Chapter 4

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in the radial direction. We enclose the charge by an imaginary sphere of radius r called the “Gaussian surface.” Figure 4.2.1 A spherical Gaussian surface enclosing a charge Q . In spherical coordinates, a small surface area element on the sphere is given by (Figure 4.2.2) 2 dA = r sin θ dθ dφ rˆ Figure 4.2.2 A small area element on the surface of a sphere of radius r. Thus, the net electric flux through the area element is ⎛ 1 Q ⎞ 2 Q dΦ E = E⋅ dA= EdA=⎜ 2 ⎟( r sin θ dθ dφ) = sin θ dθ dφ 4 0 r 4πε 0 ⎝ πε ⎠ The total flux through the entire surface is Q π 2π Q Φ E = ∫∫ E⋅ dA= sin θ dθ dφ = 4πε ∫ ∫ ε S 0 0 0 0 (4.2.1) (4.2.2) (4.2.3) The same result can also be obtained by noting that a sphere of radius r has a surface area 2 A = 4π r , and since the magnitude of the electric field at any point on the spherical 2 surface is E = Q/4πε r , the electric flux through the surface is 0 3
E A ⎛ 1 Q ⎞ Q ∫∫ ∫∫ (4.2.4) 2 Φ E = ⋅ d = E dA= EA= ⎜ 4π r 2 ⎟ = 4πε S S 0 r ε0 ⎝ ⎠ In the above, we have chosen a sphere to be the Gaussian surface. However, it turns out that the shape of the closed surface can be arbitrarily chosen. For the surfaces shown in Figure 4.2.3, the same result ( Φ E = Q / ε0 ) is obtained. whether the choice is S 1 , 2 or . S S3 Figure 4.2.3 Different Gaussian surfaces with the same outward electric flux. The statement that the net flux through any closed surface is proportional to the net charge enclosed is known as Gauss’s law. Mathematically, Gauss’s law is expressed as q Φ = ⋅ = E enc ∫∫ E dA (Gauss’s law) (4.2.5) ε S 0 where q enc is the net charge inside the surface. One way to explain why Gauss’s law holds is due to note that the number of field lines that leave the charge is independent of the shape of the imaginary Gaussian surface we choose to enclose the charge. To prove Gauss’s law, we introduce the concept of the solid angle. Let 1 1ˆ A ∆ A =∆ r be an area element on the surface of a sphere of radius , as shown in Figure 4.2.4. r S1 1 Figure 4.2.4 The area element ∆ A subtends a solid angle ∆Ω . The solid an ∆Ω 1 1ˆ A ∆ A =∆ r gle subtended by at the center of the sphere is defined as 4
Page 1 and 2: Chapter 4 Gauss’s Law 4.1 Electri
Page 3: Note that with the definition for t
Page 7 and 8: ∆A ∆A cosθ ∆Ω = = (4.2.11)
Page 9 and 10: Figure 4.2.7 Gaussian surface for a
Page 11 and 12: (4) The total flux through the Gaus
Page 13 and 14: T he charge enclosed by the Gaussia
Page 15 and 16: which is proportional to the volume
Page 17 and 18: Figure 4.3.3 Normal and tangential
Page 19 and 20: Consider a hollow conductor shown i
Page 21 and 22: ⎛ q ⎞ dWext = Vdq =⎜ ⎟dq 4
Page 23 and 24: E patch ⎧ σ ˆ ⎪ + k, z > 0
Page 25 and 26: 4.6 Appendix: Tensions and Pressure
Page 27 and 28: Figure 4.6.3 An electric charge in
Page 29 and 30: 2 across the surface of this sphere
Page 31 and 32: System Figure Identify the symmetry
Page 33 and 34: Figure 4.8.3 Electric field of two
Page 35 and 36: ∫ ∫ dx x = ( x + a ) a ( x + a
Page 37 and 38: 4.9 Conceptual Questions 1. If the
Page 39: 4.10.5 Electric Potential Energy of

Chapter 4

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