Physics 122B Electricity and Magnetism

1. Physics 122B Electricity and Magnetism Lecture 2Coulomb’s Law and the Electric FieldMarch 28, 2007 Martin Savage

2. Recall : Coulomb’s Law Like charges repel. Charles Augustine de Coulomb (1736-1806). Opposite charges attract. ( Magnitude of force ) Coulomb’s Law: Physics 122B - Lecture 2

3. Recall : Units of Charge (Note that in Newton’s law of gravitation, G, which plays a role similar to K, has the valueG = 6.67 ´10-11 N m2/kg2.) Physics 122B - Lecture 2

4. Recall : Using Coulomb’s Law • Coulomb’s Law applies only to point charges. • (This is particularly important because charge are free to move around on conductors.) • Strictly speaking, Coulomb’s Law applies only to electrostatics (non-moving charges). • (However, it is usually OK provided v<<c). • Electrostatic forces (vectors!) can be superposed. • Linear superposition !!!! Physics 122B - Lecture 2

5. Example: Sum of Two Forces Two +10 nC charged particles are 2 cm apart on the x axis. (1) What is the net force on a +1.0 nC particle midway between them? (2) What is the net force if the charged particle on the right is replaced by a -10 nC charge? -9 9 Physics 122B - Lecture 2

6. Example: Point of Zero Force Two positively charged particles q1 and q2=3q1 are placed 10. 0 cm apart. Where (other than infinity) could another charge q3 be placed so as to experience no net force? Need force vectors to be co-linear, so location must be on x axis. Need force vectors to be in opposite directions, so location must be between 0 and d. Need force vectors equal in magnitude, so F=Kq1q3/x2=Kq2q3/(d-x)2. Therefore,(d-x)2=3x2orx=d/(1±√3) = 10 cm/(1±1.732); x+=10.0 cm/2.732 = 3.66 cm; x-=10.0 cm/-0.732=-13.66 cm, which does not satisfy the 2nd criterion. Physics 122B - Lecture 2

7. Example: Three Charges (1) Three charges with q1 = -50 nC, q2 = +50 nC, and q3 = +30 nC, are placed at the corners of a 10 cm x 5 cm rectangle as shown. What is the net force on q3 due to the other two charges? Physics 122B - Lecture 2

8. Example: Three Charges (2) Cos q Sin q Physics 122B - Lecture 2

9. Example: Lifting a Glass Bead A small plastic sphere is charged to -10 nC. It is held 1.0 cm above a small glass bead that rests on a table. The bead has a mass of 15 mg and a charge of +10 nC. Will the glass bead “leap up” to the plastic sphere? Therefore, F1 on 2 exceeds w by a factor of 60. Therefore, the sphere should indeed leap upward. (Note that we have neglected electrical forces between the bead and table, which could be significant.) Physics 122B - Lecture 2

10. Motion and Response Question: If charge A moves, how long does it take for the force vector on B to respond by changing direction? Physics 122B - Lecture 2

11. Isaac Newton vs.Michael Faraday Isaac Newton: Forces like gravity and electricity are examples of action at a distance. The “influence” of A reaches across space to affect B instantly and without the need for contact. Michael Faraday: Forces like gravity and electricity are mediated by a “field” that alters the space around the mass or charge. Particle A produces a field, and particle B responds to this field by experiencing a force. Changes in the field require time to propagate. Physics 122B - Lecture 2

12. Example:The Gravitational Field (1) Physics 122B - Lecture 2

13. Example:The Gravitational Field (2) • The field strength g is proportional to the mass. g~m1. Larger masses create stronger gravitational fields. • The field strength is inversely proportional to the square of the distance. g~1/r2. • The field produced by mass m1 points directly to m1. Physics 122B - Lecture 2

14. Relative Force Strengths The electric force, with a force constant of K = 8.99 x 109 N m2/C2, is much stronger than the gravitational force, with a force constant of G = 6.67 x 10-11 N m2/kg2. To illustrate this, let’s consider the ratio of electric to gravitational forces between an identical pair of protons and an identical pair of electrons. Proton: mass = 1.67 x 10-27 kg, charge = +1.60 x 10-19 C, Strength ratio = (K q1q2/r2)/(G m1m2/r2) = K e2/G mp2 = 1.24 x 1036 Electron: mass = 9.11 x 10-31 kg, charge = -1.60 x 10-19 C, Strength ratio = (K q1q2/r2)/(G m1m2/r2) = K e2/G me2 = 4.16 x 1042 How is it that we notice the gravitational force at all? Because the universe contains only positive mass, but equal amounts of positive and negative electric charge, which tends to cancel. Physics 122B - Lecture 2

15. Fields • Fields: • The electric field permeates all of space • Charges interact with the electric field inducing a non-zero value of E (subject to boundary conditions). • A charge in an electric field experiences a force F exerted by the field. • This relation assigns a field vector to every point in space. • If q is positive, the electric field vector points in the same direction as the force vector. • Does E depend on q (the charge that detects it)? No, because the force is proportional to q. Physics 122B - Lecture 2

16. Question An electron (q =-e, m = me) is placed at the position shown by the black dot. The force on the electron is: (a) Zero. (b) To the right. (c) To the left.(d) Insufficient information to tell. Physics 122B - Lecture 2

17. The Electric Fieldof a Point Charge Physics 122B - Lecture 2

18. Electric Field Diagrams • Points about field diagrams: • The diagram is only a representative sample of the electric field. The field exists at all points in space. • The arrow indicates the direction and strength of the field at the point to which the vector tail is attached. The length of the vector is a relative measure of the field strength. • The electric field is not a quantity that ``stretches’’ from one point to another. Each vector represents the field at that one point in space. Physics 122B - Lecture 2

19. Unit Vector Notation Positive q Negative q Physics 122B - Lecture 2

20. Example:Calculating the Electric Field A 1.0 nC charged particle is located at the origin. Points 1, 2, and 3 have (x,y) coordinates in cm of (1,0), (0,1), and (1,1), respectively. Determine the electric field E at each of these points. Physics 122B - Lecture 2

21. Example:The Electric Field of a Proton • In a hydrogen atom, the electron orbits the proton at a radius of 0.053 nm. Both have charges of magnitude e = 1.60 x 10-19 C. • What is the proton’s electric field strength at the position of the electron? • What is the magnitude of the electric force on the electron? Physics 122B - Lecture 2

22. Question 2 Which system has the largest electric field at the location of the numbered point? (a) (b) (c) (d) Physics 122B - Lecture 2

23. Summary: Chapter 25 (1) Physics 122B - Lecture 2

24. Summary: Chapter 25 (2) Physics 122B - Lecture 2

25. Typical Electric Field Strengths Physics 122B - Lecture 2

26. End of Lecture 2 • Before the next lecture, read Knight, Chapters 26.2 through 26.3 • This weeks homework is on Tycho, and ready for you to think about.