Chapter 24

# Chapter 24 - Chapter 24 Gauss Law In this chapter we will...

This preview shows pages 1–9. Sign up to view the full content.

Chapter 24 Gauss’ Law In this chapter we will introduce the following new concepts: The flux (symbol Φ ) of the electric field Symmetry Gauss’ law We will then apply Gauss’ law and determine the electric field generated by: An infinite, uniformly charged insulating plane An infinite, uniformly charged insulating rod A uniformly charged spherical shell A uniform spherical charge distribution We will also apply Gauss’ law to determine the electric field inside and outside charged conductors.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
ˆ n ˆ n Consider an airstream of velocity that is aimed at a loop of area . The velocity vector is at angle with respect to the ˆ loop normal . The product cos is know n as v A v n vA θ θ Φ = Flux of a Vector. r r the . In this example the flux is equal to the volume flow rate through the loop (thus the name flux). depends on . It is maximum and equal to for 0 ( perpendicular to the loop vA v θ θ Φ = flux Note 1: r plane). It is minimum and equal to zero for 90 ( parallel to the loop plane). cos . The vector is parallel to the loop normal and has magnitude equal to . v vA v A A A θ θ = = Note 2 : r r r r
ˆ n ˆ n ˆ n Consider the closed surface shown in the figure. In the vicinity of the surface assume that we have a known electric field . The flux of the electric field thro h ug E Φ Flux of the Electric Field. r the surface is defined as follows: 1. Divide the surface into small "elements" of area . 2. For each element calculate the term cos . 3. Form the sum . 4. Take the limit of the sum a A E A EA E A θ = Φ = P r r r r 2 Flux SI unit: N m / C s the area 0. The limit of the sum becomes the integral: The circle on the integral sign indicates that the surface is closed. When we apply Gauss' la A E dA Φ = Note 1: r r ° w the surface is known as "Gaussian." is proportional to the net number of electric field lines that pass through the surface. Φ Note 2 : E dA Φ = r r °

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Example 24-1: Net Flux for a Uniform Field.
Example 24-2: Flux for a Nonuniform Field.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
ˆ n ˆ n ˆ n 0 enc 0 enc The flux of through any closed surface net charge enclosed by the surface. Gauss' law can be formulated as follows: In equ ation form: Equivalently: E q q ε ε = Φ = Gauss' Law r 0 enc ε E dA q = P r r ° 0 enc q ε Φ = 0 enc ε E dA q = P r r ° Gauss' law holds for closed surface. Usually one particular surface makes the problem of determining the electric field very simple. When calculating the net charge inside a c Note 1: any Note 2 : losed surface we take into account the algebraic sign of each charge. When applying Gauss' law for a closed surface we ignore the charges outside the surface no matter how large they are . Note 3 : Examp 1 0 1 2 0 2 3 0 3 4 0 4 1 2 3 4 Surface : , Surface : Surface : 0, Surface : 0 We refer to , , , as "Gaussian surfaces." S q S q S S q q S S S S ε ε ε ε Φ = + Φ = - Φ = Φ = - + = le : Note :
ˆ n ˆ n ˆ n EXAMPLE: Figure shows five charged lumps of plastic and an electrically neutral coin. The cross section of a Gaussian surface S is indicated. What is the net flux through the surface if q 1 = q 4 = +3.1nC, q 3 = q 5 = -5.9 nC, and q 2 = -3.1nC.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
ˆ n ˆ n ˆ n PROBLEM 4: Four charges: 2q, q, -q, and -2q.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern