chapter 22 gauss s law l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 22 Gauss’s Law PowerPoint Presentation
Download Presentation
Chapter 22 Gauss’s Law

Loading in 2 Seconds...

play fullscreen
1 / 28

Chapter 22 Gauss’s Law - PowerPoint PPT Presentation


  • 135 Views
  • Uploaded on

Chapter 22 Gauss’s Law. Electric charge and flux (sec. 22.2 & .3) Gauss’s Law (sec. 22.4 & .5) Charges on conductors (sec. 22.6). C 2012 J. Becker. Learning Goals - we will learn: CH 22

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chapter 22 Gauss’s Law' - tracey


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chapter 22 gauss s law
Chapter 22 Gauss’s Law
  • Electric charge and flux (sec. 22.2 & .3)
  • Gauss’s Law (sec. 22.4 & .5)
  • Charges on conductors (sec. 22.6)

C 2012 J. Becker

slide2

Learning Goals - we will learn: CH 22

• How you can determine the amount of charge within a closed surface by examining the electric field on the surface!• What is meant by electric flux and how you can calculate it.• How to use Gauss’s Law to calculate the electric field due to a symmetric distribution of charges.

slide4

The volume (V) flow rate (dV/dt) of fluid through the wire rectangle (a) is vA when the area of the rectangle is perpendicular to the velocity vector v and (b) is vA cosfwhen the rectangle is tilted at an anglef.

We will next replace the fluid velocity flow vector v with the electric field vector E to get to the concept of electric fluxFE.

Volume flow rate through the wire rectangle.

slide5

(a) The electric flux through the surface = EA.

(b) When the area vector makes an anglefwith the vector E, the area projected onto a plane oriented perpendicular to the flow is A perp.   = A cosf.  The flux is zero whenf= 90o because the rectangle lies in a plane parallel to the flow and no fluid flows through the rectangle

A flat surface in a uniform electric field.

slide6

FE=òE . dA=òE dA cosf=ò E dA = EòdA = E (4p R2) = (1/4peo) q /R2) (4pR2) = q /eo. So the electric flux FE= q /eo.

Now we can write Gauss's Law:FE=òE . dA =ò|EdA| cosf =Qencl /eo

Electric FLUX through a sphere centered on a point charge q.

slide7

The projection of an element of area dA of a sphere of radius R UP onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by 2, so the area element on the larger sphere is 4 dA. The same number of lines of flux pass thru each area element.

Flux FEfrom a point charge q.

slide8

The projection of the area element dA onto the spherical surface is dA cos f.

Flux through an irregular surface.

slide12

Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E

slide13

Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field. 

slide14

Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA

slide15

Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral.

slide16

Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E . dA4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral.5. Determine the value of Qencl from your figure and insert it into Gauss's equation.

slide17

Use the following recipe for Gauss’s Law problems:1. Carefully draw a figure - location of all charges, and direction of electric field vectors E2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.  3. Write Gauss Law and perform dot product E o dA4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the |E| out of the integral.5. Determine the value of Qencl from your figure and insert it into Gauss's equation.6. Solve the equation for the magnitude of E.

C 2012 J. F. Becker

slide18

Gaussian surface

Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

slide19

Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at its center.

Electric field = zero (electrostatic) inside a solid conducting sphere

slide20

A coaxial cylindricalGaussian surface is used to find the electric field outside an infinitely long charged wire.

slide21

Q22.9

For which of the following charge distributions would Gauss’s law not be useful for calculating the electric field?

A. a uniformly charged sphere of radius R

B. a spherical shell of radius R with charge uniformly distributed over its surface

C. a right circular cylinder of radius R and height h with charge uniformly distributed over its surface

D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface

E. Gauss’s law would be useful for finding the electric field in all of these cases.

slide22

A cylindricalGaussian surface is used to find the electric field of an infinite plane sheet of charge.

slide23

(Ignoring

edge effects)

Electric field between two (large) oppositely charged parallel plates.

slide24

“Volume charge density“: r= charge / unit volume is used to characterize the charge distribution.

The electric field of a uniformly charged INSULATING sphere.

slide25

+q

The solution of this problem lies in the fact that the electric field inside a conductor is zero and if we place our Gaussian surface inside the conductor (where the field is zero), the charge enclosed must be zero (+ q – q) = 0.

Find the electric charge q on surface ofa hole in the charged conductor.

slide26

A Gaussian surface drawn inside the conducting material of which the box is made must have zero electric field on it (field inside a con-ductor is zero).  If the Gaussian surface has zero field on it, the charge enclosed must be zero per Gauss's Law.

E

The E field inside a conducting box (a “Faraday cage”) in an electric field.

chapter 23 electric potential
Chapter 23 Electric Potential
  • Electric potential energy (sec. 23.1)
  • Electric potential (sec. 23.2)
  • Calculating elec. potential (sec. 23.3)
  • Equipotential surfaces (sec. 23.4)
  • Potential gradient (sec. 23.5)

C 2012 J. Becker

slide28

Learning Goals - we will learn: ch 23

• How to calculate the electric potential energy (U) of a collection of charges.• The definition and significance of electric potential (V).• How to use the electric potential to calculate the electric field (E).