Chapter 23: Gauss s Law

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1 Chapte 3: Gauss s Law Homewok: Read Chapte 3 Questions, 5, 1 Poblems 1, 5, 3

2 Gauss s Law Gauss s Law is the fist of the fou Maxwell Equations which summaize all of electomagnetic theoy. Gauss s Law gives us an altenative to Coulomb s Law fo calculating the electic field due to a given distibution of chages.

3 Gauss s Law: The Geneal Idea The net numbe of electic field lines which leave any volume of space is popotional to the net electic chage in that volume. E lines leaving V volume V E lines enteing V

4 Flux The flux Φ of the field E though the suface S is defined as Φ E da S The meaning of flux is just the numbe of field lines passing though the suface.

5 Best Statement of Gauss s Law The outwad flux of the electic field though any closed suface equals the net enclosed chage divided by ε.

6 Gauss s Law: The Equation da S E Q /ε enc S is any closed suface. Q enc is the net chage enclosed within S. da is an element of aea on the suface of S. da is in the diection of the outwad nomal. ε SI units

7 Flux Examples Assume two chages, +q and q. Find fluxes though sufaces. Remembe flux is negative if lines ae enteing closed suface. Φ 1 +q/ε Φ q/ε Φ 3 Φ 4 (q q )/ε

8 Anothe Flux Example Given unifom field E, find flux though net.

9 Must have closed suface, so let S be (net+cicle). Then Q enc so Gauss s law says Φ S. But Φ cicle π a E Φ net π a E

10 Gauss Coulomb Given a point chage, daw a concentic sphee and apply Gauss s Law: E Φ Gauss E( ) ˆ E da π E( ) 4 Φ q / ε E ( ) q /(4 ) kq / π ε!

11 If a vecto field v passes though an aea A at an angle θ, what is the name fo the poduct v A vacosθ? Q.3-1 (1) cul () enegy (3) flux (4) gadient

12 Q.3-1 What is the name fo the poduct v A vacosθ? 1. Cul. Enegy 3. Flux 4. Gadient

13 Q.3-1 Φ v A vacosθ flux fom the Latin to flow. (1) cul () enegy (3) flux (4) gadient

14 Q.3- (1) () (3) (4) (5) Φ Φ Φ Φ Φ ( Q + Q ) k 1 ( Q + Q ) 1 Q1 + Q 4 πε a Q1 Q 4 πε a Two chages Q 1 and Q ae inside a closed cubical box of side a. What is the net outwad flux though the box? / ε / a

15 Q.3- Two chages Q 1 and Q ae inside a closed cubical box of side a. What is the net outwad flux though the box? Φ Φ Φ Φ Φ ( Q + Q ) k Q 4 Q 4 ( Q + Q ) πε πε Q a Q a / ε / a

16 Q.3- Φ Φ Φ Φ Φ ( Q + Q ) k Q 4 Q 4 ( Q + Q ) πε πε Q a Q a / Two chages Q 1 and Q ae inside a closed cubical box of side a. What is the net outwad flux though the box? ε / a Gauss: The outwad flux of the electic field though any closed suface equals the net enclosed chage divided by ε.

17 Application of Gauss s Law We want to compute the electic field at the suface of a chaged metal object. This gives a good example of the application of Gauss s Law. Fist we establish some facts about good conductos. Then we can get a neat useful esult: E σ /ε

18 Fields in Good Conductos Fact: In a steady state the electic field inside a good conducto must be zeo. Why? If thee wee a field, chages would move. Chages will move aound until they find the aangement that makes the electic field zeo in the inteio.

19 Chages in Good Conductos Fact: In a steady state, any net chage on a good conducto must be entiely on the suface. Why? If thee wee a chage in the inteio, then by Gauss s Law thee would be a field in the inteio, which we know cannot be tue.

20 Field at the Suface of a Conducto Constuct closed gaussian suface, sides pependicula to metal suface, face aea A. Flux though left face is zeo because E. Field is pependicula to suface o chages would move, theefoe flux though sides is. So net outwad flux is EA.

21 Field at the Suface of a Conducto Let σ stand fo the suface chage density (C/m ) Then Q enc σ A. Now Gauss s Law gives us Φ Q enc / ε EA σ A / ε E σ / ε

22 Lage Sheet of Chage σ chage /aea suface chage density Φ Q Gauss law ε Q σa Q A Φ E da EA EA σa ε E σ ε

23 Long Line of Chage λ λ chage/length linea chage density l Φ Q ε Q lλ Gauss law Φ E d A EA E Top view A πl E λ πε

24 Summay fo diffeent dimensions Point chage, d Φ E 4π E Q /(4πε Q ) / ε Line chage, d1 E 1 d Φ E π le λ l λ /(πε ) / ε Suface chage, d Φ AE E σ /(ε ) σa/ ε

25 Potential

26 Outline fo today Potential as enegy pe unit chage. Thid fom of Coulomb s Law. Relations between field and potential.

27 Potential Enegy pe Unit Chage Just as the field is defined as foce pe unit chage, the potential is defined as potential enegy pe unit chage: F qe and U qv The SI unit fo potential is the volt. (1V1J/C) Potential is often casually called voltage. As with potential enegy, it is eally the potential diffeence which is impotant.

28 Potential Enegy Diffeence If a chage q is oiginally at point A, and we then move it to point B, the potential enegy will incease by the amount of wok we have done in caying the chage fom A to B. A U q B ds U A W AB B A B F ext dw F ds ext ds

29 Potential Diffeence The potential diffeence between point A and point B is this wok pe unit chage. W AB qv AB B A F ext ds B A qe ds V AB V B V A B E A ds

30 Relation between E and V E x V E x dx And E x dv dx etc.

31 Potential Relative to Infinity We have defined potential diffeence V. But to have a value fo V itself we need to decide on a zeo point. In cicuits, we define the eath to have V, and often gound the cicuit. In electostatics we nomally define V fa away fom all chages. Then at point P we wite V(P) to mean V(P) - V( ).

32 Potential at a point The potential at point P is the wok equied to bing a one-coulomb test chage fom fa away to the point P. P q ds P F ds V ( P) E ds q P

33 Example 1: Unifom Field V d E Edx Ed V () V ( d) Ed Wok equied to push a 1-coulomb test chage against the field fom xd to x. x Wok foce X distance

34 Q.4-1 Suppose, using an xyz coodinate system, in some egion of space, we find the electic potential is V ( x) Ax whee A is a constant. What is the x-component of the electic field in this egion? A. B. E x E x Ax Ax C. E Ax x

35 Q. 4-1 Given V ( x) Ax whee A is has the constant value A 5 V / m. What is the x-component of the electic field? Solution: d dx E V A Ax x dx dx (1) E Ax () E Ax (3) E Ax x x x

36 Futhemoe What ae the y and z components of the electic field in the pevious question? E y d dy V A dx dy and also E z d V dz

37 Case of a Single Point Chage Q R P q ds V ( P) E ds E( ) d P R d V ( R) E( ) d kq R kq R R 1 kq 1 R kq R

38 Coulomb s Law fo V So we now have a thid fom of Coulomb s Law: 1. F kqq /. E kq / 3. V kq /

39 Potential is not a Vecto Adding foces and fields means adding vectos: finding the esultant vecto. Adding potentials means adding numbes, and taking account of thei signs. But it is much simple than adding vectos. Thus the thid fom of Coulomb s Law is the simplest!

40 Example 1: Adding Potentials P a a Q a -Q V ( P) V + V kq + kq 1 Note that the field at point P is not zeo! a a

41 Q. 4- Unifomly chaged od with chage of Q bent into ac of 1 with adius R. What is V(P), the electic potential at the cente? kq kq kq ( 1) + () (3) (4) R R R kq R

42 Q.4- Unifomly chaged od with chage of Q bent into ac of 1 with adius R. What is V(P), the electic potential at the cente? A. +kq/r B. -kq/r C. +kq/r D. -kq/r

43 Q. 4- What is V(P), the electic potential at the cente? Solution: All bits of chage ae at the same distance fom P. Thus V kdq k dq k ( Q) R R kq kq kq ( 1) + () (3) (4) R R R kq R

44 Potential Enegy of Some Chages The potential enegy U of a goup of chages is the wok W equied to assemble the goup, binging each chage in fom infinity. We can show that the esult is U U + U + U L Whee the potential enegy of each pai is of the fom U kq q / 1 1 1

45 Binding Enegy If the total potential enegy U of a goup of chages is negative that means we have to do wok to pull them apat. The magnitude of this negative potential enegy is called the binding enegy. Examples: Removing an electon fom an atom to fom a positive ion. Removing a space pobe fom eath s gavitational field.

46 Example 3: Chaged Ring In Ch. the E field of a chaged ing is calculated. Hee we compute V fist and then use it to get E. Get V at P on axis of cicula ing, a distance z fom cente: V ( P) kdq

47 Example 3 Continued Key point: All bits of chage ae the same distance fom point P! kdq k kq V ( P ) dq R + z Note we have no stuggling with angles o adding many little vectos as we do if we compute E.

48 Using V to get E Now that we have V on the axis we can get E z on the axis by diffeentiation: E z This is slightly messy, but we just need to n emembe that df n 1 df nf dz dz whee hee 1/ f n ( R + z ) dv dz

49 So finally we get d kq kqz E( z) dz R ( ) 3/ + z R + z which is exactly the esult the textbook gets in Ch. by diect integation of the field. And also: E E y z d dy d dz V V A dx dy

50 Futhemoe What ae the y and z components of the electic field in the pevious question? E y d dy V A dx dy and also E z d V dz

51 Summay of Basics U qv V kq/ W AB q ( V B V A ) V E x dv dx E x dx etc.

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