F E = Electric field q T (+ test charge)

Transcription

1 Summary Page week 1 charges opposite attract like repel F = k Q Q r T + Cancelation/screening 1 Coulombs Law vector sum of forces Superposition Principle solution (A+B)= solution (A)+ solution (B) F E = Electric field q T (+ test charge) q E Field Lines E Field behavior in/atsurfaceof metals surf E A = q inside ε o Gausses' Law k = 1 4 o 10

2 11

3 micro coulomb 11a

4 Examples F= force on q 1 (due to q ) F= force on q (due to q 1 ) q 1 q 6 C +4 C F= F by Newton s 3 rd Law (and later, Coulomb s Law) F= force on q 1 (due to q ) q 1 q +6 C +4 C F= force on q (due to q 1 ) Note: forces along straight line between charges 11b

5 Where does charge come from? 1

6 point charges Electrostatic force F = k q 1 Coulomb s Law q q q 1 q q = r 4 r Nm k =8.99 (10) C 1 C ε =8.85 (10) Nm 0 Force depends on magnitude of both charges (q 1 q like m 1 m in Newton s Gravity) Force decreases with distance like 1/r (again like Newton s Gravity) Force acts along straight line between point charges (again like Newton s Gravity) Force very big bigger than Gravity 13

7 Note: 1e 1.6 (10) 19 C or 1C 6.3 (10) 18 e Consider the He nucleus: p+ n r About 1015 m p+ n What is the repulsion force between protons? k q 8.99 (10) [1.6 (10) ] F = = 15 r (10 ) 9 19 F ~ 30 N repulsive Nm C [ ][ ] C m Strong nuclear force (see end of sem.) balances this repulsion. 14

8 Example Forces and directions on Q 1 &Q F= kq Q r F 1 r = 1m F Q 1 = 6(10) 6 C Q = +4(10) 6 C 6 μc +4 μc = 1 = 16(10) 3 N in magnitude (10) 6(10) 4(10) l Nm C CC m F 1 =F on1 due to = +.16N (attractive = toward other charge) F =F on due to 1 =.16N (attractive = toward other [1] charge) Force reaction force pair NEWTON s 3 rd LAW + signs come from + definition + def. 14a

9 14b

10 Net force superposition of forces due to individual charges 15

11 Example: 3 colinear charges Consider force on Q Consider total force F on Q + direction def. 16 to right 16

12 Electric Field = Force per unit charge Define E F E = q 0 or F = 0 q E 17

13 Electric field of one point charge magnitude! Direction = direction of force on + test charge! on straight line between points Group of point charges: add electric fields of point charges like vectors!! 18

14 Example: E at point, P, between point charges E= 3000 m q 1 + q 3 C test 4 C charge 9 3 E 1 =8.99 (10) (3000) 3 =8.99 (10) 3 3 E 1 =3(10) N/C m P + dir Nm C N = C m C E = E + E = [ E + E ] xˆ 1 1 xˆ 4 E =8.99 (10) (000) 9 4 = 8.99 (10) 3 E =8.99(10) N / C 3 q k r 3 E = E1 + E = [ ] (10) x ˆ N/ C 3 E = 5.99 (10) x N/C ˆ last slide 9/6/17 lec 18a

15 Vector addition of E fields: use symmetry to simplify when possible Q. Electric field at center cancel E q k r +q q cancel + r (place + test charge at center) q +q = q s the same r s the same E s the same E fields cancel so E tot =0 r a/ r a a/ a a = + homework problem help 19

16 Vector addition of E fields: symmetry continued Q. Electric field at center +q +q + r homework problem help E q = k r r a/ r a a/ 45 o a a = + q q E x E y cancel E x q 45 o o q E Ey y = k sin(45 )=k similarly superimpose E y r q s the same r s the same E s the same 1 r E 4 E tot = y 1 E (y) ˆ y 1 45 o 45 o y direction 19.1

17 E Field Lines direction = force direction on + test charge (away from +) (toward ) line density field magnitude repulsive attractive 110

18 Static (no charge motion) condition 111

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20 minimize force (energy) 5 E concentrate at sharp points (low radius of curvature) 111. for later flat plane Electric field constant Electric field surface!!!! would cause motion (violates static) E

21 E concentrate at sharp points (low radius of curvature) can move apart decrease density can t move apart (surf. contains) enough E can actually rip charges from surface

22 Faraday Bucket can accumulate charge!! (e) (f) (g) 111.4

23 Van de Graff generator continuous charge accumulation!! + pulled off belt + charge left on rubber belt sharp points big electric field charge on rubber belt drained static charge induced on inside of rubber belt sharp points big electric field charge transferred metal pulley plastic pulley 111.5

24 Electric Flux 111.6

25 Gauss s Law E = Q inside o surf Q E A inside = ε = o E 111.7

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27 Surface charge density & E field for metal Q= tot. surface charge E Electric field surface!!!! Electric field constant air metal A =surface area = Q A units C/m Q. What is relation between E and? 113

28 Electric field above charged metal surface (pill box shape) Gassian surface Q= tot. surface charge Gauss = tot = tot Q sides top E air in ε0 A top ε 0 113a bottom metal = + + tot through surface A top E is top = E A top top top bottom sides 0 E=0 in metal A tot =surface area = Atop = E A top ε E = 0 Q A Q = A units C/m 0 E=0 in metal E side in air ε 0

29 Consider // plates connected to charge reservoirs. 1. The E field, by symmetry, must run straight across the gap.. Apply Gauss's Law 3. Metal Plates 114

30 115abc

31 115d summary

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33 117extra Brute force electric force vector addition

34 117.1extra