24 - Gauss's Law

24 - Gauss's Law - Chapter 24 Gausss Law CHAPTE R OUTLI N E...

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739 Gauss’s Law CHAPTER OUTLINE 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions 24.4 Conductors in Electrostatic Equilibrium 24.5 Formal Derivation of Gauss’s Law ± In a table-top plasma ball, the colorful lines emanating from the sphere give evidence of strong electric ﬁelds. Using Gauss’s law, we show in this chapter that the electric ﬁeld surrounding a charged sphere is identical to that of a point charge. (Getty Images) Chapter 24

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740 I n the preceding chapter we showed how to calculate the electric ﬁeld generated by a given charge distribution. In this chapter, we describe Gauss’s law and an alterna- tive procedure for calculating electric ﬁelds. The law is based on the fact that the fundamental electrostatic force between point charges exhibits an inverse-square behavior. Although a consequence of Coulomb’s law, Gauss’s law is more convenient for calculating the electric ﬁelds of highly symmetric charge distributions and makes possible useful qualitative reasoning when dealing with complicated problems. 24.1 Electric Flux The concept of electric ﬁeld lines was described qualitatively in Chapter 23. We now treat electric ﬁeld lines in a more quantitative way. Consider an electric ﬁeld that is uniform in both magnitude and direction, as shown in Figure 24.1. The ﬁeld lines penetrate a rectangular surface of area A , whose plane is oriented perpendicular to the ﬁeld. Recall from Section 23.6 that the number of lines per unit area (in other words, the line density ) is proportional to the magnitude of the electric ﬁeld. Therefore, the total number of lines penetrating the surface is proportional to the product EA . This product of the magnitude of the electric ﬁeld E and surface area A perpendicular to the ﬁeld is called the electric ﬂux ± E (uppercase Greek phi): (24.1) From the SI units of E and A , we see that ± E has units of newton-meters squared per coulomb (N ² m 2 /C.) Electric ﬂux is proportional to the number of electric ﬁeld lines penetrating some surface. ± E ³ E A Example 24.1 Electric Flux Through a Sphere What is the electric ux through a sphere that has a radius of 1.00 m and carries a charge of ´ 1.00 µ C at its center? Solution The magnitude of the electric eld 1.00 m from this charge is found using Equation 23.9: ³ 8.99 10 3 N/C E ³ k e q r 2 ³ (8.99 10 9 N ² m 2 /C 2 ) 1.00 10 · 6 C (1.00 m) 2 The eld points radially outward and is therefore every- where perpendicular to the surface of the sphere. The ux through the sphere (whose surface area A ³ 4 ¸ r 2 ³ 12.6 m 2 ) is thus ³ 1.13 10 5 N ² m 2 /C ± E ³ EA ³ (8.99 10 3 N/C)(12.6 m 2 ) Figure 24.1 Field lines representing a uniform electric eld penetrating a plane of area A perpendicular to the eld. The electric ux ± E through this area is equal to .
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This note was uploaded on 02/24/2011 for the course PHYS 102 taught by Professor Wang during the Spring '11 term at Nanjing University.

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24 - Gauss's Law - Chapter 24 Gausss Law CHAPTE R OUTLI N E...

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