SUBJECT – EOE, 4 TH SEM EE

1. SUBJECT – EOE, 4TH SEM EE • Coulomb’s Law • The Electric Field - Examples • Gauss Law - Examples 4. Conductors in Electric Field

2. Coulomb’s Law Coulomb’s law quantifies the magnitude of the electrostatic force. Coulomb’s law gives the force (in Newtons) between charges q1 and q2, where r12 is the distance in meters between the charges, and k=9x109 N·m2/C2.

3. Coulomb’s Law Force is a vector quantity. The equation on the previous slide gives the magnitude of the force. If the charges are opposite in sign, the force is attractive; if the charges are the same in sign, the force is repulsive. Also, the constant k is equal to 1/40, where 0=8.85x10-12 C2/N·m2.

4. Coulomb’s Law • If q1 and q2 are located at points having position vectors r1 and r2 then the Force F12 is given by;

5. + - + - + - The equation is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed, r12 is the distance between the centers of the spheres. r12 - + If more than one charge is involved, the net force is the vector sum of all forces (superposition). For objects with complex shapes, you must add up all the forces acting on each separate charge (turns into calculus!).

6. Solving Problems Involving Coulomb’s Law and Vectors Example: Calculate the net electrostatic force on charge Q3 due to the charges Q1 and Q2. y Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm

7. Step 0: Think! This is a Coulomb’s Law problem (all we have to work with, so far). We only want the forces on Q3. Forces are additive, so we can calculate F32 and F31 and add the two. If we do our vector addition using components, we must resolve our forces into their x- and y-components.

8. F32 F31 Step 1: Diagram y Draw a representative sketch. Draw and label relevant quantities. Draw axes, showing origin and directions. Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm Draw and label forces (only those on Q3). Draw components of forces which are not along axes.

9. F32 F31 Step 2: Starting Equation y Q3=+65C 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm “Do I have to put in the absolute value signs?”

10. F32 F31 Step 3: Replace Generic Quantities by Specifics y Q3=+65C r31=60 cm r32=30 cm =30º x Q1=-86C Q2=+50C 52 cm (from diagram) F32,y = 330 N and F32,x = 0 N.

11. F32 F31 Step 3 (continued) y Q3=+65C r31=60 cm r32=30 cm =30º x Q1=-86C Q2=+50C (+ sign comes from diagram) 52 cm (- sign comes from diagram) You would get F31,x = +120 N and F31,y = -70 N.

12. F32 F3 F31 You know how to calculate the magnitude F3 and the angle between F3 and the x-axis. Step 3: Complete the Math y Q3=+65C The net force is the vector sum of all the forces on Q3. 30 cm 60 cm =30º x Q1=-86C Q2=+50C 52 cm F3x = F31,x + F32,x = 120 N + 0 N = 120 N F3y = F31,y + F32,y = -70 N + 330 N = 260 N

13. Many Point Charges • If we have N point charges Q1, Q2… QN, located at r1, r1… rN the vector sum of the forces exerted on Q by each of the charges is given by

14. Coulomb’s Law: The Big Picture Coulomb's Law quantifies the interaction between charged particles. r12 - + Q1 Q2 Coulomb’s Law was discovered through decades of experiment. By itself, it is just “useful." Is it part of something bigger?

15. The Electric Field Coulomb's Law (demonstrated in 1785) shows that charged particles exert forces on each other over great distances. How does a charged particle "know" another one is “there?”

16. The Electric Field Faraday, beginning in the 1830's, was the leader in developing the idea of the electric field. Here's the idea: F12  A charged particle emanates a "field" into all space. + F13  Another charged particle senses the field, and “knows” that the first one is there. F31 + - F21 unlike charges attract like charges repel

17. The Electric Field We define the electric field by the force it exerts on a test charge q0: By convention the direction of the electric field is the direction of the force exerted on a POSITIVE test charge. The absence of absolute value signs around q0 means you must include the sign of q0 in your work.

18. The Electric Field If the test charge is "too big" it perturbs the electric field, so the “correct” definition is You won’t be required to use this version of the equation. Any time you know the electric field, you can use this equation to calculate the force on a charged particle in that electric field.

19. The units of electric field are Newtons/Coulomb. Later you will learn that the units of electric field can also be expressed as volts/meter: The electric field exists independent of whether there is a charged particle around to “feel” it.

20. + Remember: the electric field direction is the direction a + charge would feel a force. A + charge would be repelled by another + charge. Therefore the direction of the electric field is away from positive (and towards negative).

21. The Electric Field Due to a Point Charge Coulomb's law says ... which tells us the electric field due to a point charge q is …or just… This is your third starting equation.

22. We define as a unit vector from the source point to the field point: source point + field point The equation for the electric field of a point charge then becomes: You may start with either equation for the electric field (this one or the one on the previous slide). But don’t use this one unless you REALLY know what you are doing!

23. Many Point Charges • If we have N point charges Q1, Q2… QN, located at r1, r1… rN the vector sum of the Electric Field E by each of the charges is given by

24. If E is constant, then a is constant, and you can use the equations of kinematics. Motion of a Charged Particle in a Uniform Electric Field A charged particle in an electric field experiences a force, and if it is free to move, an acceleration. - - - - - - - - - - - - - If the only force is due to the electric field, then - E F + + + + + + + + + + + + +

25. Example: an electron moving with velocity v0 in the positive x direction enters a region of uniform electric field that makes a right angle with the electron’s initial velocity. Express the position and velocity of the electron as a function of time. y - - - - - - - - - - - - - x - E v0 + + + + + + + + + + + + +

26. unit vector from q to wherever you want to calculate E The Electric Field Due to a Collection of Point Charges The electric field due to a small "chunk" q of charge is The electric field due to collection of "chunks" of charge is As qdq0, the sum becomes an integral.

27. Charge Distributions • Continious charge distributions are given by; • Do not confuse ρ with radial distance

28. Line Charge If charge is distributed along a straight line segment parallel to the z-axis, the amount of charge dq on a segment of length dz is ρdz. The Charge elemen is given by; The total charge is given by;

29. Line Charge The electric field at point P due to the charge dq is The field point is denoted by (x,y,z) while the source point is denoted by (x’,y’,z’) I’m assuming positively charged objects in these “distribution of charges” slides.

30. Line Charge Definitions

31. Line Charge • Using given definitions we have; • For a finite line charge;

32. Line Charge • For an infinite line chargeα1andα2areequaltoπ/2 and - π/2 respectively, the z component is canceledandtheequationbecomes

33. Surface Charge If charge is distributed over a two-dimensional surface, the amount of charge dq on an infinitesimal piece of the surface is ρsdS, where ρs is the surface density of charge (amount of charge per unit area). y charge dQ = ρsdS ρs x area = dS

34. Surface Charge y dE P r’ x The electric field at P due to the charge dq is

35. Surface Charge dE P r’ x The net electric field at P due to the entire surface of charge is

36. Surface Charge • Definitions • And the E field becomes;

37. Surface Charge • After evaluating the previous integration; • Or a more general formula is • Note that the Electric Field is independent of the distancebetween the sheet and the point of observation P an is a unit vector normal to the sheet

38. Volume Charge The net electric field at P due to a three-dimensional distribution of charge is… z E P r’ x y

39. Volume Charge • For a sphere with radius a centered at the origin • The E field due to elementary volume charge is given by;

40. Volume Charge • Due to the symmetry of the charge distribution, the contributionsto Ex or Ey add up to zero

41. Volume Charge • Definitions • And the resulting expression for E field is;

42. Summarizing: Charge distributed along a line: Charge distributed over a surface: Charge distributed inside a volume: If the charge distribution is uniform, then , , and  can be taken outside the integrals.

43. The Electric Field Due to a Continuous Charge Distribution (worked examples)

44. Example: A rod of length L has a uniform charge per unit length  and a total charge Q. Calculate the electric field at a point P along the axis of the rod at a distance d from one end. y x P d L Let’s put the origin at P. The linear charge density and Q are related by Let’s assume Q is positive.

45. Note: dE is in the –x direction. dE is the magnitude of dE. I’ve used the fact that Q>0 (so dq=0) to eliminate the absolute value signs in the starting equation. y dQ =  dx dx x dE x P d L The electric field points away from the rod. By symmetry, the electric field on the axis of the rod has no y-component. dE from the charge on an infinitesimal length dx of rod is

46. y dQ =  dx dx x dE x P d L

47. By symmetry, the y- and z-components of E are zero, and all points on the ring are a distance r from point P. Example: A ring of radius a has a uniform charge per unit length and a total positive charge Q. Calculate the electric field at a point P along the axis of the ring at a distance x0 from its center. dQ r a P  x x0  dE

48. dQ No absolute value signs because Q is positive. r a P  x x0  dE For a given x0, r is a constant for points on the ring. Or, in general, on the ring axis

49. Example: A disc of radius R has a uniform charge per unit area . Calculate the electric field at a point P along the central axis of the disc at a distance x0 from its center. dQ The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr). r P x x0 R We can use the equation on the previous slide for the electric field due to a ring, replace a by r, and integrate from r=0 to r=R. Caution! I’ve switched the “meaning” of r!

50. dQ r P x x0 R