Gauss’ law (chap. 23)

1. Gauss’ law(chap. 23) • In principle one can simply apply the Coulomb’s law and find out the electrostatic field generated by any charge distributions. • But it may be a very complex task for many situations. • However, for certain charge distributions involving symmetry, we can save far more work by using a law called Gauss’ law, developed by C.F. Gauss (1777-1855). • To begin with, we need to learn about the idea of • electric flux電通量 .

2. consider the case of a uniform air stream flow through a square area. • if the velocity vector of the air makes an angle with the loop, the volume flow rate (i.e. the volume of air flow through the loop per unit time) is given by: This is an example of flux.

3. Using a vector notation for area A, its direction is the normal the area A. Then the flux can be rewritten as: • For the case of electric field, the flux is simply the electric field passes through a certain area A. • if we are considering a small area ΔA, the flux passes through it is Note that ΔΦ can be positive, negative or zero depends on the direction of the vectors. For an imaginary surface, which we call Gaussian surface, the total flux passes through the whole surface will the sum of all ΔΦ:

4. or more accurately It is surface integral which takes all over the whole closed surface. • Note that the flux is a scalar and has an unit of Nm2/C. • Since the number of electric field lines is proportional to the electric field density, the electric flux through a Gaussian surface is then proportional to the net number of field lines passing through that surface. • The total flux on the left side of the Gaussian surface shown here is negative while those on the right side is positive. • The total flux through this surface is in fact equal to zero.

5. To compute this integral is in general difficult , but for surfaces with high symmetry, it becomes a simple task. Example: The flux passes through a cylindrical surface in a uniform field can be calculated by taking the surface integral. Note that dA and E is always perpendicular to each other on the surface b, therefore the surface integral must be zero. surface integral on surface a

6. Since the field is uniform that it is a constant on the surface a and b, the surface integral on the surfaces is simply: This is not surprise because every field lines enter the cylinder from one surface will eventually leave it from another surface. Therefore the net flux must be zero.

7. Here we consider the case of non-uniform field with a Gaussian cube: Ey=4 The x component of the field changes with x while Eyis constant. The flux through the right side of the cube is: Ex=3x = surface area=4m

8. Similar calculation applies to the left surface. The only difference is now x=1m, and dA=-dA i. So: dA=dAj For the top surface, dA=dAj.

9. Gauss’ Law • One can easily notice that every field lines enter a Gaussian surface must also leave the surface, therefore the number field lines enter the closed surface must equal to number of field lines that leave it, i.e. the net flux = 0. • UNLESS there are field lines terminated or created inside the surface, which means there are charges (positive if field lines are created and negative if field lines are terminated) inside the Gaussian surface. • This observation leads to the Gauss’ law: for the case in vacuum (or air).

10. The Gauss’ law is very powerful because it contains no details of the surface or the charge distribution. • Therefore it is very general and applies to all cases. • For example, we can apply the Gauss’ law to the surfaces S1, S2, S3, and S4. • The net flux through S1 is positive and is equal to +q/ε0. • The net flux through S2 is negative. • The net flux through S3 is zero and there is no charge inside. • The net flux through S4 is also zero because the total charge is zero. There is as many lines entering the surface as leaving it.

11. zero charge q=5ε0 q=-3ε0

13. Gauss’ law and Coulomb’s law These two law must be related to each other because both describe the electric field distribution around a system of charges. In fact, we can derive the Coulomb’s law starting from the Gauss’s law. • Consider a spherical Gaussian surface around a point charge q. • apply the Gauss’ law and we have note that E and dA has the same direction on every point of the surface. because of symmetry, EdA is the same on every point and the integral becomes: It gives the Coulomb’s law at the end:

14. A charged isolated conductor • First we notice that there is no electric field inside a conductor, otherwise free electrons in the conductor have to flow and lead to current. But it shouldn’t be the case for an isolated conductor. • Then we can draw a Gaussian surface very close to the surface of the conductor. • Since there is no net flux (no field inside) through the Gaussian surface, by Gauss’ law, total charge inside the conductor must be zero. • Therefore all charges must sit on the surface instead. • For (b), because there is no flux through the Gaussian surface, the net charge enclosed by it must be zero, i.e. there is also no charge on the inner surface.

15. While there is no field inside a conductor, the charges on the surface generate an electric field outside the surface: • consider a flat surface with uniform charge density σ • one can draw a cylindrical Gaussian surface as shown in (a). • the net charge enclosed by the Gaussian surface is σA and the flux through the outside surface is EA • note that the field on the left is zero and hence there is no flux on the left surface • by Gauss’ law Using symmetry, Gauss’ law provides an easy way to find out the field around a system of charges.

16. Since there is no field inside the conductor, there is no flux through the Gaussian surface. By Gauss’ law, the net charge enclosed by the Gaussian surface must be zero, so the induced charges on the inner surface = +5μC.

17. Cylindrical Symmetry圓柱對稱 Find the electric around a line charge: There is no flux through the top and bottom of the Gaussian surface and the total flux through the curve surface is: The result is same as we have found before using Coulomb’s law, but the calculation here is much simpler.

18. Planar symmetry平面對稱 • non-conducting plane • this time, the field on the left side is not zero. • the total flux is now 2EA.

19. Spherical symmetry球對稱 For a charged spherical shell, the electric field outside the shell is: by Gauss’ law on S2 make use of spherical symmetry Inside the shell: by Gauss’ law, there is no charge enclosed by S1, so there is no field inside the shell

20. For a charged sphere with a uniformcharge density ρ: The electric field outside of the sphere, of course, is: When r < R, then where q’ is given by