How do most objects interact with each other?

1. How do most objects interact with each other? Electric Charge (q or Q) An intrinsic property of matter (as is mass) S.I. unit: coulombs (C), (also: μC) Two types (+) or (-): add algebraically to give a net charge. Like charges repel; unlike charges attract. Coulomb’s Law: The force between two charges (q1) and q2) separated by a distance r is: Fe=k q1 q2 / r2. k is Coulomb’s constant: k = 9 × 109 Nm2/C2 Force by q1 on q2is equal and opposite to the force by q2 on q1.

2. More about Charge. • Quantization of charge (e): charge comes in discrete amounts of e = 1.6×10-19C • e is the “elementary charge”; we always treat e as positive. • Electrons have a charge of –e • Protons have a charge of +e and about 2000 times the mass of electrons. • Conservation of charge: In any reaction, net charge remains the same. (Is this different from conservation of mass?)

3. Examples • Book sitting on a table • Pushing off • A hard wall • A soft wall • Electric Force is a measure of hardness in repulsion. • Sodium Metal vs. Aluminum Metal. • Electric Force is a measure of hardness in attraction. • What happens if we hold a • Negatively charged rod held near a neutral atom? • Positively charged rod held near a neutral atom?

4. How can charge be transferred between objects? • Friction • Electrons transferred by rubbing. • Rubbing plastic with fur; fur loses electrons. • Rubbing glass with silk; silk gains electrons. • Induction • A charged object causes electrons to move in a neutral object. • Atoms become polarized; they turn into dipoles. • Charged rod (+) or (-) near a neutral object; what happens?

5. How can charge be transferred between objects? (cont’d) • Conduction • Electrons can move freely through some materials, called conductors. • Conductors can be charged directly through contact with charged objects (not friction). • Conductors can be charged through induction • Electrons move freely to one side or another; object as a whole becomes polarized.

6. Van de Graf Generator • Device to concentrate positive charge on a conducting sphere. • Operation: • Lower roller (G) rubs e-off belt. • Belt becomes (+) • (+) belt attracts e- from brush (B), bleeding them off the conducting sphere. • Belt is neutralized as it travels back to lower roller. • Sphere (A) becomes (+) more and more.

7. Electrical Properties of Materials • Conductors: Outer electrons can move nearly freely through the material. (What is it about metallic bonding that allows this?) • Insulators: Outer electrons are bound tightly to the atoms. Cannot move freely.

8. Electrical Properties of Materials(cont’d) • Semiconductors: Outer electrons are bound, but not so tightly. (e.g. silicon, germanium) • Can be made to conduct electricity in very controlled ways. • Doping: impurities to increase or decrease electrons in crystal structure (e.g. transistors) • Photoelectric Effect: electrons can be energized by absorbing light ot leave their atoms and travel through the material. (e.g. photocells)

9. Using Coulomb’s Law • A charge (Q) of +2µC is fixed in the ground. A charge (q) of +3µC having a mass of 0.24 kg is held 0.15m directly above the fixed charge and released. How does the charge q move? Give the magnitude and direction of its acceleration. Use (g = 10 m/s2 and k = 9 × 109 Nm2/C2)

10. Using Coulomb’s Law (cont’d) • Two protons are separated by 5×10-15m. • Calculate the force on one proton due to the other. (qp= + e = 1.6×10-19 C). Does the force seem small or large as felt by: a person? a fly? a proton? b. Calculate the acceleration of the proton. Is this a large number? (mp= 1.67×10-27 kg)

11. Using Coulomb’s Law (cont’d) • Three charges (q1 and q3 are fixed): q1 = -86 µC at (52 cm, 0) q2 = +50 µC at (0, 0) q3 = +65 µC at (0,30 cm) What is the net force on q2? (magnitude and direction) What is its acceleration if it has a mass of 0.5 kg?

12. Electric Field • How does one charge know that another charge is there? (“Action at a distance”). • A charge actually changes the space around it; a force field is caused. • Other charges interact with this “Electric Field”. • The Electric Field due to a “source” charge (qs): • E = k qs /r2 ř (a vector) • The direction of E at a point is in the same direction that a positive point charge would move if placed at that point. (BY DEFINITION)

13. Electric Field (cont’d) • Suppose we havemany charged objects creating an E-field in space. • At some point (r) (position vector from origin) in space: • Put a small positive test charge (qT) at r. • The magnitude of E at r equals the Force on qT divided by qT. E = Fe/qT (By definition) • The direction of E at r is in the direction of the force on qT. (By definition)

14. Advantages of the E-Field • The Electric fields caused by several charges add up as vectors. • For a system of charges: • Find the resultant (net) E-Field • Now you can easily find the force on any charge (q) you place in the system: Fe= q E • If q is (+), then Fe and E are in the same direction, otherwise they are opposite to each other. • If you change q, you don’t need to recalculate the E-field. • The E-field is physical; energy is actually stored in space.

15. Common Types of E-Field Problems • Given acollection of point charges: • Find the resulting E-field at a specified location • Find the location where the E-field has a specified magnitude and direction (such as zero.) • Given the E-field: • find the forces and/or accelerations produced on a charge or charged system.

16. Common Mistake to Avoid • E = k q /r2 ONLY when it is CAUSED by a POINT CHARGE we call “q”. • If you have are given an E-field due to some other source or sources, the above equation is FALSE. • Example: A uniform E-field has the same value everywhere in space. For such a field, the above equation is FALSE.

17. Using the E-field • For charges: q1 = -3 μC placed at (0,0) q2 = +6 μC placed at (2 m, 0) a. Find the E-field at (-1m, 0) b. Find the force on an electron placed at (-1 m , 0) c. Where on the x-axis is E = 0?

18. Using the E-field (cont’d) • A pendulum with: charge (+Q) and mass (m) is in equilibrium in a uniform E-field applied in the +x-direction. In terms of Q, m, and θ, • Find the magnitude of the E-field. • Find the tension in the string.

19. Using the E-field (cont’d) • A charge of (+q) and mass (m) is at the origin and moving in the +x-direction with a speed v. A uniform E-field (with magnitude: E) acts in the –x-direction. • Find the acceleration of the charge. • Find the maximum value of +x reached by the charge. • Find the time for the charge to return to its start point.

20. Electric Field Lines • Used to visualize an electric field’s • Direction: Directed path along which a positive test charge would move. • Magnitude: Proportional to the number of field lines coming through an area (line density).

21. Drawing Electric Field Lines To draw the electric field lines for a single charge • Draw lines directed away from (+) charges. • Draw lines directed towards (-) charges. • The number of lines entering or leaving the charge is proportional to the amount of charge. • Examples to do: Using 4 lines for a (+1 C) charge, draw the field lines for: • +1C charge, +2C charge, -1C charge, -2C charge.

22. Drawing Electric Field Lines (cont’d) To draw electric field lines for a system of charges: • Draw field lines near each charge in the system. • Connect field lines smoothly between (+) and (-) charges. • Field lines between like charges can go to infinity. • Draw field lines perpendicular (normal) to conducting surfaces (why?).

23. Examples • Dipole (+ -) • Two (+) charges (+ +) • Quadrupole + - - + 4. Oppositely charged plates that are close together.(*Important*) Check answers at Electric Field Applet

24. How does a Conductor behave in an E-Field? • In electrostatics, charges don’t move. (Fields due to stationary charges are called electrostatic fields.) • Recall what happens when we put a charged rod near a conductor: • Transient current • Polarization of entire object • Electrons moved so as to cancel the E-field inside the conductor. (Why must this be so?) • Excess charges move to surface and distribute themselves out of mutual repulsion. • And so ....

25. How does an E-field behave in a Conductor? • No electrostatic field can exist in the material of a conductor. • At the surface of a charged conductor: • E-field is perpendicular to the surface. • Component tangential to the surface is zero. • Higher concentrations of charge occur at surfaces of higher curvature. • In an empty cavity within a conductor • E-field is zero • Region is shielded from any external charges.

26. Charge densities at curved surfaces • On flat surfaces of low curvature, repulsive forces are directed mostly parallel to surface, keeping charges further apart. • On highly curved surfaces, parallel components of the repulsive forces are smaller, allowing charges to be closer together. • Result: Strong E-fields near highly curved surfaces. (Ex: Lightning Rods)

27. Electric Field Lines near a Conductor

28. Conductor in an E-field(summary) Charge concentrates at high curvature. Field is normal to surface Field is NOT zero in a cavity containing charge. Field is ZERO in an empty cavity.

29. How do we describe the ability of an Electric Field to do Work? • We want to discuss the ability of a field to do work without regard to the material being worked on. • Approach: • Develop an intuitive gravitational analog. • Extend the concept to the electrical case. • Apply the concept to different combinations of positive/negative source/test charges.

30. What is the Gravitational Equivalent to an E-field? • For uniform fields: • Fg = mg Fe = q E • For fields due to a point mass or charge: • Fg = m GM/r2 ř Fe = q kqs/r2 ř So E-field is to charge as ____________ is to mass.

31. Gravitational Potential Energy b _______________ a _______________ • Mass falling from b to a loses P.E.: ΔUg = -mgh • Work done by the gravitational field: Wfield = mgh • It takes work to lift mass from a to b: Wexternal = mgh • More mass will lose more P.E. • More mass will require more work to lift.

32. Gravitational Potential • Define Gravitational Potential: Vg = Ug/m (gravitational potential energy per unit mass). • Gravitational Potential Difference: ΔVg = -gh in going from b to a. • ΔVgis a measure of • How fast and how far material will fall from b to a. • How difficult it is to lift material from a to b. • ΔVgdescribes the terrain through which we can climb or fall without regard to our mass. • For a uniform field (such as near Earth’s Surface) Vg = gh or Vg = -GM/r

33. Topographical Maps • Topo map: maps contours or lines of equal gravitational potential. • No work is done to move objects along these lines. • Moving across lines requires work • Closely spaced lines indicates steepness. • Steepness indicates the strength of the field in the direction of your movement. (e.g. “field strength” here would be analogous to an an inclined plane: g sinθ; direction of fall would be down the plane.)

34. Topo map

35. Electrical Potential Energy Let’s extend the concept to the E-field: • A (+q) charge “falls” from high (P.E.)b to low (P.E.)a : ΔUe = Uea - Ueb < 0 • Work done by the field in this falling: Wfield = F d = qE d • Change in P.E. : ΔUe = -qEd • More charged material means • More work done by the field in falling • More P.E. lost in falling • More external work to push the charge “uphill” against the field. + - | | | | (b) (a) E

36. Electric Potential • Define Electric Potential: V= Ue/q (Electric potential energy per unit charge). (Units: joule/coulomb or “volt”; J/C or V) • Potential Difference or (“Voltage”): (going from b to a) ΔV = Va-Vb=ΔUe/q = -qEd/q = -Ed • ΔVis a measure of • How fast and how far charged material will fall from b to a. • How difficult it is to lift charged material from a to b. • ΔUe = q ΔV (the signs of q and ΔV are important) • ΔV describes the electrical terrain through which we can climb or fall without regard to our charge. • BUT: The direction of your fall depends on the sign of your charge.

37. Electric Potential and Charged Material + - | | | | (b) (a) High V Low V E • E-field direction specifies region of high V to region of low V. • (+) charges fall to: • ___________ potential • _________ potential energy • (-) charges fall to: • ___________ potential • _________ potential energy ΔUe = q ΔV

38. Some Examples • An electron is accelerated across a voltage of 5000 volts between two plates. • What is the change in potential energy of the electron? • What is the work done on the electron by the E-field between the two plates? • What speed does the electron attain? • Who works with these kinds of problems? • What is an “electron-volt (eV)”?

39. The Electron Volt • The amount of work needed to push an electron across a potential difference of 1 volt. • Allows MUCH easier calculations of energies. • Allows MUCH easier grading of the calculations. • Use them when charges are given in units of “e”. • 1 eV = (1.60×10-19C)(1 J/C) = 1.60×10-19J • Notice: instead of multiplying voltage by the electronic charge in coulombs, you just put an “e” in front of it. • If you’re asked for speeds; you’re back to S.I. units.

40. Examples (cont’d) 2. Two plates are charged to a voltage 50V. If their separation is 5.0 cm, what is the E-field between them? The E-field is uniform between the plates. + - | | | | | | | | +50 V 0V | d = 5 cm |

41. Visualizing Electric Potential • Draw Equipotentials:Lines along which the electric potential is constant. • Similar to contours on a topo map. • Questions • How would you calculate the work done in moving a charge along an equipotential? • How are equipotentials related to electric field lines? They are perpendicular to each other at every point.

42. Mapping Equipotentials • What do equipotentials look like for • A point charge? • A dipole? • Two like charges? • Applet as before: Electric Field Applet • For a charge distribution, how could you experimentally map • Equipotentials? • Electric field lines?

43. Finding the Electric Potential due to a Point Charge • Suppose we have a positive charge: Q • How can we calculate (estimate) the work done by Q’s E-field as a (+) charge q falls away from rA to rB? • WTOT = Σ Wi = Σ Fi Δr = Σ q Ei Δr = q Σ kQ/ri2 Δr = q (kQ/ rA - kQ/ rB) (By Calculus) • What is the resulting change in potential energy of q? • ΔUe = - WTOT = q (kQ/rB - kQ/rA) • What is the potential difference (voltage) between rA and rB? • ΔV = ΔUe /q = kQ/rB - kQ/rA

44. Electric Potential due to a Point Charge • Due to a point charge (Q) • V = kQ/r at a point a distance r from Q • V  0 as r  ∞ • V increases as you get closer to a (+) charge. • V decreases as you get closer to a (-) charge. • Be able to plot V vs. r for a point charge. • Potential Difference or Voltage between two points (rA and rB) (due to a point charge): • ΔV = kQ/rB - kQ/rA • Work done in moving a charge from one point to another is independent of the path taken between the two points. • ΔV between two points is independent of any path between the two points.

45. Electric Potential Energy(of two point charges) • If we have a (+q1) charge fixed in space • If we push another (+q2) charge towards it from ∞, how does the P.E. of the system of two charges change? • If a (-q2) charge falls towards it from ∞, how does the P.E. of the system change? • Change in P.E. of two point charges (q1 and q2) (The q’s include the signs) • ΔUe = q2 ΔV = q2 (V1 - V∞) = q2 (k q1 /r12 - 0) = k q1 q2/r12 = U12 • How is this different from gravitational P.E. ?

46. Electric Potential Energy(for various charge configurations) • For several charges, the potential energies add (same as in the gravitational case): • For three charges: • Utotal = U12 + U13 + U23 • For four charges • Utotal = U12 + U13 + U23 + U14 + U24 + U34 • The potentials at a point due to a collection of charges (q1, q2, q3, …) also add: • Vtotal = V1 + V2 + V3 + ….

47. Examples Ex 1: Looking at some charge configurations (Two charges at different spacing as shown) • +q -q • +q -q • +q +q • Which set has a positive P.E.? • Which set has the most negative P.E.? • Which set requires the most work to separate the charges to ∞?

48. Examples (cont’d) Q1 is at (0, 26 cm) Q2 is at (0, -26 cm) 2. Potential above two charges: B 40 cm 40 cm A 30 cm Find the potential at points A and B due to charges Q1 and Q2. 60 cm Q2=+50μC Q1=-50μC At A: VA = VA1 + VA2 = k(Q1/rA1 + Q2/rA2) = At B: VB = VB1 + VB2 = k(Q1/rB1 + Q2/rB2) =

49. Examples (cont’d) • For the configuration of charges shown • Calculate the energy necessary to build the configuration. • Calculate the potential at the center of the triangle. q1=+4µC S=.2m q3= -4µC q2=+4µC

50. How can we store Electric Energy? (How can we store Electric Charge?)