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Chapter 24 Gauss’s Law (cont.) Dr. Jie Zou PHY 1361 1 Outline Gauss’s law (24.2) Application of Gauss’s law to various charge distributions (24.3) Dr. Jie Zou PHY 1361 2 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. Dr. Jie Zou PHY 1361 3 Let’s begin with one example. A spherical gaussian surface of radius r surrounding a point charge q. E E dA EdA E dA Dr. Jie Zou The magnitude of the electric field everywhere on the surface of the sphere is E = keq/r2. The electric field is to the surface at every point on the surface. Net electric flux through such gaussian surface is ke q 1 q 2 4 r 4 k q 4 q e r 2 PHY 1361 40 0 4 A non-spherical closed surface surrounding a point charge Dr. Jie Zou 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 S1 is equal to the number of lines through the nonspherical surfaces S2 and S3. 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. PHY 1361 5 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 Dr. Jie Zou PHY 1361 6 Example: Problem #14, P. 762 Calculate the total electric flux through the paraboloidal surface due to a constant electric field of magnitude E0 in the direction shown in the figure. Dr. Jie Zou PHY 1361 7 Now let’s consider a more general case. Dr. Jie Zou The net electric flux through S is E = q1/0. The net electric flux through S’ is E = (q2 + q3)/ 0. The net electric flux through S” is E = 0. Charge q4 does not contribute to the flux through any surface because it is outside all surfaces. PHY 1361 8 Gauss’s law Gauss’s law: the net flux through any closed q surface is E E dA in 0 qin = 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. Dr. Jie Zou PHY 1361 9 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 = keQ/r2, r > a; (B) E k e Dr. Jie Zou PHY 1361 Q r , r < a. a3 10 Application of Gauss’s law to various charge distributions Dr. Jie Zou 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: E 2k e 20 r PHY 1361 r 11 Homework Ch. 24, P. 762, Problems: #12, 14, 29. Dr. Jie Zou PHY 1361 12