The Electric Field EM-L2-1

Transcription

1 The EM-L2-1

2 Review of Lecture 1 Electric charge is quantised and conserved in interactions The force between two charges is given by Coulomb s law Force F 1,2 exerted by a charge q 1 on another charge q 2 at distance r: F 1,2 = k q1 q 2 r 2 1,2 ˆr 1,2 where ˆr 1,2 = r 2 r 1 r 2 r 1 EM-L2-2

3 Overview of Lecture 2 Force from a system of charges The electric field: - motivation - definition Electric field lines Summary EM-L2-3

4 Forces by a system of charges The net force F j on a charge q j is the vector sum of all individual forces F i,j exerted by the other charges q i on q j. F j = i j F i,j = i j k q i q j r 2 i,j ˆr i,j Forces by a system of charges EM-L2-4

5 Example: net force Find the resultant force on charge q 0 Forces by a system of charges EM-L2-5

6 Example: net force ( ) 25 nc 20 nc cos 45 F 1,0 = k 8 m 2 sin 45 ( ) = cos 45 N sin 45 Find the resultant force on charge q 0 Forces by a system of charges EM-L2-6

7 Example: net force ( ) 25 nc 20 nc cos 45 F 1,0 = k 8 m 2 sin 45 ( ) = cos 45 N sin 45 ( 15 nc 20 nc 0 F 2,0 = k 4 m 2 1 = N ( 0 1 ) ) Find the resultant force on charge q 0 Forces by a system of charges EM-L2-7

8 Example: net force ( ) 25 nc 20 nc cos 45 F 1,0 = k 8 m 2 sin 45 ( ) = cos 45 N sin 45 ( 15 nc 20 nc 0 F 2,0 = k 4 m 2 1 = N ( 0 1 ) ) Find the resultant force on charge q 0 F net = F 1,0 + F ( 2, cos = 5.62 sin ( ) 3.97 = N ( = cos ( 34.9 N ) sin ( 34.9 ) ) 10 7 N ) Forces by a system of charges EM-L2-8

9 EM-L2-9

10 : Motivation Coulomb s law implies a mysterious interaction at a distance. F 1,2 = k q1 q 2 r 2 1,2 ˆr 1,2 Electromagnetic interactions are clearly not instantaneous. Changes propagate at speed of light c. c = m s Solution: Introduce an intermediary. A charge q j at r j interacts instantaneously with a field E( r j ). EM-L2-10

11 Electric field: definition Definition The Electric field E( r j ) at point r j generated by a charge q i at r i is defined as Measurement E( r j ) = k q i r 2 i,j ˆr i,j The total electric field can be found by measuring the force on a small positive test charge, q, and dividing by q. F E = lim q 0 q It is important that we take the limit q 0 so the test charge does not disturb the electric field (and, for example, move the other charges). EM-L2-11

12 System of point charges The electric field due to a system of point charges is the vector sum of the electric field due to each charge. For a system of point charges q i where each charge is the source of an electric field E i the total electric field at r j is E( r j ) = i E i ( r j ) = i kq i r 2 i,j ˆr i,j EM-L2-12

13 Example: electric dipole field Find the electric field on the x-axis at an arbitrary point x > a. EM-L2-13

14 Example: electric dipole field E = kq (x a) 2 î + k( q) (x + a) 2 î Find the electric field on the x-axis at an arbitrary point x > a. EM-L2-14

15 Example: electric dipole field kq E = (x a) 2 î + k( q) (x + a) 2 î [ 1 = kq (x a) 2 1 (x + a) 2 ] î Find the electric field on the x-axis at an arbitrary point x > a. EM-L2-15

16 Example: electric dipole field kq E = (x a) 2 î + k( q) (x + a) 2 î [ 1 = kq (x a) 2 1 (x + a) 2 [ (x + a) 2 (x a) 2 = kq (x a) 2 (x + a) 2 4ax = kq (x 2 a 2 ) 2 î ] ] î î Find the electric field on the x-axis at an arbitrary point x > a. EM-L2-16

17 Example: continued For q = 1nC and a = 1cm: For large distances x a: E = 4kqa x 3 î EM-L2-17

18 Drawing electric field lines The Rules: 1. Field lines begin on positive charges (or infinity) and end on negative charges (or infinity). 2. The number of lines entering or leaving a source is proportional to the magnitude of the charge on a source. 3. The density of lines at any point is proportional to the magnitude of the field at that point. 4. Field lines do not cross. EM-L2-18

19 Electric field lines for a single charge Left figure shows lines of force for a positive test charge. In the right figure the same lines show in bits of threads in oil (electric dipoles) with a charged object in the center. EM-L2-19

20 Example: Dipole field For simulations of electric fields, see: EM-L2-20

21 Summary EM-L2-21

22 Summary Superposition of Coulomb forces F j = i j F i,j Electric field Electric field lines E = i k q i r 2 i ˆr i Recommended reading: Tipler, sections 21-4 and 21-5 Next lecture: Tipler, sections 22-1, 23-1, 23-3 Summary EM-L2-22