2/20/2009. Lecture notes Electrostatics

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1 PHYS0 SPRING 009 Lectue notes Electostatics 1

2 What is Electicity Static effects known since ancient times. Static chages can be made by ubbing cetain mateials togethe. Descibed dby Benjamin Fanklin as a fluid. Excess of the fluid was descibed as positive. Deficit of the fluid was descibed as negative. Like chages epel, unlike chages attact.

3 A Quantitative Appoach Chales Augustin de Coulomb fist descibed elationship of foce and chages. Coulomb invented the tosion balance to measue the foce between bt chages. Basic unit of chage in SI system is called the Coulomb One of the basic ieducible measuements. Like mass, length and time. 3

4 Fundamental Chage Due to atomic stuctue. Positivelychaged heavy nuclei and negativechaged electons. Chaged objects have excess o deficit of electons. Potons and electons have equal and opposite chage: e = 1.60 x C Rubbing some objects causes electons to move fom one object to anothe. Negatively chaged objects have an excess of electons. Positively chaged objects have a deficit of electons. Nt h i h i l d t t Net change in chage in a closed system must be zeo. 4

5 Fundamental Chage Poblem 18.6 Wate has a mass pe molecule of 18 g/mol and each molecule has 10electons. How many electons ae in one lite of wate? electons molecules 1mol 1g 10 ml N = 1molecule 1mol 18g 1ml 1lite N = What is the net chage of the electons? q 6 electons lite ( electons)( C / electon ) = Ne= q = C 5

6 Consevation of Chage Net change in chage in any closed system is zeo. O, the change of chage in anyvolumeequals equals thenet chage enteing and leaving the system: ( Chage) = ( Oiginal Chage) + ( Chage in) ( Chage out) We denote the amount of chage by the lette Q o q If Chage is ceated, it must be equal and opposite. q = q0 + qin -qout 6

7 Consevation of Chage Two identical metal sphees of chage 5.0 C and 1.6 C ae boughttogethe togethe and then sepaated. What is the chage on each sphee? q =3.4C total = q 1+ q Since the chage is distibuted equally, each must have an equal chage: q total q = 1.7C 1 = q = 7

8 Insulato Conductos and Insulatos Chage stays whee it is deposited. Wood, mica, Teflon, ubbe. Conducto Chage moves feely within the object. Metals exhibit high conductivity Due to atomic stuctue. Is the outemost electon(s) () tightly bound? 8

9 By fiction Chaging Objects Rubbing diffeent mateials Example: fu and ebonite By contact Bi Binging i a chaged object into contact with ihanothe (chaged o unchaged) object. By induction Binging a chaged object nea a conducto tempoally causes opposite chage to flow towads the chaged object. Gounding that conducto dains the excess chage leaving the conducto oppositely chaged. 9

10 Foce Between Chages Coulomb found that the foce between two chages is popotional to the poduct of thei chages and invesely popotion to the squae of the distance between them: F qq 1 1 Whee q 1 and q epesent the chages and 1 the distance between them. 10

11 Coulomb s Law By using the tosion balance, Coulomb discoveed the constant of popotionality: k = 8.99 N m /C qq F = k 1 1 If both q 1 and q ae both positive, the foce is positive. If both q 1 and q ae both negative, the foce is positive. If both q 1 is positive and q is negative, the foce is negative Hence, the sign of the foce is positive fo epulsion and negative fo attaction. ti 11

12 Coulomb s Law Pinciple of Supeposition Electic foce obeys the law of linea supeposition the net foce on a chage is the vecto sum of the foces due to each chage: F i qq i j k ˆ ij i j ij = Add the vectos: 3 1 The net foce on chage 4 is F 14 +F 4 +F 34 =F net Red ae positive chages Blue ae negative chages F 4 F 34 4 F 14 1

13 Coulomb s Law Pinciple of Supeposition The foces add fo as many chages as you have: F i qq i j k ˆ ij i j ij = Red ae positive chages Blue ae negative chages 13

14 Coulomb s Law Pinciple of Supeposition Electic foce obeys the law of linea supeposition the net foce on a chage is the vecto sum of the foces due to each chage: F i qq i j k ˆ ij i j ij = The net foce on chage 5 is the vecto sum of all the foces on chage 5. 8 Red ae positive chages Blue ae negative chages 7 14

15 Coulomb s Law Two tiny conducting sphees ae identical and cay chages of 0.0 μc and μc. They ae sepaated by a distance of.50 cm. (a) What is the magnitude of the foce that each sphee expeiences, and is the foce attactive o epulsive? qq F = k = ( N m /C )( C)( C) ( m) N The net foce is negative so that the foce is attactive. F = (b) The sphees ae bought into contact and then sepaated to a distance of.50 cm. Detemine the magnitude of the foce that each sphee now expeiences, and state whethe the foce is attactive o epulsive. Since the sphees ae identical, i the chage on each afte bi being sepaated is onehalf the net chage, so q 1 + q = +15μC. ( 9 )( N m /C C)( C) ( ) qq 1 3 F = k = = N m The foce is positive so it is epulsive. 15

16 Coulomb s Law 18.1 An electically neutal model aiplane is flying in a hoizontal cicle on a 3.0 m guideline, which is nealy paallel to the gound. The line beaks when the kinetic enegy of the plane is 50.0 J. Reconside the same situation, except that now thee is a point chage of +q on the plane and a point chage of q at the othe end of the guideline. In this case, the line beaks when the kinetic i enegy of the plane is 51.8 J. Find the magnitude of the chages. Fom the definition of kinetic enegy, we see that mv = (KE), so that Equation fo the centipetal foce becomes Centipetal t foce F mυ = = ( ) KE c So, kq ( KE) chaged ( KE) unchaged Tmax + () 1 T = = 1443 p 1443 max Centipetal foce ( ) 16

17 Centipetal foce Coulomb s Law 18.1 (cont.) kq ( KE) chaged ( KE) unchaged Tmax + () 1 Tmax = 1443 = 1443 Centipetal foce Subtacting Equation () fom Equation (1) eliminates T max and gives ( ) kq Solving fo q gives = ( ) ( ) KE KE chaged unchaged q ( ) ( ) ( )( ) KE KE chaged unchaged 3.0 m 51.8 J 50.0 J 5 = = = C k N m / C 17

18 The Electic Field Electic chages ceate and exet and expeience the electic foce. qq i j q j F = k ˆ F = q k ˆ i ij i i ij i j ij i j ij We define the Electic Field as the foce a test chage would feel if placed at that point: q j k ˆ E = F = qe i j The electic field exists aound all chaged objects. 18

19 Electic Foce Lines Michael Faaday saw the foce caused by one chage to be fom lines emanating fom the chages. Lines begin on Positive chages and end on Negative chages, flowing smoothly though space, neve cossing. 19

20 Electic Foce Lines Faaday imagined these lines and the foce was geate whee the lines ae dense, weake whee they ae spead out. 0

21 Electic Field Lines Go out fom positive chages and into negative chages. Positive Negative Neutal 1

22 The Invese Squae Law Many laws based on the geomety of thee dimensional i space: I sound Gm1m Fg = P L = I = 4π 4π qq 1 Fe = k light

23 The Electic Field 18.8 Fou point chages have the same magnitude of.4 x 10 1 C and ae fixed to the cones of a squae that is 4.0 cm on a side. Thee of the chages ae positive and one is negative. Detemine Dt the magnitude of the net electic field that t exists it at the cente of the squae. B + + C The dawing at the ight shows each of the field contibutions at the cente of the squae (see black E D dot). Each his diected d along a diagonal of the squae. Note that E D and E B point in opposite diections and, E A theefoe, cancel, since they have the same magnitude. E C E B In contast E A and E C point in the same diection towad A cone A and, theefoe, combine to give a net field that - + D is twice the magnitude of E A o E C. In othe wods, the net field at the cente of the squae is given by the following vecto equation: Σ E= EA + EB + EC + ED = EA + EB + EC EB = EA + EC = E A 3

24 The Electic Field B + + C 18.8 (cont.) The magnitude of the net field is Σ E = EA = kq In this esult is the distance fom a cone to the cente of the squae, which is one half of the diagonal distance d. Using L fo the length of a side of the squae and tki taking advantage of the Pythagoean Pth theoem, we have = L/. With this substitution fo, the magnitude of the net field becomes A E D E A E E B C - + D Σ E = k q 4k q Σ E = = L / L ( ) 9 1 ( )( ) N m /C.4 10 C ( m) Σ E = 54 N/C 4

25 Electic Field Lines Review Conceptual Example 1 befoe attempting to wok this poblem. The magnitude of each of the chages in Figue 18 1 is 8.6 x 10 1 C. The lengths of the sides of the ectangles ae cm and cm. Find the magnitude of the electic field at the cente of the ectangle in Figues 18 1a and b. The figue at the ight shows the configuation given in text Figue 18.1a. The electic field at the cente of the ectangle is the esultant 1 of the electic fields at the cente due to each of the fou chages. As discussed in Conceptual Example 11, the magnitudes of the electic field at the cente due to each of the fou chages ae equal. Howeve, the fields poduced by the chages in cones 1 and 3 ae in opposite 4 4 diections. Since they have the same magnitudes, they combine to give zeo esultant. The fields poduced by the chages in cones and 4 point in the same diection (towad cone ). Thus, E C = E C + E C4, whee E C is the magnitude of the electic field at the cente of the ectangle, and E C and E C4 ae the magnitudes of the electic field at the C C4 cente due to the chages in cones and 4 espectively. Since both E C and E C4 have the same magnitude, we have E C = E C. + q - q q + q 5

26 Electic Field Lines (cont.) The distance, fom any of the chages to the cente of the ectangle, can be found using the Pythagoean theoem: E C d = (3.00 cm) +(5.00 cm) = 5.83 cm d Theefoe, = =.9 cm =.9 10 m d θ 3.00 cm kq (899 ( N m /C )( C) = EC = = = N/C (.9 10 m) The figue at the ight shows the configuation given in text - q - q Figue 18.1b. All fou chages contibute tib t a non zeo component 1 1 to the electic field at the cente of the ectangle. As discussed E in Conceptual Example 11, the contibution fom the chages in 13 E 4 C cones and 4 point towad cone and the contibution fom the chages in cones 1 and 3 point towad cone q + q 5.00 cm 6

27 Electic Field Lines (cont.) Notice also, the magnitudes of E4 and E13 ae equal, and, fom the fist pat of this poblem, we know that E 4 = E 13 = N/C The electic field at the cente of the ectangle is the vecto sum of E 4 and E 13. The x components of E 4 and E 13 ae equal in magnitude and opposite in diection; hence (E 13 ) x (E 4 ) x = 0 Theefoe, 13 x 4 x E = ( E ) + ( E ) = ( E ) = ( E )sinθ C 13 y 4 y 13 y 13 - q - q 1 1 E 13 E 4 C q + q Fom Figue, we have that t 5.00 cm 5.00 cm sinθ = = = d 5.83 cm C 13 ( ) ( )( ) E = E sinθ = N/C = N/C 7

28 The Electic Field The electic field exists aound all chaged objects. This is shown by electic field lines. Constucted identically i to the foce lines. 8

29 Dawing Electic Field Lines Numbe of lines at a chage is popotional to the chage. 9

30 Electic Field Lines The chage with 3q must have thee times as many lines as the chage with +q and 1.5 times the lines fo +/ q. 30

31 Paallel Plates of Chage Adding fields fom each point on suface causes field lines paallel to the plates to cancel. Positively Chaged Negatively Chaged Away fom the edges, the field is paallel and pependicula to the plate(s). 31

32 Electic Field & Conductos In conductos excess electons will move when an electic field is pesent. Fo steady state (i.e. no moving chages), the net electic field must be zeo. F = qe F = E = Theefoe the chages move to the suface(s) of the conducto until they cancel the electic field inside the conducto. Field lines can stat and end on a conducto at ight angles to the suface of the conducto. 0 3

33 Electic Field Lines Two paticles ae in a unifom electic field whose value is N/C. The mass and chage of paticle 1 ae m 1 = 1.4 x 10 5 kg and q 1 = 7.0 μc, while the coesponding values fo paticle ae m 6 5 =.6 x 10 5 kg and q = +18 μc. C Initially the paticles ae at est. The paticles ae both located on the same electic field line, but ae sepaated fom each othe by a distance d. When eleased, they acceleate, but always emain at this same distance fom each othe. Find d. d E x +x q 1 = 7.0 μc m 1 = kg q = +18 μc m = kg The dawing shows the two paticles in the electic field E x. They ae sepaated by a distance d. If the paticles ae to stay at the same distance fom each othe afte being eleased, they must have the same acceleation, so a x,1 = a x,. Accoding to Newton s second law, the acceleation a x of each paticle is equal to the net foce ΣF x acting on it divided by its mass m, o a = Σ F / m. x x 33

34 18.44 (cont.) Electic Field Lines d q 1 = 7.0 μc q = +18 μc E x +x m 1 = kg m = kg Note that t paticle 1 must be to the lft left of paticle. If paticle 1 wee to the ight iht of paticle, the paticles would acceleate in opposite diections and the distance between them would not emain constant. Σ qq F 1 = q E qq x,1 1 x k Σ F 1 = q E + k x, x d d qq 1 qq ΣF qe x,1 1 x k 1 a d ΣF qe, x + k x x,1 = = a d x, = = m m m m 1 1 Note the signs indicated the diection of the vecto pointing fom chage to chage 1 fo the foce on chage 1. 34

35 18.44 (cont.) Electic Field Lines d q 1 = 7.0 μc q = +18 μc E x +x m 1 = kg m = kg The acceleation of each chage must be the same if the distance bt between the chages emains unchanged: a = a 1, x, x qq qq qe 1 x k q Ex + k d = d m m Solving fo d, d x 1 ( 1+ ) ( ) kq q m m = = E q m q m m 35

36 Electic Flux Flux (fom Latin meaning to flow) Electic flux measues amount of electic field passing though a suface. E φ E Flux though suface is Aea of suface times the field flowing though that suface. Φ = E EA So, to calculate, l find the nomal to the suface. Then find the angle between the field and thatnomalvecto: φ Φ = ( cos ) E E φ A 36

37 Electic Flux Φ = ( cos ) E E φ A This can be expessed as the vecto dot poduct: Φ E = E A E A = E A cos φ o E A = EA x x + EA y y + EA z z E So, to calculate, l find the nomal to the suface. Then find the angle between the field and thatnomalvecto: φ 37

38 Electic Flux A ectangula suface (0.16 m x 0.38m) is oiented in a unifom electic field of 580 N/C. What is the maximum possible electic flux though the suface? The maximum possible flux occus when the electic field is paallel to the nomal of the ectangula suface (that is, when the angle between the diection of the field and the diection of the nomal is zeo). Then, ( ) ( )( ) Φ = ( Ecos φ) A= 580 N / C (cos 0 ) 0.16 m 0.38 m E Φ = E 35 N m / C 38

39 Gauss Law Electic Flux though a closed suface is popotional to the chage enclosed by that suface: qenc Φ = E The fundamental constant ε 0 is called the pemittivity of fee space. It is elated to 1 Coulomb s law by k =, ε 0 = 8.85 x 10 1 C /(N m ) 4πε 0 Gauss Law is most useful when the electic field on the entie suface is constant. Take a point chage. Constuct a sphee centeed on the point chage (all points on the suface equidistant fom the chage) the electic field is constant. ε 0 Since the field is adial, it is pependicula to the concentic spheical suface eveywhee so φ = 0, cos φ = 1 qenc q Φ E = EAcosφ = ε ε enc 0 0 The diection of the field is adially outwad fom the chage (o inwad towads the chage if q is negative). qenc q E( 4π ) = E = ε0 4πε 0 39

40 Gauss Law fo a plate of chage In this case we take ou closed suface to be a box oiented such that two faces ae paallel to the plate of chage and the othes ae pependicula to the plate. Since the electic field fom a plate of chage is pependicula to the suface of the plate, the sides that ae pependicula to the plate have no flux. E The chage enclosed by the box is the suface chage density multiplied by the aea the box passes though the plate: σ A. The aea of the box with flux is A: q σ A σ Φ E = EA = E = ε ε ε

41 Gauss Law fo a line of chage A long, thin, staight wie of length L has a positive chage Q distibuted unifomly along it. Use Gauss law to find the electic field ceated by this wie at a adial distance. As the hint suggests, we will use a cylinde as ou Gaussian suface. The electic field must point adially outwad fom the wie. Because of this symmety the field must be the same stength pointing adially outwad on the suface of the cylinde egadless of the point on the cylinde, thus we ae justified in using Gauss Law. Long staight wie Gaussian cylinde E Gausssian cylinde L End view 41

42 Gauss Law fo a line of chage (cont.) Let us define the flux though the suface. Note that the field is paallel to the endcaps, thus the angle bt between the nomal fom the suface (which his paallel l to the wie) is pependicula to the field, thus φ = π/, so, cos(φ)=0 fo the ends. Fo the side if the cylinde, the flux is E side ( π ) Φ = EA = E L Long staight wie Gaussian cylinde Thus, fom Gauss Law, q ( π L) Q enc Φ E = = ε 0 ε 0 E = 0 E Q πε L It is customay to efe to chage density λ = Q/L, λ E = πε 1 0 L E 3 End view Gausssian cylinde 4

43 Wok Let us take an object with chage q in a constant electic field E. If we elease this chage, the electic field causes a foce to move the chage. q F= qe a= E m Let us define the field as E= Eˆ x, so thee is only one dimension to conside. Let us also set the initial velocity and position to be both zeo. Let the object move though a distance d. The equations of motion ae 1 x= at υ = at 1 So, afte a time t, the object moves x = d, d = at t = d / a t = υ / a υ d = υ = ad qe 1 a a υ= d m υ = qed m 43

44 Consevation of Enegy 1 mυ = qed The left hand side is the kinetic enegy the paticle has afte moving a distance d. That means, the ight hand side must epesent some fom of potential enegy the object had pio to being eleased. Types of Enegy aleady coveed: Tanslational kinetic enegy Rotational kinetic enegy Gavitational potential enegy Elastic (sping) potential enegy m I gh PE) Etotal = υ + ω + m + kx + ( E Electic Potential Enegy Cutnell & Johnson use (EPE), howeve, it is often moe convenient to use U. 44

45 Electic Potential Enegy Like gavitational potential enegy, EPE is defined with espect to the enegy at some location. Impotance lies in the diffeence between electic potential enegy between two points. 45

46 Example Poblem 19.4 A paticle has a chage of +1.5 μc and moves fom point A to point B, a distance of 0.0 m. The paticle expeiences a constant electic foce, and its motion is along the line of action of the foce. The diffeence between bt the paticle s electic potential enegy at A and B is U A U B = +9.0 x 10 4 J. Find the magnitude and diection of the electic foce that acts on the paticle and the magnitude and diection of the electic field that the paticle expeiences. The wok done mustbe equal to the change in potential enegy: Wok = U U FΔ x = U U A B A B 4 UA UB J 3 = = = N F (fom A to B) Δx ( 0.0m) But, we know that fo a paticle in an electic field: F = qe UA UB UA UB qe = E = Δ x q Δ x J 3 E = = N / C C 0.0 m ( )( ) 46

47 Electic Potential The electic potential enegy of an object depends on the chage of an object. Once again, we can find a measuement independent of the objects chage: V The SI unit fo V is measued in Volts ( = Joules/Coulomb) = ( EPE ) q A common unit of enegy is the electon Volt. Specifically, How much potential enegy does an electon get if moved though a potential of 1 Volt. 19 1eV = J 47

48 Electic Potential Like the Electic field, the electic potential is independent of any test chage placed to measue a foce. Unlike the electic field, the potential is a scale, not a vecto. By using calculus methods it can be shown that the electic potential due to a point chage is V kq = Note that the value of the electic potential ildepends d on the sign of the chage. 48

49 Electic Potential fo a point chage Example: What was the potential diffeence due to a chage q A = 10.0 nc, at the distances 0 = 10.0 cm and 1 = 0.0 cm? kq kq A A Δ V = V0 V1 = Δ V = kq A Δ V = ( N m / C )( 10 C) 0.1m 0.m ΔVV = 450 V 49

50 Electic Potential fo a point chage Example: A chage q A is held fixed. A second chage q B is placed a distance 0 away and is eleased. What is the velocity of q B when it is a distance 1 away? We use the consevation ofenegy: Enegy = mυ + U mυ + U = mυ + U 1 mυ1 = U0 U1 = qb ( V0 V1) υ q ( V V ) = m

51 Electic Potential fo a point chage υ = q V m ( V ) Find the electic potential at 0 and 1 : kqa kq V 0 = V 1 = υ q kqa kq A kqaq B 1 1 = = m 0 1 m 0 1 A O, υ1 = qδv / m 51

52 Electic Potential fo a point chage Fo a single chage, we set the potential at infinity to zeo. kq V = So what is the potential enegy contained in two chages? Bing in the chages fom infinity. Thefistchage isfee asitdoes notexpeience anyfoces. The second moves though the field ceated by the fist. So, the enegy in the system must be Δ U = q ΔV Δ V = V V kq kq kq Δ V = = U = kq q 1 1 5

53 Potential Enegy of a collection of chages Let s add a thid chage to this goup. Δ U = q Δ V + q Δ V + q ΔV kq kq Δ V = Δ V = kq q kq q kq q Δ U = Each successive chage expeiences the field (i.e. potential) ceated by the pevious chages. Be caeful not to double count! 53

54 Example Poblem An electon and a poton ae initially vey fa apat (effectively an infinite distance apat). They ae then bought togethe to fom a hydogen atom, in which the electon obits the poton at an aveage distance of x m. What is the change in the electic potential enegy? The fist chage comes in fo fee, the second chage expeiences the field of the fist, so, Δ = Δ Δ = kq1 kq1 kq1 Δ V = = U q V V V V ( ) ( N m /C )( C ) C 11 Δ U = Δ U = J m 54

55 Electic Field and Potential Let a chage be moved by a constant electic field fom points A to B (let Δd be that distance). W = F Δ d W = qe Δ d Now, we know that the wok done must be equal to the negative change in electic potential enegy: But potential enegy is simply: Δ U = qeδd AB qδ V = qeδd AB Thus: ΔV E = Δd 55

56 Electic Field and Potential ΔV E = Δ d This implies that the field lines ae pependicula to lines (sufaces) of equal potential. This equation only gives the component of the electic field along the displacement Δd. To tuly find the electic vecto field, you must find the components in each diection: Δ V Δ V Δ V Ex = Ey = Ez = Δx Δy Δz 56

57 Electic Field and Potential If the potentials at the following points (x, y) ae V(1,1) = 30V, V(0.9,1)=15V, 15V, and V(1,0.9)=40V. V. What is the (appoximate) electic field at (1,1)? E x 30V 15V = = 150 V / m 1.0m 0.9m ΔV ΔV ΔV Ex = Ey = Ez = Δx Δy Δz E y 30V 40V = = 100 V / m 1.0m 0.9m E θ E ( ) ( ) = 150V + 100V = 180V 100V = = 150V 1 o tan 34 E y (1,1) x 57

58 Example Poblem Detemine the electic field between each of the points. E AB V E = Δ Δx x 5.0V 5.0V = = 0V/m 0.0m 0.0m E BC E CD 3.0V 5.0V = = 10.0 V / m 0.40m 0.0m 1.0V 3.0V = = 5.0 V / m 0.80m 0.40m 58

59 Equipotential Sufaces Electic field lines ae pependicula to equipotential sufaces. Electic field lines ae pependicula to sufaces of conductos. Thus, sufaces of conductos ae equipotential sufaces. The inteio of conductos have no electic field and ae theefoe equipotential. The Electic potential of a conducto is constant thoughout the conducto. 59

60 Example An electon is eleased at the negative plate of a paallel plate capacito and acceleates to the positive plate (see the dawing), (a) As the electon gains kinetic enegy, does its electic potential enegy incease o decease? Why? The electic potential enegy deceases. The electic foce F is a consevative foce, so the total enegy (kinetic enegy plus electic potential enegy) emains constant as the electon moves acoss the capacito. Thus, as the electon acceleates and its kinetic enegy inceases, its electic potential enegy deceases. (b) The diffeence in the electon s electic potential enegy between the positive and negative plates is U pos U neg. How is this diffeence elated to the chage on the electon ( e) and to the diffeence in the electic potential between the plates? Thechange in the electon s electic potentialenegy enegy isequal to the chage on the electon ( e) times the potential diffeence between the plates. (c) How is the potential diffeence elated to the electic field within the capacito and the displacement of the positive plate elative to the negative plate? 60

61 Example (cont.)() (c) How is the potential diffeence elated to the electic field within the capacito and the displacement of the positive plate elative to the negative plate? The electic field E is elated to the potential diffeence between the plates and the displacement Δs by ( Vpos Vneg ) E = Δs Note that (V pos V neg ) and Δs ae positive numbes, so the electic field is a negative numbe, denoting that it points to the left in the dawing. Poblem The plates of a paallel plate capacito ae sepaated by a distance of 1. cm, and the electic field within the capacito has a magnitude of.1 x 10 6 V/m. An electon stats fom est at the negative plate and acceleates to the positive plate. What is the kinetic enegy of the electon just as the electon eaches the positive plate? Use consevation of enegy: KEpos + U pos = KEneg + Un eg KEpos + evpos = KEneg + evneg Total enegy at positive plate Total enegy at negative plate 61

62 19.68 (cont.) KE + ev = KE + ev Example pos pos neg neg But the electon stats fom est on the positive plate, so, the kinetic i enegy at the positive ii plate is zeo. ev pos = KE + ev neg neg KE KE neg neg ( ) = ev = ev pos Vpos Vneg But, E =, so, Δs ev neg ( pos Vn eg ) KEneg 19 6 ( EΔ s) = ( C)( V / m)( 0.01 m) = e KE neg 15 = J 6

63 Lines of Equal Potential Lines of equal elevation on a map show equal potential enegy (mgh). 63

64 Example 19.9 Two equipotential sufaces suound a +1.5 x 10 8 C point chage. How fa is the 190 V suface fom the 75.0 V suface? The electic potential ti lvv at a distance fom a point chage q is V = kq/. / The potential is the same at all points on a spheical suface whose distance fom the chage is =kq/v. The adial distance 75 fom the chage to the 75.0 V equipotential suface is 75 = kq/v 75, and the distance to the 190 V equipotential suface is 190 = kq/v 190. The distancebetween these twosufaces is kq kq = = kq V V V V ( ) N m 1 1 = C 1.1 m C + = 75.0 V 190 V

65 Capacitos A capacito consists of two conductos at diffeent potentials being used to stoe chage. Commonly, these take the shape of paallel plates, concentic sphees o concentic cylindes, but any conductos will do. q = CV, whee C is the capacitance, measued in SI units of Faads (Coulomb / Volt o C / J ), V is the voltage diffeence between the conductos and q is the chage on each conducto (one is positive, the othe negative). Capacitance is depends only upon the geomety of the conductos. 65

66 Paallel Plate Capacito Field in a paallel plate capacito is E q = ε A But, the a constant electic field can be defined in tems of electic potential: V E = V = Ed d Substituting into the equation fo capacitance, q= CV = CEd q A ε 0 q = C d C ε0a = d 0 66

67 Othe Capacitos Two concentic spheical shells 4πε C = a b Cylindical capacito has a capacitance pe unit length: C πε = 0 L b ln a 67

68 Enegy stoed in a capacito A battey connected to two plates of a capacito does wok chaging the capacito. Each time a small amount of chage is added to the capacito, the chage inceases, so the voltage between the plates inceases. V q/c V =q/c Δq q q 68

69 Enegy stoed in a capacito Potential Enegy is U = qv We see that adding up each small incease in chage Δq x V is just the aea of the tiangle. 1 U = qv q= CV V q/c Such that, Δq q 1 U = U = CV V V =q/c q q So the enegy density (i.e.enegy enegy pe unit volume) must be U 1 κε A 1 d 0 = ( Ed) ( ) U = κε 0 E Ad Ad is the aea of the capacito! U 1 u = = κε 0E Ad 69

70 Example How much enegy is stoed in a paallel plate capacito made of 10.0 cm diamete cicles kept.0 mm apat when a potential diffeence of 10 V is applied? Find the capacitance: C ε0a επ 0 = = d d The enegy stoed is U 0 = CV = = 1 ( C / (N m )) π ( 0.05m) ( 10V) 1 επv d 0.00m U = J ( ) 70

71 Electic Polaization in Matte Electic chage distibution in some molecules is not unifom. Example: Wate both H atoms ae on one side. In the pesence of an Electic Field, wate will align opposite to the field. 71

72 Electic Polaization in Matte In the absence of a field. Molecules l will be oiented andomly. In an electic field, they line up opposite to the field. 7

73 Electic Polaization in Matte The esulting electic field is the sum of the extenal field and the fields geneated by the molecules. The electic field stength though the matte is educed. The eduction of the field depends on the enegy in the molecules, the stength of the field each one makes and the density of the molecules. This can be measued and the mateial is said to have a dielectic constant κ. 73

74 Dielectics This dielectic constant changes the value of the pemittivity in space (ε) Coulombs law in a dielectic is F = qq 1 4πε 1 ˆ Whee ε is elated to ε 0 by a constant κ: ε = κε 0 The dielectic constant fo a vacuum is κ = 1. All othe mateials have a dielectic constant value geate than 1. Wate at about 0 C has κ = Ai at about 0 C has κ =

75 Dielectics ε = κε 0 How does this affect the capacitance of paallel plates? C d ε A κε 0 A = = d d C = κc d 0 75

76 Dielectic Beakdown If the field in the dielectic becomes too stong (i.e. voltage is too geat), the mateial begins to conduct cuent this is called dielectic beakdown. When this happens the chage flows though When this happens the chage flows though the mateial to the othe plate and neutalizes the capacito. 76

77 Example Two hollow metal sphees ae concentic with each othe. The inne sphee has a adius of m and a potential of 85.0 V. The adius of the oute sphee is m and its potential is 8.0 V. If the egion between the sphees is filled with Teflon, find the electic enegy contained in this space. The electic enegy stoed in the egion between the metal sphees is 1 1 u = κε 0E U = κε 0E ( Volume) The volume between the sphees is the diffeence in the volumes of the sphees: ( Volume) = π π = π ( ) Since E = ΔV/Δs and Δs = 1, so, E = ΔV/( 1 ) 1 ΔV U = 0 ( ) J κε 1 3 π = 77