Gauss s Law. Lecture 3. Chapter Course website:

Transcription

1 Lecture 3 Chapter 24 Gauss s Law Course website:

2 Today we are going to discuss: Chapter 24: Section 24.2 Idea of Flux Section 24.3 Electric Flux Section 24.4 Gauss s Law

3 Gauss s Law

4 The idea behind Gauss s law It looks like number of lines passing through a closed surface are related to the amount of charge inside (Gauss s Law) But, first, we need to learn how to count lines Flux

5 Before stating Gauss s Law, we need to introduce the electric flux

6 The Basic Idea of Flux Imagine holding a rectangular wire loop of area A in front of a fan. The volume of air flowing through the loop each second depends on the angle between the loop and the direction of flow. The flow is maximum through a loop that is perpendicular to the airflow. No air goes through the same loop if it lies parallel to the flow.

7 The Area Vector Before defining the electric flux, we need to introduce the area vector. Let s define an area vector to be a vector in the direction of, perpendicular to the surface, with a magnitude A equal to the area of the surface. Vector has units of m 2.

8 The Electric Flux Consider 1) a uniform electric field E 2) a flat surface The electric flux through a surface of area A can be defined as the dot-product: θ

9 ConcepTest The electric flux through the shaded surface is Electric Flux A) 0 B) 200 N m 2 /C C) 400 N m 2 /C D) Some other value

10 ConcepTest The electric flux through the shaded surface is Electric Flux A) 0 B) 200 N m 2 /C C) 400 N m 2 /C D) Some other value

11 ConcepTest Electric Flux The electric flux through the shaded surface is A) 0 B) 400cos20 N m 2 /C C) 400cos70 N m 2 /C D) 400 N m 2 /C E) Some other value

12 The Electric Flux (general case) Consider 1) a non-uniform electric field E 2) area is not flat (curved) Divide the surface into many small pieces of area A. a) Each piece is small enough that it is essentially flat b) The field is nearly uniform over each piece c) Thus, the formula from the previous slide can be used The electric flux through each small piece is: The electric flux through the whole surface is the surface integral:

13 The Electric Flux through a closed surface Consider 1) a non-uniform electric field E 2) a closed surface NOTE: For a closed surface, we use the convention that the area vector da is defined to always point toward the outside. The electric flux through a closed surface: closed surface

14 Gauss s Law q 4 For any closed surface enclosing total charge Q in, the net electric flux through the surface is: q 1 q 3 q 2 da E q 5 Φ Gaussian surface Gaussian surface Properties: 1) It works for any closed surface called Gaussian (A Gaussian surface is an imaginary, mathematical surface) 2) Q in is the net charge enclosed by the Gaussian surface (charges outside must not be included) 3) Distribution of Q in doesn't matter This result for the electric flux is known as Gauss s Law. Both, Gauss s law and Coulomb s law, help to find electric fields based on distribution of charges.

15 Gauss s Law/Symmetry Φ Gaussian surface Evaluation of this surface integral is often difficult. However, when the charge distribution has sufficient symmetry (spherical, cylindrical, planar), evaluation of the integral becomes simple.

16 ConcepTest Which spherical Gaussian surface has the larger electric flux? Gauss s Law A) Surface A B) Surface B C) They have the same flux D) Not enough information to tell Flux depends only on the enclosed charge, not the radius.

17 Example Electric Flux Determine the electric flux through each surface Φ Flux depends only on the enclosed charge, not the radius. S1: Φ=(+Q-3Q)/ε 0 S2: Φ=(+Q+2Q-3Q)/ε 0 S3: Φ=(+2Q-3Q)/ε 0 S4: Φ=0 (no charge inside) S5: Φ=(+2Q)/ε 0

18 Example E field outside of a charged dielectric sphere Example 24.3 A total charge Q is spread uniformly throughout a dielectric sphere of radius R. What is the electric field outside the sphere (r>r)? Notice a remarkable feature of this result: The field outside the sphere is exactly the same as it would have been if all the charge had been concentrated at the center.

19 Thank you See you next time

20 Example Electron moving parallel to E An electron is released from rest at the surface of the negative plate. How long does it take the electron to cross to the positive plate? Example 23.8

21 Electron moving perpendicular to E An electron with a horizontal initial velocity v 0 enters a parallel-plate capacitor where the electric field is E. Write the equation of its path. Example