Notes 11 Gauss s Law II

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1 ECE 3318 Applied Electicit and Magnetism ping 218 Pof. David R. Jackson Dept. of ECE Notes 11 Gauss s Law II Notes pepaed b the EM Goup Univesit of Houston 1

2 Eample Infinite unifom line chage Find the electic field vecto [ ] l = l C/m 2

3 Eample (cont.) h D nˆ d = Q Assume D= ˆ ( ) D encl = [ ] l l C/m 3

4 Eample (cont.) t nˆ = ˆ ˆ ˆ top side bottom D nˆ d = Q encl b ide view c h LH = D nˆ d ( D ˆ) = ˆ d c t b ( D ˆ ) + ˆ d ( D ˆ ) ( ˆ ) + d 4

5 Eample (cont.) ( D ˆ) ˆ d D d D d D ( πh) LH = = = = 2 c c c = Q = h RH encl l Hence, t ( 2π ) = D h h D = l l 2π c h We then have b E ˆ l = 2πε [ V/m] 5

6 Eample (cont.) umma E ˆ l = 2πε [ V/m] [ ] l = l C/m 6

7 Note About Clindical Coodinates Note: In clindical coodinates, the LH is alwas: LH = D 2 ( πh) Assume: ( ) ( ) φ,, = f ( a function of onl, not φ and ) v ( ) D = ˆ D ( in the ˆ diection, and a function of onl, not φ and ) ( π ) LH = D nˆ d = D ˆ d = D d = D 2 h 7

8 Eample h This eample illustates when Gauss s Law is not useful. D nd ˆ = Q but D ˆ encl D E has moe than one component! h l E is not a function of onl! Finite unifom line chage Note: Although Gauss s law is still valid, it is not useful in helping us to solve the poblem. We must use Coulomb s law. 8

9 Eample Infinite clinde of non-unifom volume chage densit a Find the electic field vecto evewhee Note: This poblem would be ve difficult to solve using Coulomb s law! = < 2 3 v 3 C/m, a 9

10 Eample (cont.) a (a) < a D nˆ d = Q encl h so LH = D 2 ( ) ( πh) D π h = Q 2 encl = < 2 3 v 3 C/m, a 1

11 Eample (cont.) RH = Q = encl V dv v h = = = h/2 2π h/2 h ( 2 ) 2 ( ) 2πh 3 d ddφd v π d v so 4 3 = 2π h 4 ide view RH = Q = encl 3 4 πh 2 11

12 Eample (cont.) LH = D 2 ( πh) 3 4 RH = πh 2 a Hence 3 D h h D = 4 4 ( 2π ) = π so E 3 3 = ˆ [ V/m], 4ε a a 12

13 Eample (cont.) (b) > a LH = D 2 ( πh) a 4 3 π RH = Qencl = 2 h 4 h so Qencl = 3 π ha 2 4 LH = RH 3 D πh π ha 2 4 ( 2 ) = 13

14 Eample (cont.) 3 D h ha a 4 4 ( 2π ) = π D = a Hence, we have E 4 3a = ˆ [ V/m], 4ε a Note: Outside the clinde, the electic field is the same as that coming fom an equivalent line chage located at the cente. a 14

15 Eample (cont.) umma a E 3 3 = ˆ [ V/m], 4ε a E 4 3a = ˆ [ V/m], 4ε a 15

16 Eample Infinite sheet of unifom suface chage densit Find the electic field vecto evewhee 2 s = s C/m 16

17 Eample (cont.) 2 s = s C/m Assume ( ) D= D ˆ Conside fist > A D nˆ d = Q encl 17

18 Eample (cont.) ( ˆ ) ˆ D n d = Q encl LH ( D ˆ) = ˆd top bottom ( D ˆ) ( ˆ) + d A Assume We then have so D A D A= Q D + encl = D + 2AD + = Qencl D + D 18

19 Eample (cont.) Fo the chage enclosed we have RH = Qencl = s A Hence, fom Gauss s law we have LH = RH A 2AD + = s A so A D + = = 2A 2 s s We then also have: Theefoe + s E = ˆ 2ε s E = ˆ 2ε 19

20 Eample (cont.) umma 2 s = s C/m s E = ± ˆ [V/m]; 2ε + fo >, fo < 2

21 Eample = A s = h B s Fom supeposition: (a) > h (b) < < h (c) < A B s s E ˆ = + 2ε 2ε A B s s E ˆ = 2ε 2ε A B s s E ˆ = + 2ε 2ε 21

22 Eample (cont.) A B =, = Choose: s s s s = s = h s (a) > h (b) < < h (c) < E = + E = 2ε 2ε A B ˆ s s A B s s s E = ˆ E = ˆ 2ε 2ε ε E = + E = 2ε 2ε A B ˆ s s 22

23 Eample (cont.) Ideal paallel-plate capacito Metal plates h s Note: The metal plates suppot the chage, but the themselves do not poduce an electic field. s E ˆ ε = s < < h D = ˆ s 23

24 Eample Infinite slab of unifom volume chage densit Find the electic field vecto evewhee d v 3 [ ] C/m 24

25 Eample (cont.) d E v 3 [ ] C/m Assume ˆ ( ) ( ) = ( ) E= E E E ( ) = E (since E () is a continuous function) 25

26 Eample (cont.) d (a) > d / 2 top ( ˆ) D nˆ d = Q ( ˆ) ˆ ( ˆ) ( ˆ) D d + D d = Q bottom ( ) ( ) ( ) encl encl A D A D A= Q = Ad ( / 2) encl v D = d v Altenative choice: Anothe choice of Gaussian suface would be a smmetical suface, smmetical about = (as was done fo the sheet of chage). 3 [ ] C/m v /2 E = d v ˆ d [ V/m ], ( / 2) 2ε 26

27 Eample (cont.) eff Note: If we define = s v d eff E = 2ε s then ˆ [ V/m] (sheet fomula) Q Note: so Q = Ad = A eff s = v v d eff s Q d v eff s A A 27

28 Eample (cont.) (b) < < d / 2 d = = v 3 [ ] C/m ( ) ( ) = = D A D A Q A encl v D = v E = [ ] d ε ˆ v V/m, / 2 28

29 Eample (cont.) umma d v 3 [ ] C/m E ( ε ) d v / 2 E = [ V/m] d 2ε ˆ d v, ( / 2) E = [ V/m] d 2ε ˆ d v, ( / 2) d /2 d /2 ( ) ε d v / 2 E = [ V/m] d d ε ˆ v, /2 /2 Note: In the second fomula we had to intoduce a minus sign, while in the thid one we did not. 29

F = net force on the system (newton) F,F and F. = different forces working. E = Electric field strength (volt / meter)

F = net force on the system (newton) F,F and F. = different forces working. E = Electric field strength (volt / meter) All the Impotant Fomulae that a student should know fom. XII Physics Unit : CHAPTER - ELECTRIC CHARGES AND FIELD CHAPTER ELECTROSTATIC POTENTIAL AND CAPACITANCE S. Fomula No.. Quantization of chage Q =