# Example 1: THE ELECTRIC DIPOLE

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1 Example 1: THE ELECTRIC DIPOLE 1

2 The Electic Dipole: z + P + θ d _ Φ = Q 4πε + Q = Q 4πε 4πε

3 The Electic Dipole: d + _ z + Law of Cosines: θ A B α C A 2 = B 2 + C 2 2ABcosα P ± = 2 ( + d ) 2 2 d 2 2 cosθ 3

4 The Electic Dipole: z + P d + θ Impotant pacbcal appoximabon: d << _ ± = 2 ( + d ) 2 d cosθ 2 4

5 The Electic Dipole: d << ± = 2 ( + d ) 2 d cosθ 2 = 1+ d d cosθ 1 d cosθ x << 1 1± x 1± x 2 1 d 2 cosθ = d 2 cosθ 5

6 The Electic Dipole: z d 2 cosθ d << + d θ _ + d 2 cosθ 6

7 The Electic Dipole: Φ = Q 1 1 4πε + d<< = Q 4πε 1 1 d 2 cosθ 1 1+ d 2 cosθ Q 4πε 1+ d 2 cosθ 1 d 2 cosθ = x << 1 1 1± x 1 x Qd 4πε 2 cosθ 7

8 The Electic Dipole: Φ Qd 4πε 2 cosθ Define and note p Qd ( )ẑ cosθ = ẑ i ˆ ẑ θ ˆ Φ p i ˆ 4πε 2 8

9 The Electic Dipole: p i ˆ Φ 4πε 2 E = Φ = ˆ Φ ˆθ 1 = ˆ Φ θ Qd 4πε cosθ 2 ˆθ 1 θ Qd 4πε cosθ 2 = ˆ 2 Qd 4πε cosθ 3 ˆθ 1 Qd 4πε sinθ 2 = Qd 4πε 3 ( 2 ˆ cosθ + ˆθ sinθ ) ϕ = 0 9

10 The Electic Dipole: E = E = Qd ( 2 ˆ cosθ + ˆθ sinθ ) 4πε 3 Qd 4cos 2 θ + sin 2 θ = Qd 4πε 3 4πε cos 2 θ 10

11 Example 2: FINITE LENGTH LINE OF CHARGE (again) Ealie we found the E- field on the z- axis. Doing anything else would have equied difficult integabons. Hee is a case whee it is easie to find the potenbal and then compute the electic field. 11

12 a dq = ρ d z z ( 0,0, z ) Note the φ- independence ( ) 2 = z ẑ R =, R = 2 + z z ρ 0 z P de de z d E P a dφ = ρ d z 4πε o 12

13 dφ = Φ = = ρ d z 4πε o a a a a ρ d z 4πε o ρ d z 4πε o 2 + ( z z ) 2 ( ) ( z a) 2 = ρ ln z a z a 4πε o z + a + { } dx = ln x + x 2 + a 2 x 2 + a 2 13

14 ( ) ( z a) 2 Φ = ρ ln z a z a 4πε o z + a + E = Φ E z = Φ z = ρ 1 4πε o 2 + z a E = Φ = ρ 4πε o ( ) ( z + a) ( z a) 2 + ( z a) 2 + ( z a) ( z + a) 2 + ( z + a) 2 + ( z + a) 2 14

15 fo z = 0 E z = ρ 1 4πε o a E = ρ 4πε o = ρ 4πε o = ρ a 2πε o ( ) ( 0 + a) 2 = ( z a) 2 + ( z a) 2 + ( z a) ( z + a) 2 + ( z + a) 2 + ( z + a) a 2 a 2 + a a 2 ρ a 2πε o 2 + a 2 + a 2 + a 2 z = 0 Agees with ou ealie esults 15

16 Example 3: INFINITELY LONG LINE OF CHARGE via Gauss s Law 16

17 Infinitely long line chage: Note that the fields MUST be independent of both z and φ Gaussian Suface E = E ˆ z No contibubon ove end caps since ˆ i ẑ = 0 D i da = ε o E ˆ i ˆ dϕ dz + ε o E ˆ i ẑ d dϕ S 2π 0 Cylinde End Caps = ε o E dϕ = 2πε o E = Q enc = ρ E = ρ 2πε o = ρ 2πε o =0 When the necessay symmety exists, Gauss s Law is geneally MUCH simple than Coulomb s Law. 17

18 Example 4: A SPHERICAL CLOUD OF CHARGE 18

19 Spheical cloud of (unifom) chage: Since the chage density is unifom: Gaussian Suface 2 Q Total = ρ 4 3 π 3 Gaussian Suface 1 ρ v a Note that the fields must be independent of both θ and φ, thus E = E ˆ 19

20 Spheical cloud of (unifom) chage: Gaussian Suface 1 ρ v Gaussian Suface 2 a Fo Gaussian Suface 1: ε oe i da = ε o E ˆ i ˆ4π 2 S 1 = Q Enclosed = Q Total a E = Q Total 4πε o 2 3 a 3 = Q Total 4πε o a 3 < a 3 Q Total = ρ 4 3 π 3 E = E ˆ 20

21 Spheical cloud of (unifom) chage: Gaussian Suface 2 Fo Gaussian Suface 2: Gaussian Suface 1 ε o E i d a = 4πε o E 2 S 2 = Q Enclosed = Q Total ρ v a E = Q Total 4πε o 2, > a Q Total = ρ 4 3 π 3 E = E ˆ 21

22 Spheical cloud of (unifom) chage: Gaussian Suface 2 E Q Total 4πε o a 2 Gaussian Suface 1 ρ v a a What is the potenbal? 22

23 Spheical cloud of (unifom) chage: Φ = E ˆ i ˆ d = = Q Total 4πε o 2 d a Q Total 4πε o d Q Total 2 4πε o a d 3 Q Total 4πε o Q Total 4πε o a + Q Total a 2 2 4πε o a 3 2 a > a < a > a < a 23

24 Example 5: AN INFINITE SHEET OF CHARGE 24

25 Infinite sheet of chage: a simple yet impotant esult fo the study of the paallel plate capacito Note how the fields must be independent of x, y, and z Gaussian Suface ρ s z y x E = ẑe z z > 0 ẑe z z < 0 25

26 Infinite sheet of chage: ε o ε o ε o E i da = Q = ρ ( s π 2 ) S E i da = ε o E i da + ε o E i da + ε o E i da S Top Suface Bottom Suface ẑe z i ẑ da + ε ( o ẑe z )i ẑda Top Suface Bottom Suface +ε o ( ±ẑe z )i ˆa dϕ dz Cylindical Side da = ddϕ = π 2 ( ) Cylindical Side 2ε o E z π 2 = π 2 ρ s E z = ρ s 2ε o 26

27 Infinite sheet of chage: E = ẑ ρ s 2ε o z > 0 ẑ ρ s 2ε o z < 0 Since the sheet extends of infinity we would expect touble finding the potenbal: z Φ = ẑe z i ẑ dz = ρ s dz 2ε o z = 27

28 Infinite sheet of chage: Howeve, Φ ab = Φ( b) Φ( a) b = ẑe z i ẑ dz = ρ s dz 2ε o a b a b = ρ s 2ε o a = ρ s 2ε o ( b a) = ρ s 2ε o ( a b) 28

29 Example 6: TWO COAXIAL SHELLS OF CHARGE 29

30 Two coaxial shells of chage: b a z Once again, neglecbng end effects, E = E ˆ h Q ρ sa Note : ρ sa = Q 2π ah, ρ = Q sb 2πbh Q ρ sb 2π ahρ sa = 2πbhρ sb ρ sb = a b ρ sa 30

31 Two coaxial shells of chage: b Gaussian Suface 1 a z Once again, neglecbng end effects, Gaussian Suface 3 E = E ˆ Gaussian Suface 2 The chage enclosed by sufaces one and thee is zeo, hence E = 0 inside the inne cylinde and outside the oute cylinde. Also, the top and bofom sufaces do not contibute to the integal as usual, since ( ) = 0 ˆ i ±ẑ 31

32 Two coaxial shells of chage: S 2 ( E ˆ )i( ˆ dϕdz) = Q ε o S 2 E dϕ dz = Q ε o E ( ) ( 2πh) = ρ 2π ah s ε o E = ρ s a ε o 32

33 Two coaxial shells of chage: E = 0 < a ρ s a a < < b ε o 0 > b Φ ( ) = E ˆ = ρ a s ε o Φ( ) = ( )i( ˆ d) = ρ a s ε o d ( ln) ρ = s a b ε o ρ s a ε o ln b b ( ln lnb) = ρ a s ε o a < < b 0 othewise ln b, a < < b 33

34 Two coaxial shells of chage: Also, Φ ba = Φ( a) Φ( b) = ρ a s ε o ln a b, a < < b Φ( ) = Φ ba ln b a < < b ln a b 0 othewise 34

35 Two coaxial shells of chage: Gauss s Law was deived fom: i D = ρ D i da = Q enclosed pointwise S ove a volume in space At a point whee thee is no chage (i.e., inside the cylinde) the divegence should equal zeo. Let s veify this fo this example: D = ε o E = ρ s a ε o, i D = 1 D ( ) = 1 a < < b ρ s a ε o 0 As an execise, veify that the divegence of the dipole field found ealie is also zeo. 35

36 Two coaxial shells of chage: As an execise, veify that the divegence of the dipole field found ealie is also zeo. i.e., show that D = ε E = Qd 4π 3 ( 2 ˆ cosθ + ˆθ sinθ ) i D = 0 36

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