1. Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Fall 2004 Coulomb’s Law ECE 2317: Applied Electricity and Magnetism Charles Coulomb

2. Coulomb’s Force Law proportional to the product of the charges, inversely proportional to the square of the distance between them, directed along the line joining them, and repulsive for like charges and attractive for unlike charges. Charles A. Coulomb, a French military engineer, measured the forces between electric charges, reporting his results in 1785. He found that the electric force between two charged particles was

3. Coulomb’s Law Experimental law: Note: c = speed of light = 2.99792458  10 8 [m/s] (exactly) x y z q1q1 q2q2

4. Electric Field Hence: But Note: no self-force on charge 2 due to its own electric field x y z q1q1 q2q2 ( : electric field due to q 1 )

5. Generalization ( q 1 not at the origin): Generalizing Coulomb’s Law r 1 = (x 1, y 1, z 1 ) q 2 (x 2,y 2,z 2 ) x y z q 1 (x 1,y 1,z 1 ) r 2 = (x 2, y 2, z 2 )

6. Example q 1 =0.7 [mC] located at (3,5,7) [m] q 2 =4.9 [  C] located at (1,2,1) [m] q2q2 q1q1 E 1 = electric field due to charge q 1, evaluated at point r 2 Find: F 1, F 2 For F 2 : F 1 = force on charge q 1 F 2 = force on charge q 2

7. Example (cont.) q 1 =0.7 [mC] located at (3,5,7) [m] q 2 =4.9 [  C] located at (1,2,1) [m] R q2q2 q1q1

8. Example (cont.) (Newton’s Law)

9. x y z General Case: Multiple Charges q 1 : r 1 = (x 1,y 1,z 1 ) q2q2 q1q1 q3q3 qNqN q 2 : r 2 = (x 2,y 2,z 2 )... q N : r N = (x N,y N,z N ) (superposition) r

10. x y z Field from Volume Charge r= (x,y,z)  V (r´) =  V (x´,y´,z´)

11. x y z Field from Volume Charge (cont.) r = (x,y,z) R dQ =  V (r´) dV´ dV´

12. r = (x,y,z) dQ =  S (r´) dS´ Field from Surface Charge x y z R dS dS´

13. dQ =  l (r´) dl´ Field from Line Charge x y z r = (x,y,z) R dl´

14. Example r = (0,0,1) x y z R1R1 q 2 = -20 [nC] R2R2 q 1 = +20 [nC] q 1 = +20 [nC] located at (1,0,0) q 2 = -20 [nC] located at (0,1,0) Find E (0,0,1) Solution: R 1 = (0,0,1) - (1,0,0) R 2 = (0,0,1) - (0,1,0)

15. Example (cont.)

16. x y z Example r = (0,0,h) R  l =  lo [C/m] r´= (0, 0, z´ ) Find E(0,0,h) h>0 semi-infinite uniform line charge

17. Example (cont.)

18. Example x a ´´ d´d´ dl´= a d  ´  l =  lo [C/m] (uniform) Find E(0,0,z) y x y z a R r = (0,0,z) r´= (a,  ´, 0 ) d  ´ > 0 so

19. Example (cont.) x ´´ Note: y x y z a R r = (0,0,z) r´= (a,  ´, 0 )

20. Example (cont.) R a z x

23. Limiting case: a  0 (while the total charge remains constant) (point-charge result)