Chapter 24

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

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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.

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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 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 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 h as magnitude equal to . v vA v A A A = = 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 &

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

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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 = Φ = Gauss' Law r 0 enc ε E dA q = P r r & 0 enc q εΦ = 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 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

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## This note was uploaded on 06/01/2008 for the course PHYS 232 taught by Professor Goksu during the Spring '08 term at Millersville.

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

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