1. Chapter 21 Electric Field and Coulomb’s Law (again) Electric fields and forces (sec. 21.4) Electric field calculations (sec. 21.5) Vector addition (quick review) C 2012 J. Becker

2. Learning Goals - we will learn: How to use Coulomb’s Law (and vector addition) to calculate the force between electric charges. How to calculate the electric field caused by discrete electric charges. How to calculate the electric field caused by a continuous distribution of electric charge.

3. Coulomb’s Law Coulomb’s Law lets us calculate the FORCE between two ELECTRIC CHARGES.

4. Coulomb’s Law Coulomb’s Law lets us calculate the force between MANY charges. We calculate the forces one at a time and ADD them AS VECTORS. (This is called “superposition.”) THE FORCE ON q 3 CAUSED BY q 1 AND q 2.

5. Figure 21.14 SYMMETRY!

6. Recall GRAVITATIONAL FIELD near Earth: F = G m 1 m 2 /r 2 = m 1 (G m 2 /r 2 ) = m 1 g where the vector g = 9.8 m/s 2 in the downward direction, and F = m g. ELECTRIC FIELD is obtained in a similar way: F = k q 1 q 2 /r 2 = q 1 (k q 2 /r 2 ) = q 1 (E) where is vector E is the electric field caused by q 2. The direction of the E field is determined by the direction of the F, or the E field lines are directed away from positive q 2 and toward -q 2. The F on a charge q in an E field is F = q E and |E| = (k q 2 /r 2 )

7. Fig. 21.15 A charged body creates an electric field. Coulomb force of repulsion between two charged bodies at A and B, (having charges Q and q o respectively) has magnitude: F = k |Q q o |/r 2 = q o [ k Q/r 2 ] where we have factored out the small charge q o. We can write the force in terms of an electric field E: Therefore we can write for F = q o E the electric field E = [ k Q / r 2 ]

8. Calculate E 1, E 2, and E TOTAL at point “C”: q = 12 nC See Fig. 21.23: Electric field at “C” set up by charges q 1 and q 1 (an “electric dipole”) At “C” E 1 = 6.4 (10) 3 N/C E 2 = 6.4 (10) 3 N/C E T = 4.9 (10) 3 N/C in the +x-direction A C See Lab #2 Need TABLE of ALL vector component VALUES. E1E1 E2E2 ETET

9. Fig. 21.24 Consider symmetry! E y = 0 XoXo dq o dE x = dE cos a =[k dq /(x o 2 +a 2) ] [x o /(x o 2 + a 2 ) 1/2 ] E x = k x o  dq /[x o 2 + a 2 ] 3/2 where x o and a stay constant as we add all the dq’s (  dq = Q) in the integration: E x = k x o Q/[x o 2 +a 2 ] 3/2 |dE| = k dq / r 2 cos a = x o / r  dE x = E x

11. Fig. 21.25 Electric field at P caused by a line of charge uniformly distributed along y-axis. Consider symmetry! E y = 0 XoXo y |dE| = k dq / r 2 dq

12. |dE| = k dq / r 2 and r = (x o 2 + y 2 ) 1/2 cos a = x o / r and cos a = dE x / dE dE x = dE cos a E x =  dE x =  dE cos a E x =  [k dq /r 2 ] [x o / r] E x =  [k dq /(x o 2 +y 2 )] [x o /(x o 2 + y 2 ) 1/2 ] Linear charge density = = charge / length = Q / 2a = dq / dy dq = dy

13. E x =  [k dq /(x o 2 +y 2 )] [x o /(x o 2 + y 2 ) 1/2 ] E x =  [k dy /(x o 2 +y 2 )] [x o /(x o 2 + y 2 ) 1/2 ] E x = k  x o  [dy /(x o 2 +y 2 )] [1 /(x o 2 + y 2 ) 1/2 ] E x = k  x o  [dy /(x o 2 +y 2 ) 3/2 ] Tabulated integral : (Integration variable “z”)  dz / (c 2 +z 2 ) 3/2 = z / c 2 (c 2 +z 2 ) 1/2  dy / (c 2 +y 2 ) 3/2 = y / c 2 (c 2 +y 2 ) 1/2  dy / ( X o 2 +y 2 ) 3/2 = y / X o 2 ( X o 2 +y 2 ) 1/2

14. E x = k  x o -a  [dy /(x o 2 +y 2 ) 3/2 ] E x = k(Q/2a) X o [ y / X o 2 ( X o 2 +y 2 ) 1/2 ] -a a E x = k (Q /2a) X o [( a –(-a)) / X o 2 ( X o 2 +a 2 ) 1/2 ] E x = k (Q /2a) X o [ 2a / X o 2 ( X o 2 +a 2 ) 1/2 ] E x = k (Q / X o ) [ 1 / ( X o 2 +a 2 ) 1/2 ] a

15. Fig. 21.47 Calculate the electric field at the proton caused by the distributed charge +Q. Tabulated integral :  dz / (c-z) 2 = 1 / (c-z) b is uniform (= constant) +Q

16. Fig. 21.48 Calculate the electric field at -q caused by +Q, and then the force on –q: F=qE Tabulated integrals:  dz / (z 2 + a 2 ) 3/2 = z / a 2 (z 2 + a 2 ) ½ for calculation of Ex  z dz / (z 2 + a 2 ) 3/2 = -1 / (z 2 + a 2 ) ½ for calculation of Ey is uniform (= constant)

17. An ELECTRIC DIPOLE consists of a +q and –q separated by a distance d. ELECTRIC DIPOLE MOMENT is p = q d ELECTRIC DIPOLE in E experiences a torque:  = p x E ELECTRIC DIPOLE in E has potential energy: U = - p E

18. Fig. 21.32 Net force on an ELECTRIC DIPOLE is zero, but torque (  ) is into the page.  = r x F  = p x E ELECTRIC DIPOLE MOMENT is p = qd

19. see www.physics.sjsu.edu/Becker/physics51 Review

20. Vectors are quantities that have both magnitude and direction. An example of a vector quantity is velocity. A velocity has both magnitude (speed) and direction, say 60 miles per hour in a DIRECTION due west. (A scalar quantity is different; it has only magnitude – mass, time, temperature, etc.)

21. A vector may be decomposed into its x- and y-components as shown:

22. Note: The dot product of two vectors is a scalar quantity. The scalar (or dot) product of two vectors is defined as

23. The vector (or cross) product of two vectors is a vector where the direction of the vector product is given by the right-hand rule. The MAGNITUDE of the vector product is given by:

24. Right-hand rule for DIRECTION of vector cross product.

25. PROFESSIONAL FORMAT