Chapter 22 Gauss s Law

Transcription

1 Chapter 22 Gauss s Law Lecture by Dr. Hebin Li

2 Goals for Chapter 22 To use the electric field at a surface to determine the charge within the surface To learn the meaning of electric flux and how to calculate it To learn the relationship between the electric flux through a surface and the charge within the surface To use Gauss s law to calculate electric fields To learn where the charge on a conductor is located

3 Calculation of electric field What is the electric field due to a particular charge distribution? In general, it is difficult to calculate the electric field due to an arbitrary charge distribution.

4 The relationship between electric charge and field Determine the charge from field. Distribution of electric charges Coulomb s law Gauss s law Distribution of electric field Determine the field from charge. Gauss s law is an alternative to Coulomb s law They are completely equivalent They provide different solutions to the same problem It is sometimes more convenient to use Gauss s law, especially when the system is symmetric.

5 Charge in a closed box? Consider an imaginary box (a closed surface) which does not affect the charge or field in any way. What can you know about the charge inside if the electric field at the surface is known? What about the electric field at the surface if the charge inside is known?

6 Charge in a box: some examples Positive charge inside box, electric field lines go outward Negative charge inside box, electric field lines go inward Flux represents electric field lines passing through a surface The net flux is zero if there is no charge or zero net charge inside box, regardless if there is electric field or not.

7 Charge in a box: what determines the flux? Doubling the enclosed charge doubles the flux. Doubling the box dimensions does NOT Change the flux. Whether there is a net outward or inward electric flux through a closed surface depends on the sign of the enclosed charge. Charges outside the surface do not give a net electric flux through the surface. The net electric flux is directly proportional to the net amount of enclosed charge but is otherwise independent of the size of the closed surface.

8 Calculating electric flux Consider a flat area A perpendicular to a uniform electric field E. The electric flux through this area is defined as Φ E = EA If the flat area is not perpendicular to the field, then fewer field lines pass through it and the electric flux is smaller. Φ E = EA cos φ The electric flux for a uniform field E and a flat surface A is given by Φ E =E A

9 Formal definition of electric flux In general, if the electric field E is not uniform and A is a curved surface. Then we need to divide A into small elements da. We calculate the electric flux through each element and integrate the results to obtain the total flux: Φ E = E cos φ da = E da = E da The integral is a surface integral.

10 Example: calculating electric flux A flat sheet of paper of area m 2 is oriented so that the normal to the sheet is at an angle of 60 to a uniform electric field of magnitude 14 N/C. (a)find the magnitude of the electric flux through the sheet. (b)does the answer to part (a) depend on the shape of the sheet? (c)for what angle φ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest?

11 Positive and negative flux The flux is positive if the enclosed charge is positive, and negative if the charge is negative.

12 Electric flux through a cube An imaginary cubical surface is in a region of uniform electric field E. The electric field flux through the cube is always zero.

13 Example: calculating electric flux A hemispherical surface with radius r in a region of uniform electric field E has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

14 Electric flux through a sphere The electric flux through a sphere centered on a point charge q. The electric field anywhere on the surface is The electric flux is E = 1 4πε 0 q r 2 Φ E = EdA = E da = EA = 1 4πε 0 q r 2 4πr2 = q ε 0 Φ E = q ε 0

15 Gauss s law Gauss s Law: The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface, divided by ε 0. Carl Friedrich Gauss ( ) Φ E = E da = Q encl ε 0

16 Applications of Gauss s law Although Gauss s law is equivalent to Coulomb s law, it is sometimes more convenient to use Gauss s law if the system under consideration has some sort of symmetry and a proper Gaussian surface is chosen. Example: Use Gauss s law to find the electric field due to a point charge. 1. Choose a Gaussian surface: a sphere with a radius r, centered at the charge q. 2. Symmetry argument: due to the symmetry, the electric field at any point on the Gaussian surface is perpendicular to the surface, and has the same magnitude. 3. Calculate the flux and apply Gauss s law: Φ E = E 4πr 2 = q ε 0 So, E = 1 4πε 0 q r 2 Coulomb s law

17 Applications of Gauss s law Example: We place a total positive charge q on a solid conducting sphere with radius R. Find the electric field at any point inside or outside the sphere.

18 Applications of Gauss s law Example: Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. Find the magnitude of the electric field at any point inside or outside the sphere.

19 Applications of Gauss s law Example: Use Gauss s law to find the electric field caused by a thin, flat, infinite sheet with a uniform positive surface charge density σ.

20 Applications of Gauss s law Example 3: Electric charge is distributed uniformly along an infinitely long, thin wire. The charge per unite length is λ (assumed positive). Find the electric field using Gauss s law.

21 Charges on conductors The charges are distributed on the surface. The electric field inside the conductor is zero. If there is a cavity inside, there is no charge field on the surface of the cavity. There is no electric field inside the conductor or cavity If there is a positive charge inside the cavity, there are negative charges on the surface of the cavity and there is electric field inside the conductor.

22 Field at the surface of a conductor The electric field at the surface of any conductor is always perpendicular to the surface and has magnitude / 0. E A = σa ε 0 E = σ ε 0 E = 0

23 The Van de Graaff generator The Van de Graaff generator, shown in the figure below, operates on the same principle as in Faraday s icepail experiment.

24 Electrostatic shielding A conducting box (a Faraday cage) in an electric field shields the interior from the field.

25 Example: Concentric Spherical Shells. A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducing spherical shell with inner radius c and outer radius d. The inner shell has total charge +2q, and the outer shell has charge +4q. (a) Calculate the electric field in terms of q and the distance r from the common center of the two shells for (i) r < a; (ii) a < r < b; (iii) b < r < c; (iv) c < r < d; (v) r > d. Show the results in a graph of the radial component of electric field as a function of r. (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell.