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PHY 102: Lecture Symmetry 3.2 Concept of Flux

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1 PHY 102: Lecture 3 3.1 Symmetry 3.2 Concept of Flux
3.3 Calculating Electric Flux 3.4 Gauss’ Law

2 PHY 102: Lecture 3 Electric Field
3.1 Symmetry

3 What Good is Symmetry Symmetry lets us find the shape of the electric field Symmetry arguments allow us to rule out many conceivable field shapes as simply being incompatible with the symmetry of the charge distribution Symmetry does not tell us about the strength (magnitude) of the field or how it changes with distance

4 Symmetry - 1 A charge distribution is symmetric if there are a group of geometrical transformations that don’t cause any physical change Suppose you close your eyes while a friend transforms a charge distribution in one of the following three ways Translate (displace) the charge parallel to an axis Rotate the charge about and axis Reflect the charge in a mirror

5 Symmetry - 2 You open your eyes
You might be able to tell if the charge distribution changed by observing a visual difference Or results of an experiment with charged particles could reveal that distribution has changed If nothing you can see or do reveals any change then we say the distribution is symmetric under that particular transformation

6 Symmetry - 3 A charge distribution that is symmetrical under these three groups of geometrical transformations is said to be cylindrically symmetric Other charge distributions have other types of symmetry

7 Symmetry - 4 Interest in symmetry can be summed up in a single statement The symmetry of the electric field must match the symmetry of the charge distribution

8 Symmetry - 5 What can we learn about the electric field of an infinitely long, uniformly positive line charge (cylindrical symmetry) Could the field look like this?

9 Symmetry – 6 This field looks the same if:
It’s translated parallel to the cylinder axis (right or left movement) If up and down are exchanged by reflecting in a plane coming out of the page

10 Symmetry – 7 Reflect the field in a plane perpendicular to the axis (exchange right and left) This reflection does not change charge distribution However, the field becomes the diagram on the right This change in the field is detectable because the field now has a component to the left instead of to the right

11 Symmetry – 8 Reflect the field in a plane perpendicular to the axis (exchange right and left) This reflection does not change charge distribution However, the field becomes the diagram on the right This change in the field is detectable because the field now has a component to the left instead of to the right

12 Symmetry – 9 The field which makes a distinction between left and right is not cylindrically symmetric and thus is not a possible field The electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the cylinder axis

13 Symmetry – 10 Could the field look like this?
Looking down the axis of cylinder Electric field vectors are restricted to planes perpendicular to the cylinder and thus do not have any components parallel to the cylinder axis This field is symmetric for rotations about the axis

14 Symmetry – 11 This field is not symmetric for a reflection in a plane containing the axis (figure b) The electric field of a cylindrically symmetric charge distribution cannot have a component tangent to the circular cross section

15 Symmetry – 12 This is the only remaining possible field shape
Side View This is the only remaining possible field shape The electric field is radial, pointing straight out from the cylinder This is the one electric field shape that matches the symmetry of the charge distribution

16 Types of Symmetry The field is radial toward or away from the center

17 PHY 102: Lecture 3 Electric Field
3.2 Concept of Flux

18 Concept of Flux - 1 An opaque box surrounds a region of space
(a) The field is coming out of each face of the box. There must be a positive charge in the box. An opaque box surrounds a region of space We can’t see what’s in the box Electric field vectors come out of each face of the box Can we figure out what’s in the box?

19 Concept of Flux - 2 Electric fields point away from positive charges
(a) The field is coming out of each face of the box. There must be a positive charge in the box. Electric fields point away from positive charges Electric field comes out of every face of box (a) It seems clear that the box contains a positive charge or charges Similarly, box (b) contains a negative charge

20 Concept of Flux - 3 Electric field points into box (c) on the left
(a) The field is coming out of each face of the box. There must be a positive charge in the box. Electric field points into box (c) on the left An equal electric charge points out on the right This might be the electric field between a large positive somewhere out of sight on the left and a large negative charge off to the right

21 Gaussian Surface - 1 Surround a region of space with a closed surface
Surface that divides space into inside and outside regions A closed surface through which an electric field passes is called a Gaussian surface This is an imaginary surface, not a physical surface Figure (a) is a spherical Gaussian surface

22 Gaussian Surface – 2 The spherical symmetry of the electric field vectors poking through the surface tells us that the positive charge inside must be spherically symmetric And centered at the center of the sphere Electric field is everywhere perpendicular to the spherical surface Electric field has the same magnitude at each point on the surface

23 Gaussian Surface - 3 Another example
Electric field emerges from four sides of cube (a) But not from the top or bottom We can’t be sure what charge is inside the box Figure (b) uses a different Gaussian surface, a closed cylinder

24 Gaussian Surface - 4 With a better choice of surface, we can tell that the cylindrical Gaussian surface surrounds some kind of cylindrical charge distribution, such as a charged wire Electric field is everywhere perpendicular to the cylindrical surface Electric field has the same magnitude at each point on the surface

25 Gaussian Surface - 5 Consider the spherical surface in figure (a)
The protruding electric field tells us there’s a positive charge inside It might be a point charge located on the left side, but we can’t say A Gaussian surface that doesn’t match the symmetry of the charge distribution isn’t useful

26 Gaussian Surface - 6 The nonclosed surface in figure (b) doesn’t provide much help either What appears to be a uniform electric field to the right could be due to a large positive charge on the left, a large negative charge on the right, or both A nonclosed surface doesn’t provide enough information

27 Gaussian Surface - 7 The electric field “flows” out of a closed surface surrounding a region of space containing a net positive charge “Flows” into a closed surface surrounding a region of space containing a net negative charge The electric field may flow through a closed surface surrounding a region of space in which there is no net charge, but the net flow is zero The electric field pattern through the surface is particularly simple if the closed surface matches the symmetry of the charge distribution inside The amount of electric field passing through a surface is called the electric flux

28 Gaussian Surface - 8 There is an outward flux through a closed surface around a net positive charge There is an inward flux through a closed surface around a net negative charge There is no net flux through a closed surface around a region of space in which there is no net charge

29 PHY 102: Lecture 3 Electric Field
3.3 Calculating Electric Flux

30 Calculating Electric Flux
Figure shows a surface of area A in a uniform electric field E n is the unit vector perpendicular to the surface Only the component E = Ecosq passes through the surface

31 Definition of Electric Flux
Electric Flux is Fe Measures the amount of electric field passing through a surface of area A If the perpendicular to the surface is titled at an angle q from the field Fe = EAcosq

32 Example–Parallel Plate Capacitor
Two 0.01 m2 parallel plates One is charged +5x10-9 C, the other is charged -5x10-9 C A 0.01 m x 0.01 m surface between the plates is tilted so its normal makes a 450 angle with the electric field What is electric flux through the surface? Fe = EAsurfacecosq = (q/e0Aplate)Asurfacecosq Fe = (5 x 10-9/8.85 x x 0.01)1.0 x 10-4 x cos45 Fe = 4.00 Nm2/C

33 Example–Cylindrical Charge-1
Electric field of a cylindrical charge distribution is E = E0(r2/r02) Calculate the electric flux through a closed cylinder of length L and radius R that is centered along the z-axis The electric field extends radially outward from the z-axis with cylindrical symmetry The z-component is Ez = 0 The cylinder is a Gaussian surface

34 Example–Cylindrical Charge-2
Figure (a) is a view of the electric field looking along the z-axis Field strength increases with increasing radial distance Field is symmetric around the z-axis Figure (b) is the closed Gaussian surface for which we need to calculate the electric flux We can place the cylinder anywhere along the z-axis because extends to infinity in that direction

35 Example–Cylindrical Charge-3
To calculate the flux, divide the closed cylinder into three surfaces: the top, the bottom, and the cylindrical wall The electric field is tangent to the surface at every point on the top and bottom surfaces For the top and bottom cosq = cos90 = 0 The flux through top and bottom surfaces is zero For the cylindrical wall the electric field is perpendicular to the surface everywhere cosq = cos0 = 1

36 Example–Cylindrical Charge-4
The electric field has constant magnitude E = E0(r2/r02) at every point on the cylindrical wall Fe = EAwall Net flux through the closed surface is Fe = Ftop + Fbottom + Fwall = EAwall Awall = 2pRL

37 Example–Cylindrical Charge-5
Notice the important role played by symmetry The electric field was perpendicular to the wall and of constant value at every point on the wall because the Gaussian surface had the same symmetry as the charge distribution

38 PHY 102: Lecture 3 Electric Field
3.4 Gauss’ Law

39 Purpose of Gauss’ Law Gauss’ law allows the electric field of some continuous distributions of charge to be found much more easily than does Coulomb’s law

40 Gaussian Surface – Point Charge - 1
Figure shows a spherical Gaussian surface of radius r centered on a positive charge q This is an imaginary surface There is a net flux through this surface because the electric field points outward at every point on the surface

41 Gaussian Surface – Point Charge - 2
The electric field is perpendicular to the surface at every point on the surface From Coulomb’s law it has the same magnitude E = q/4pe0r2 at every point on the surface This simple situation arises because the Gaussian surface has the same symmetry as the electric field

42 Gaussian Surface – Point Charge - 3
This is equally valid for a negative charge In this case the flux is negative because the electric field points inward

43 Gaussian Surface – Point Charge - 4
Note that the electric flux depends on the amount of charge but not on the radius of the sphere The point charge is the only source of electric field Every electric field line passing through a small radius spherical surface also passes through a large-radius spherical surface Hence, the electric flux is independent of r

44 Gaussian Surface – Point Charge - 5
This is also the flux through any closed surface surrounding a point charge The net electric flux is zero through a closed surface that does not contain any net charge

45 Gauss’ Law – Field of Point Charge
Gaussian Surface is a sphere with the charge at the center Electric field is the same everywhere on the sphere Electric field is perpendicular to sphere Electric Flux = EA = q/e0 E(4pr2) = q/e0 E = q/4pr2e0 = kq/r2

46 Gauss’ Law – Field Outside Shell of Charge
Gaussian Surface is a sphere with the shell at the center (outside shell) Electric field is the same everywhere on the sphere Electric field is perpendicular to sphere Electric Flux = EA = q/e0 E(4pr2) = q/e0 E = q/4pr2e0 = kq/r2 (r > R) Charge q uniformly spread over spherical shell

47 Gauss’ Law – Field Inside Shell of Charge
Gaussian Surface is a sphere with the shell at the center (inside shell) Electric field is the same everywhere on the sphere Electric field is perpendicular to sphere Electric Flux = EA = q/e0 Charge is 0 inside Gaussian Surface E(4pr2) = 0/e0 E = 0 (r < R) Charge q uniformly spread over spherical shell

48 Gauss’ Law – Field in Parallel Plate Capacitor-1
Gaussian Surface is a cylinder starting inside one of the plates Electric field is the same everywhere on the sphere Electric field is perpendicular to the plates Surface 1 Electric Flux = EA = q/e0 Field is 0 inside metal plate Flux = EA = 0

49 Gauss’ Law – Field in Parallel Plate Capacitor-2
Gaussian Surface is a cylinder starting inside one of the plates Electric field is the same everywhere on the sphere Electric field is perpendicular to the plates Surface 2 Electric Flux = EA Field is parallel to surface cosf = cos90 = 0 Flux = EA = 0

50 Gauss’ Law – Field in Parallel Plate Capacitor-3
Gaussian Surface is a cylinder starting inside one of the plates Electric field is the same everywhere on the sphere Electric field is perpendicular to the plates Surface 3 Electric Flux = EA

51 Gauss’ Law – Field in Parallel Plate Capacitor-4
Electric Flux = EA = q/e0 Electric Flux = EA Electric Flux = EA Charge on plate contained within cylinder is q EA = q/e0 E = (q/A)e0 charge/area = s = q/A E = s/e0


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