Electrostatics

1. Electrostatics October 15, 2014

2. iClicker An object is placed on the axis of a converging thin lens of focal length 2 cm, at a distance of 8 cm from the lens. The distance between the image and the lens is most nearly a. 0.4 cm b. 0.8 cm c. 1.6 cm d. 2.0 cm e. 2.7 cm

3. Atomic Charges • The atom has positive charge in the nucleus, located in the protons. • The positive charge cannot move from the atom unless there is a nuclear reaction. • The atom has negative charge in the electron cloud on the outside of the atom. • Electrons can move from atom to atom without all that much difficulty.

4. Charge • Charge comes in two forms, which Ben Franklin designated as positive (+) and negative(–). • Charge is quantized • It only comes in packets • The smallest possible stable charge, which we designate as e, is the magnitude of the charge on 1 electron or 1 proton. • e is referred to as the “elementary” charge. • e = 1.602 × 10-19 Coulombs. • We say a proton has charge of e, and an electron has a charge of –e.

5. Problem • A certain static discharge delivers -0.5 Coulombs of electrical charge. How many electrons are in this discharge?

6. Problem • The total charge of a system composed of 1800 particles, all of which are protons or electrons, is 31x10-18 C. How many protons are in the system? How many electrons are in the system?

7. Coulomb’s Law andElectrical Force

8. Electric Force • Charges exert forces on each other. • Like charges (two positives, or two negatives) repel each other, resulting in a repulsive force. • Opposite charges (a positive and a negative) attract each other, resulting in an attractive force.

9. Coulomb’s Law • Coulomb’s law tells us how the magnitude of the force between two particles varies with their charge and with the distance between them. • k = 9.0 × 109 N m2 / C2 • q1, q2 are charges (C) • r is distance between the charges (m) • F is force (N) • Coulomb’s law applies directly only to spherically symmetric charges.

10. Problem • A point charge of positive 12.0 μC experiences an attractive force of 51 mN when it is placed 15 cm from another point charge. What is the other charge?

11. iClicker Two isolated charges, + q and ‑ 2q, are 2 centimeters apart. If F is the magnitude of the force acting on charge ‑2q, what are the magnitude and direction of the force acting on charge + q ? MagnitudeDirection (A) (1/2) F Toward charge ‑ 2q (B) 2 F Away from charge ‑2q (C) F Toward charge ‑ 2q (D) F Away from charge ‑ 2q (E) 2F Toward charge ‑ 2q

12. Superposition • Electrical force, like all forces, is a vector quantity. • If a charge is subjected to forces from more than one other charge, vector addition must be performed. • Vector addition to find the resultant vector is sometimes called superposition.

13. Problem

14. Electric Fields

15. The Electric Field • The presence of + or – charge modifies empty space. This enables the electrical force to act on charged particles without actually touching them. • We say that an “electric field” is created in the space around a charged particle or a configuration of charges. • If a charged particle is placed in an electric field created by other charges, it will experience a force as a result of the field. • We can easily calculate the electric force from the electric field.

16. Electric Fields • Forces exist only when two or more particles are present. • Fields exist even if no force is present.

17. Field Lines Around Charges • The arrows in a field are not vectors, they are “lines of force”. • The lines of force indicate the direction of the force on a positive charge placed in the field. • Negative charges experience a force in the opposite direction.

18. Field Line Rules

19. Single Charges

20. Double Charges

21. Plate Charges

22. Field Vectors from Field Lines • The electric field at a given point is not the field line itself, but can be determined from the field line. • The electric field vectors, and thus force vector, is always tangent to the line of force at that point.

23. Force due to Electric Field • The force on a charged particle placed in an electric field is easily calculated. • F = Eq • F: Force (N) • E: Electric Field (N/C) • q: Charge (C)

24. Problem • The electric field in a given region is 4000 N/C pointed toward the north. What is the force exerted on a 400 μg Styrofoam bead bearing 600 excess electrons when placed in the field? Ignore gravitational effects.

25. iClicker An electron e and a proton p are simultaneously released from rest in a uniform electric field E. Assume that the particles are sufficiently far apart so that the only force acting on each particle after it is released is that due to the electric field. At a later time when the particles are still in the field, the electron and the proton will have the same a. direction of motion b. speed c. displacement d. magnitude of acceleration e. magnitude of force acting on them

26. Problem • A 400 mg Styrofoam bead has 600 excess electrons on its surface. What is the magnitude and direction of the electric field that will suspend the bead in midair?

27. Problem • A proton traveling at 440 m/s in the +x direction enters an electric field of magnitude 5400 N/C directed in the +y direction. Find the acceleration.

28. Electric Fields • The Electric Field surrounding a point charge or a spherical charge can be calculated by: • E = kq / r2 • E: Electric Field (N/C) • k: 9 x 109 N m2/C2 • q: Charge (C) • r: distance from center of charge q (m)

29. iClicker A point P is 0.50 meter from a point charge of 5.0 X 10‑8 coulomb. The intensity of the electric field at point P is most nearly (A) 2.5 x 10‑8 N/C (B) 2.5 x 101 N/C (C) 9.0 x 102 N/ C (D) 1.8 x 103 N/C (E) 7.5 x 108 N/C

30. iClicker Charges + Q and ‑ 4Q are situated as shown above. The net electric field is zero nearest which point? (A) A (B) B (C) C (D) D (E) E

31. Superposition of Fields • When more than one charge contributes to the electric field, the resultant electric field is the vector sum of the electric fields produced by the various charges. • Again, as with force vectors, this is referred to as superposition.

32. iClicker The diagram above shows an isolated, positive charge Q. Point (B) is twice as far away from Q as point A. The ratio of the electric field strength at point A to the electric field strength at point B is (A) 8 to 1 (B) 4 to 1 (C) 2 to 1 (D) 1 to 1 (E) 1 to 2

33. Electric Field Lines • Electric field lines are NOT VECTORS, but may be used to derive the direction of electric field vectors at given points. • The resulting vector gives the direction of the electric force on a positive charge placed in the field.

34. Problem • A particle bearing -5.0 μC is placed at -2.0 cm, and a particle bearing 5.0 μC is placed at 2.0 cm. What is the field at the origin?

35. iClicker An electron is accelerated from rest for a time of 10-9 second by a uniform electric field that exerts a force of 8.0 x 10-15 Newton on the electron. What is the magnitude of the electric field? (A) 8.0 x 10‑24 N/C (B) 9.1 x 10‑22 N/C (C) 8.0 x 10‑6 N/C (D) 2.0 x 10-5 N/C (E) 5.0 x 104 N/C

36. iClicker An electron is accelerated from rest for a time of 10-9 second by a uniform electric field that exerts a force of 8.0 x 10-15 Newton on the electron. The speed of the electron after it has accelerated for the 10‑9 second is most nearly (A) 101 m/s (B) 103 m/s (C) 105 m/s (D) 107 m/s (E) 109 m/s

37. Problem • A particle bearing 10.0 mC is placed at the origin, and a particle bearing 5.0 mC is placed at 1.0 m. Where is the field zero?

38. Electric Potential

39. Electric Potential Energy • Electrical potential energy is the energy contained in a configuration of charges. • Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable. • The unit is the Joule.

40. Electric Potential Energy • Electrical potential energy increases when charges are brought into less favorable configurations + - - -

41. Electric Potential Energy • Electrical potential energy decreases when charges are brought into more favorable configurations. + + + -

42. Work • Workmust be done on the charge to increase the electric potential energy.

43. Electric Potential • Electric potential is hard to understand, but easy to measure. • We commonly call it “voltage”, and its unit is the Volt. • 1 V = 1 J/C • Electric potential is easily related to both the electric potential energy, and to the electric field.

44. Electric Potential • The change in potential energy is directly related to the change in voltage. • ΔU = qΔV • Δ U: change in electrical potential energy (J) • q: charge moved (C) • Δ V: potential difference (V) • All charges will spontaneously go to lower potential energies if they are allowed to move.

45. Electric Potential • Since all charges try to decrease U, and ΔU = q ΔV, this means that spontaneous movement of charges result in negative ΔU. • V = U / q • Positive charges like to DECREASE their potential (ΔV < 0) • Negative charges like to INCREASE their potential. (ΔV > 0)

46. Problem • A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

47. E-field & Potential • The electric potential is related in a simple way to a uniform electric field. • ΔV = -Ed • ΔV: change in electrical potential (V) • E: Constant electric field strength (N/C or V/m) • d: distance moved (m)

48. Problem • An electric field is parallel to the x-axis. What is its magnitude and direction of the electric field if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V?

49. More Electric Field Stuff

50. iClicker Two large, flat, parallel, conducting plates are 0.04 m apart, as shown above. The lower plate is at a potential of 2 V with respect to ground. The upper plate is at a potential of 10 V with respect to ground. Point P is located 0.01 m above the lower plate. The electric potential at point P is a. 10 V b. 8 V c. 6 V d. 4 V e. 2 V