and -σ Identify: Opposite charges attract. “Fringing” on the edges of plates. Neglect for large plates. Setup: Exploit symmetry. S1, S2, S3, S4 are cylinders with end area A Execute: Consider S1, flux through the end is EA, net charge enclosed is σA Therefore from Gauss’s Law, the total flux = enclosed charge/Єo Solve for E: Evaluate: The same result was obtained with superposition of fields . Example 21.13, p.732 E=0 E=0 Chapter 22

  16. There are practical applications • Figure 22.17 treats excess charge as residing on the surface of a conductor. • Consider Example 22.5. • Figure 22.18 illustrates Example 22.5.

  17. The field of a line or plane of charge • Consider Example 22.6 and Figure 22.19. • See also Example 22.7 and Figure 22.20.

  18. The field of a uniformly charged sphere • Consider Example 22.9. • Figure 22.22 illustrates the example. • Follow Example 22.10.

  19. Charges on conductors • The electric field within a charged conductor may be found. • Consider Figure 22.23. • Follow Example 22.11 and Figure 22.24.

  20. Excess charge on solid conductors are located on the surface The electric field at every point within a conductor is zero and any excess charge on a solid conductor is located entirely on the surface. Chapter 22

  21. Experimental tests of Gauss’s Law • Regard Figure 22.25. • A metal container on an insulating stand.

  22. The Van de Graaff generator • The source of all the static on the child’s hair in our introduction. • Consider Figure 22.27 below.

  23. Prob. 22.17 • On a Humid day E=2.00 x 104 N/C will produce a one inch spark • (a)Using Gauss’ Law what is the charge stored on the surface of the sphere. • b)Assume all charge is concentrated in the sphere center, us e Coulomb’s law to calculated the electric field. • Note : P0 = Є0 Identify: The electric field required to produce a spark 6 in. long is 6 times as strong as the field needed to produce a spark 1 in. long. Set Up:  Gauss’s law, P0 = Є0 By Gauss’s Law solve for the enclosed charge: the electric field is the same as for a point-charge, Execute: (a) The electric field for 6-inch sparks is six times the electric field of a one inch spark The charge to produce this field is (b) Using Coulomb’s law gives Evaluate: It takes only about to produce this field Chapter 22

  24. A Faraday cage blocks flow • Refer to Figure 22.28 below. • Science-fiction movies always place alien transmitters in these to prevent them from calling for help. • Follow Examples 22.12 and 22.13.

  25. Chapter-22 Important Equations Chapter 22

  26. Chapter-22 Important Concepts Chapter 22

  27. Chapter-22 Important concepts and results Chapter 22

  28. Q22.1 A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared to the electric flux through surface #1, the flux through surface #2 is +q Gaussian surface #1 Gaussian surface #2 A. greater. B. the same. C. less, but not zero. D. zero. E. not enough information given to decide

  29. Cavity Q22.4 Conductor A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity A. points generally toward theouter surface of the conductor. B. points generally away from the outer surface of the conductor. C. is zero. D. not enough information given to decide

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