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Dr. Jie ZouPHY 13611 Chapter 24 Gauss’s Law (cont.)

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Presentation on theme: "Dr. Jie ZouPHY 13611 Chapter 24 Gauss’s Law (cont.)"— Presentation transcript:

1 Dr. Jie ZouPHY 13611 Chapter 24 Gauss’s Law (cont.)

2 Dr. Jie ZouPHY 13612 Outline Gauss’s law (24.2) Application of Gauss’s law to various charge distributions (24.3)

3 Dr. Jie ZouPHY 13613 Gauss’s law Gauss’s law describes a general relationship between the net electric flux through a closed surface and the charge enclosed by the surface. Closed surface: often called a gaussian surface.

4 Dr. Jie ZouPHY 13614 Let’s begin with one example. A spherical gaussian surface of radius r surrounding a point charge q. The magnitude of the electric field everywhere on the surface of the sphere is E = k e q/r 2. The electric field is  to the surface at every point on the surface. Net electric flux through such gaussian surface is

5 Dr. Jie ZouPHY 13615 A non-spherical closed surface surrounding a point charge As we discussed in the previous section, the electric flux is proportional to the number of electric field lines passing through a surface. The number of lines through S 1 is equal to the number of lines through the nonspherical surfaces S 2 and S 3. The net flux through any closed surface surrounding a point charge q is given by q/  0 and is independent of the shape of that surface.

6 Dr. Jie ZouPHY 13616 A point charge located outside a closed surface Any electric field line that enters the surface leaves the surface at another point. The net electric flux through a closed surface that surrounds no charge is zero. Revisit Example 24.2

7 Dr. Jie ZouPHY 13617 Example: Problem #14, P. 762 Calculate the total electric flux through the paraboloidal surface due to a constant electric field of magnitude E 0 in the direction shown in the figure.

8 Dr. Jie ZouPHY 13618 Now let’s consider a more general case. The net electric flux through S is  E = q 1 /  0. The net electric flux through S’ is  E = (q 2 + q 3 )/  0. The net electric flux through S” is  E = 0. Charge q 4 does not contribute to the flux through any surface because it is outside all surfaces.

9 Dr. Jie ZouPHY 13619 Gauss’s law Gauss’s law: the net flux through any closed surface is q in = the net charge inside the gaussian surface. E = the (total) electric field at any point on the surface, which includes contributions from charges both inside and outside the surface. Pitfall prevention: Zero flux is not zero field.

10 Dr. Jie ZouPHY 136110 Application of Gauss’s law to various charge distributions Example 24.5 A spherically symmetric charge distribution: An insulating solid sphere of radius a has a uniform volume charge density  and carries a total positive charge Q. (A) Calculate the magnitude of the electric field at a point outside the sphere. (B) Find the magnitude of the electric field at a point inside the sphere. Answer: (A) E = k e Q/r 2, r > a; (B), r < a.

11 Dr. Jie ZouPHY 136111 Application of Gauss’s law to various charge distributions Example 24.7 A cylindrical symmetric charge distribution: Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length. Answer:

12 Dr. Jie ZouPHY 136112 Homework Ch. 24, P. 762, Problems: #12, 14, 29.


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