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Flux and Gauss’s Law Spring 2008. Last Time: Definition – Sort of – Electric Field Lines DIPOLE FIELD LINK CHARGE.

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Presentation on theme: "Flux and Gauss’s Law Spring 2008. Last Time: Definition – Sort of – Electric Field Lines DIPOLE FIELD LINK CHARGE."— Presentation transcript:

1 Flux and Gauss’s Law Spring 2008

2 Last Time: Definition – Sort of – Electric Field Lines DIPOLE FIELD LINK CHARGE

3 Field Lines  Electric Field

4 Last time we showed that

5 Ignore the Dashed Line … Remember last time.. the big plane?  0 E=0  0 E=0 We will use this a lot!

6 NEW RULES (Bill Maher) Imagine a region of space where the ELECTRIC FIELD LINES HAVE BEEN DRAWN. The electric field at a point in this region is TANGENT to the Electric Field lines that have been drawn. If you construct a small rectangle normal to the field lines, the Electric Field is proportional to the number of field lines that cross the small area.  The DENSITY of the lines.  We won’t use this much

7 What would you guess is inside the cube? A. A positive charge B. A negative charge C. Can’t tell D. An Alien E. A car

8 What about now? A. A positive charge B. A negative charge C. Can’t tell D. An Alien E. A car

9 How about this?? A. Positive point charge B. Negative point charge C. Large Sheet of charge D. No charge E. You can’t tell from this

10

11 Which box do you think contains more charge? A. Top B. Bottom C. Can’t tell D. Don’t care

12 All of the E vectors in the bottom box are twice as large as those coming from the top box. The top box contains a charge Q. How much charge do you think is in the bottom box? 1. Q 2. 2Q 3. You can’t tell 4. Leave me alone.

13 So far … The electric field exiting a closed surface seems to be related to the charge inside. But … what does “exiting a closed surface mean”? How do we really talk about “the electric field exiting” a surface?  How do we define such a concept?  CAN we define such a concept?

14 Mr. Gauss answered the question with..

15 Another QUESTION: Solid Surface Given the electric field at EVERY point on a closed surface, can we determine the charges that caused it??

16 A Question: Given the magnitude and direction of the Electric Field at a point, can we determine the charge distribution that created the field? Is it Unique? Question … given the Electric Field at a number of points, can we determine the charge distribution that caused it?  How many points must we know??

17 Still another question Given a small area, how can you describe both the area itself and its orientation with a single stroke!

18 The “Area Vector” Consider a small area. It’s orientation can be described by a vector NORMAL to the surface.  We usually define the unit normal vector n.  If the area is FLAT, the area vector is given by A n, where A is the area.  A is usually a differential area of a small part of a general surface that is small enough to be considered flat.

19 The “normal component” of the ELECTRIC FIELD E n EnEn

20 DEFINITION FLUX E n EnEn

21 We will be considering CLOSED surfaces  The normal vector to a closed surface is DEFINED as positive if it points OUT of the surface. Remember this definition!

22 “Element” of Flux of a vector E leaving a surface For a CLOSED surface: n is a unit OUTWARD pointing vector.

23 This flux is LEAVING the closed surface. 

24 Definition of TOTAL FLUX through a surface

25 Flux is A. A vector B. A scaler C. A triangle

26 Visualizing Flux n is the OUTWARD pointing unit normal.

27 Definition: A Gaussian Surface Any closed surface that is near some distribution of charge

28 Remember n E  A  Component of E perpendicular to surface. This is the flux passing through the surface and n is the OUTWARD pointing unit normal vector!

29 Example Cube in a UNIFORM Electric Field L E E is parallel to four of the surfaces of the cube so the flux is zero across these because E is perpendicular to A and the dot product is zero. Flux is EL 2 Total Flux leaving the cube is zero Flux is -EL 2 Note sign area

30 Simple Example r q

31 Gauss’ Law n is the OUTWARD pointing unit normal. q is the total charge ENCLOSED by the Gaussian Surface. Flux is total EXITING the Surface.

32 Simple Example UNIFORM FIELD LIKE BEFORE E AA E E No Enclosed Charge

33 Line of Charge L Q

34 From SYMMETRY E is Radial and Outward

35 What is a Cylindrical Surface?? Ponder

36 Looking at A Cylinder from its END Circular Rectangular Drunk

37 Infinite Sheet of Charge  cylinder E h We got this same result from that ugly integration!

38 Materials Conductors  Electrons are free to move.  In equilibrium, all charges are a rest.  If they are at rest, they aren’t moving!  If they aren’t moving, there is no net force on them.  If there is no net force on them, the electric field must be zero. THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO!

39 More on Conductors Charge cannot reside in the volume of a conductor because it would repel other charges in the volume which would move and constitute a current. This is not allowed. Charge can’t “fall out” of a conductor.

40 Isolated Conductor Electric Field is ZERO in the interior of a conductor. Gauss’ law on surface shown Also says that the enclosed Charge must be ZERO. Again, all charge on a Conductor must reside on The SURFACE.

41 Charged Conductors E=0 E - - - - - Charge Must reside on the SURFACE  Very SMALL Gaussian Surface

42 Charged Isolated Conductor The ELECTRIC FIELD is normal to the surface outside of the conductor. The field is given by: Inside of the isolated conductor, the Electric field is ZERO. If the electric field had a component parallel to the surface, there would be a current flow!

43 Isolated (Charged) Conductor with a HOLE in it. Because E=0 everywhere inside the conductor. So Q (total) =0 inside the hole Including the surface.

44 A Spherical Conducting Shell with A Charge Inside.

45 Insulators In an insulator all of the charge is bound. None of the charge can move. We can therefore have charge anywhere in the volume and it can’t “flow” anywhere so it stays there. You can therefore have a charge density inside an insulator. You can also have an ELECTRIC FIELD in an insulator as well.

46 Example – A Spatial Distribution of charge. Uniform charge density =  charge per unit volume (Vectors) r E O A Solid SPHERE

47 Outside The Charge r E O R Old Coulomb Law!

48 Graph R E r

49 Charged Metal Plate E is the same in magnitude EVERYWHERE. The direction is different on each side. E  ++++++++++++++++ ++++++++++++++++ E A A

50 Apply Gauss’ Law  ++++++++++++++++ ++++++++++++++++ E A A Same result!

51 Negatively Charged ISOLATED Metal Plate - --- ---  E is in opposite direction but Same absolute value as before

52 Bring the two plates together A     e e B As the plates come together, all charge on B is attracted To the inside surface while the negative charge pushes the Electrons in A to the outside surface. This leaves each inner surface charged and the outer surface Uncharged. The charge density is DOUBLED.

53 Result is ….. A     e e B  E E=0

54 VERY POWERFULL IDEA Superposition  The field obtained at a point is equal to the superposition of the fields caused by each of the charged objects creating the field INDEPENDENTLY.

55 Problem #1 Trick Question Consider a cube with each edge = 55cm. There is a 1.8  C charge In the center of the cube. Calculate the total flux exiting the cube. NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE! Easy, yes??

56 Note: the problem is poorly stated in the text. Consider an isolated conductor with an initial charge of 10  C on the Exterior. A charge of +3mC is then added to the center of a cavity. Inside the conductor. (a) What is the charge on the inside surface of the cavity? (b) What is the final charge on the exterior of the cavity? +3  C added +10  C initial

57 Another Problem   m,q both given as is  Gaussian Surface Charged Sheet

58 -2   m,q both given as is  mg qE T Free body diagram 

59 -3 (all given)

60 A Last Problem A uniformly charged cylinder.   R


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