1. Chapter 21 Electric Charge and Electric Field
Chapter 21 opener. This comb has acquired a static electric charge, either from passing through hair, or being rubbed by a cloth or paper towel. The electrical charge on the comb induces a polarization (separation of charge) in scraps of paper, and thus attracts them. Our introduction to electricity in this Chapter covers conductors and insulators, and Coulomb’s law which relates the force between two point charges as a function of their distance apart. We also introduce the powerful concept of electric field. Electrostatic Bell Demo

2. ConcepTest 21.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero? 3R +Q
+4Q R 2R 1 2 3 4 5
Answer: 2

3. ConcepTest 21.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero? 3R +Q
+4Q R 2R 1 2 3 4 5
The force on Q0 due to +Q is: F = k(Q0)(Q)/R2
The force on Q0 due to +4Q is: F = k(Q0)(4Q)/(2R)2
Since +4Q is 4 times bigger than +Q, Q0 needs to be farther from +4Q. In fact, Q0 must be twice as far from +4Q, since the distance is squared in Coulomb’s law.

4. ConcepTest 21.5c Proton and Electron III
A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet?
1) in the middle
2) closer to the electron’s side
3) closer to the proton’s side
Answer: 3 p e

5. ConcepTest 21.5c Proton and Electron III
A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet?
1) in the middle
2) closer to the electron’s side
3) closer to the proton’s side
By Newton’s 3rd law, the electron and proton feel the same force. But, since F = ma, and since the proton’s mass is much greater, the proton’s acceleration will be much smaller!
Thus, they will meet closer to the proton’s original position. p e
Follow-up: Which particle will be moving faster when they meet?

6. 21-6 The Electric Field
For a point charge:

7. 21-6 The Electric Field An electric field surrounds every charge.
Figure An electric field surrounds every charge. P is an arbitrary point.

8. 21-6 The Electric Field Force on a point charge in an electric field:
Figure (a) Electric field at a given point in space. (b) Force on a positive charge at that point. (c) Force on a negative charge at that point.

9. 21-6 The Electric Field
Example 21-6: Electric field of a single point charge.
Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x 10-6 C.
Figure Example 21–6. Electric field at point P (a) due to a negative charge Q, and (b) due to a positive charge Q, each 30 cm from P.
Solution: Substitution gives E = 3.0 x 105 N/C. The field points away from the positive charge and towards the negative one.

10. 21-6 The Electric Field
Example 21-7: E at a point between two charges.
Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?
Figure Example 21–7. In (b), we don’t know the relative lengths of E1 and E2 until we do the calculation.
Solution: a. The electric fields add in magnitude, as both are directed towards the negative charge. E = 6.3 x 108 N/C.
b. The acceleration is the force (charge times field) divided by the mass, and will be opposite to the direction of the field (due to the negative charge of the electron). Substitution gives a = 1.1 x 1020 m/s2.

11. ConcepTest 21.9b Superposition II4 3 2 1
-2 C
What is the electric field at the center of the square?
E = 0
E = 1
E = 2
E = 3
E = 4
Answer: 5

12. ConcepTest 21.9b Superposition II4 3 2 1
-2 C
What is the electric field at the center of the square?
5) E = 0
The four E field vectors all point outward from the center of the square toward their respective charges. Because they are all equal, the net E field is zero at the center!!
Follow-up: What if the upper two charges were +2 C? What if the right-hand charges were +2 C?

13. 21-7 Electric Field Calculations for Continuous Charge Distributions
A continuous distribution of charge may be treated as a succession of infinitesimal (point) charges. The total field is then the integral of the infinitesimal fields due to each bit of charge:
Remember that the electric field is a vector; you will need a separate integral for each component.

14. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-9: A ring of charge.
A thin, ring-shaped object of radius a holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. Let λ be the charge per unit length (C/m).
Solution: Because P is on the axis, the transverse components of E must add to zero, by symmetry. The longitudinal component of dE is dE cos θ, where cos θ = x/(x2 + a2)1/2. Write dQ = λdl, and integrate dl from 0 to 2πa. Answer: E = (1/4πε0)(Qx/[x2 + a2]3/2)

15. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-9: A ring of charge.
A thin, ring-shaped object of radius a holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. Let λ be the charge per unit length (C/m).
Solution: Because P is on the axis, the transverse components of E must add to zero, by symmetry. The longitudinal component of dE is dE cos θ, where cos θ = x/(x2 + a2)1/2. Write dQ = λdl, and integrate dl from 0 to 2πa. Answer: E = (1/4πε0)(Qx/[x2 + a2]3/2)
SYMMETRY!!

16. 21-7 Electric Field Calculations for Continuous Charge Distributions
Conceptual Example 21-10: Charge at the center of a ring.
Imagine a small positive charge placed at the center of a nonconducting ring carrying a uniformly distributed negative charge. Is the positive charge in equilibrium if it is displaced slightly from the center along the axis of the ring, and if so is it stable? What if the small charge is negative? Neglect gravity, as it is much smaller than the electrostatic forces.
Solution: The positive charge is in stable equilibrium, as it is attracted uniformly by every part of the ring. The negative charge is also in equilibrium, but it is unstable; once it is displaced from its equilibrium position, it will accelerate away from the ring.

17. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-11: Long line of charge.
Determine the magnitude of the electric field at any point P a distance x from a very long line (a wire, say) of uniformly distributed charge. Assume x is much smaller than the length of the wire, and let λ be the charge per unit length (C/m).
Solution: By symmetry, there will be no component of the field parallel to the wire. Write dE = (λ/4πε0) (cos θ dy)/x2 + y2). This can be transformed into an integral over θ: dE = (λ/4πε0x) cos θ dθ. Integrate θ from –π/2 to + π/2. Answer: E = (1/2πε0) λ/x.

18. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-11: Long line of charge.
Determine the magnitude of the electric field at any point P a distance x from a very long line (a wire, say) of uniformly distributed charge. Assume x is much smaller than the length of the wire, and let λ be the charge per unit length (C/m).
Solution: By symmetry, there will be no component of the field parallel to the wire. Write dE = (λ/4πε0) (cos θ dy)/x2 + y2). This can be transformed into an integral over θ: dE = (λ/4πε0x) cos θ dθ. Integrate θ from –π/2 to + π/2. Answer: E = (1/2πε0) λ/x.
SYMMETRY!!

19. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-12: Uniformly charged disk.
Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area (C/m2) is σ. Calculate the electric field at a point P on the axis of the disk, a distance z above its center.
Solution: The disk is a set of concentric rings, and we know (from example 21-9) what the field due to a ring of charge is. Write dQ = σ 2πr dr. Integrate r from 0 to R. See text for answer.

20. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-12: Uniformly charged disk.
Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area (C/m2) is σ. Calculate the electric field at a point P on the axis of the disk, a distance z above its center.
Solution: The disk is a set of concentric rings, and we know (from example 21-9) what the field due to a ring of charge is. Write dQ = σ 2πr dr. Integrate r from 0 to R. See text for answer.
SYMMETRY!!

21. 21-7 Electric Field Calculations for Continuous Charge Distributions
In the previous example, if we are very close to the disk (that is, if z << R), the electric field is:
This is the field due to an infinite plane of charge.

22. 21-7 Electric Field Calculations for Continuous Charge Distributions
Example 21-13: Two parallel plates.
Determine the electric field between two large parallel plates or sheets, which are very thin and are separated by a distance d which is small compared to their height and width. One plate carries a uniform surface charge density σ and the other carries a uniform surface charge density -σ as shown (the plates extend upward and downward beyond the part shown).
Solution: The field due to each plate is σ/2ε0. Between the plates the fields add, giving E = σ/ε0; outside the plates the fields cancel.

23. 21-8 Field Lines
The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.
Figure Electric field lines (a) near a single positive point charge, (b) near a single negative point charge.

24. 21-8 Field Lines
The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge.
The electric field is stronger where the field lines are closer together.
Schematic only – fields in 3-d, not 2-d

25. 21-8 Field Lines
Electric dipole: two equal charges, opposite in sign:

26. ConcepTest 21.12b Electric Field Lines II1) 2)
3) both the same
Which of the charges has the greater magnitude?
Answer: 2

27. ConcepTest 21.12b Electric Field Lines II1) 2)
3) both the same
Which of the charges has the greater magnitude?
The field lines are denser around the red charge, so the red one has the greater magnitude.
Follow-up: What is the red/green ratio of magnitudes for the two charges?

28. 21-8 Field Lines
The electric field between two closely spaced, oppositely charged parallel plates is constant.
Field lines Demo

29. 21-8 Field Lines Summary of field lines:
Field lines indicate the direction of the field; the field is tangent to the line.
The magnitude of the field is proportional to the density of the lines.
Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.

30. 21-9 Electric Fields and Conductors
The static electric field inside a conductor is zero – if it were not, the charges would move.
The net charge on a conductor resides on its outer surface.
Figure A charge inside a neutral spherical metal shell induces charge on its surfaces. The electric field exists even beyond the shell, but not within the conductor itself.

31. 21-9 Electric Fields and Conductors
The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.
Figure If the electric field at the surface of a conductor had a component parallel to the surface E||, the latter would accelerate electrons into motion. In the static case, E|| must be zero, and the electric field must be perpendicular to the conductor’s surface: E = E┴.

32. 21-9 Electric Fields and Conductors
Conceptual Example 21-14: Shielding, and safety in a storm.
A neutral hollow metal box is placed between two parallel charged plates as shown. What is the field like inside the box?
The field inside the box is zero. This is why it can be relatively safe to be inside an automobile during an electrical storm.

33. 21-10 Motion of a Charged Particle in an Electric Field
The force on an object of charge q in an electric field is given by:
= q
Therefore, if we know the mass and charge of a particle, we can describe its subsequent motion in an electric field.

34. ConcepTest 21.6 Forces in 2D +2Q +4Q +Q 1 2 3 4 5
Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges?
Answer: 2

35. Follow-up: What would happen if the yellow charge were +3Q?
ConcepTest Forces in 2D
+2Q
+4Q +Q 1 2 3 4 5 d
Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges?
The charge +2Q repels +Q toward the right. The charge +4Q repels +Q upward, but with a stronger force. Therefore, the net force is up and to the right, but mostly up.
+2Q
Follow-up: What would happen if the yellow charge were +3Q?
+4Q

36. 21-10 Motion of a Charged Particle in an Electric Field
Example 21-15: Electron accelerated by electric field.
An electron (mass m = 9.11 x kg) is accelerated in the uniform field (E = 2.0 x 104 N/C) between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. Assume the hole is so small that it does not affect the uniform field between the plates.
Solution. a. The acceleration of the electron is qE/m = 3.5 x 1015 m/s2. In 1.5 cm it therefore accelerates from a speed of zero to v = 1.0 x 107 m/s.
b. The electric force on the electron is qE = 3.2 x N; the gravitational force is mg = 8.9 x N. Therefore the gravitational force can safely be ignored.

37. 21-10 Motion of a Charged Particle in an Electric Field
Example 21-16: Electron moving perpendicular to .
Suppose an electron traveling with speed v0 = 1.0 x 107 m/s enters a uniform electric field , which is at right angles to v0 as shown. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity.
Solution: The acceleration is in the vertical direction (perpendicular to the motion) and is equal to –eE/m. Then y = ½ ay2 and x = v0t; eliminating t gives the equation y = -(eE/2mv02)x2, which is a parabola.
Electrostatic Wind Demo

38. 21-11 Electric Dipoles
An electric dipole consists of two charges Q, equal in magnitude and opposite in sign, separated by a distance . The dipole moment, p = Q , points from the negative to the positive charge.
Figure A dipole consists of equal but opposite charges, +Q and –Q, separated by a distance l. The dipole moment p + Ql and points from the negative to the positive charge.

39. 21-11 Electric Dipoles
An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:
Figure An electric dipole in a uniform electric field.

40. 21-11 Electric Dipoles
The electric field created by a dipole is the sum of the fields created by the two charges; far from the dipole, the field shows a 1/r3 dependence:
Figure Electric field due to an electric dipole.

41. 21-11 Electric Dipoles Example 21-17: Dipole in a field.
The dipole moment of a water molecule is 6.1 x C·m. A water molecule is placed in a uniform electric field with magnitude 2.0 x 105 N/C. (a) What is the magnitude of the maximum torque that the field can exert on the molecule? (b) What is the potential energy when the torque is at its maximum? (c) In what position will the potential energy take on its greatest value? Why is this different than the position where the torque is maximum?
Solution. a. The maximum torque occurs then the field is perpendicular to the dipole moment; τ = pE = 1.2 x N·m.
b. The potential energy is zero.
c. The potential energy is maximum when the dipole moment is in the opposite direction to the field; this is the largest angle through which the dipole can rotate to reach equilibrium. The torque is maximum when the moment and the field are perpendicular. This situation is analogous to a physical pendulum – its gravitational potential energy is greatest when it is poised above the axis, even though the torque is zero there.

42. 21-12 Electric Forces in Molecular Biology; DNA
Molecular biology is the study of the structure and functioning of the living cell at the molecular level.
The DNA molecule is a double helix:
Figure 21-47a. Section of a DNA double helix.

43. 21-12 Electric Forces in Molecular Biology; DNA
The A-T and G-C nucleotide bases attract each other through electrostatic forces.
Figure 21-47b. “Close-up” view of the helix, showing how A and T attract each other and how G and C attract each other through electrostatic forces. The + and - signs represent net charges, usually a fraction of e, due to uneven sharing of electrons. The red dots indicate the electrostatic attraction (often called a “weak bond” or “hydrogen bond”—Section 40–3). Note that there are two weak bonds between A and T, and three between C and G.

44. 21-12 Electric Forces in Molecular Biology; DNA
Replication: DNA is in a “soup” of A, C, G, and T in the cell. During random collisions, A and T will be attracted to each other, as will G and C; other combinations will not.
Figure Replication of DNA.

45. Summary of Chapter 21
Two kinds of electric charge – positive and negative.
Charge is conserved.
Charge on electron:
e = x C.
Conductors: electrons free to move.
Insulators: nonconductors.

46. Summary of Chapter 21 Charge is quantized in units of e.
Objects can be charged by conduction or induction.
Coulomb’s law:
Electric field is force per unit charge:

47. Summary of Chapter 21 Electric field of a point charge:
Electric field can be represented by electric field lines.
Static electric field inside conductor is zero; surface field is perpendicular to surface.

48. Chapter 22 Gauss’s Law
Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can figure out the integral. In this drawing, two different surfaces are shown, both enclosing a point charge Q. Gauss’s law states that the product E·dA, where dA is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface Qencl divided by ε0. Both surfaces here enclose the same charge Q. Hence ∫E·dA will give the same result for both surfaces. 48

49. Units of Chapter 22 Electric Flux Gauss’s Law
Applications of Gauss’s Law
Experimental Basis of Gauss’s and Coulomb’s Laws

50. 22-1 Electric Flux Electric flux:
Electric flux through an area is proportional to the total number of field lines crossing the area.
Figure (a) A uniform electric field E passing through a flat area A. (b) E┴ = E cos θ is the component of E perpendicular to the plane of area A. (c) A┴ = A cos θ is the projection (dashed) of the area A perpendicular to the field E. 50

51. 22-1 Electric Flux Example 22-1: Electric flux.
Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°.
Solution. The flux is EA cos θ = 3.5 N·m2/C. 51

52. 22-1 Electric Flux
Flux through a closed surface:

53. Think in terms of flux lines:

54. 22-2 Gauss’s Law
The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law:
This can be used to find the electric field in situations with a high degree of symmetry.

55. 22-2 Gauss’s Law For a point charge, Spherical Therefore,
Solving for E gives the result we expect from Coulomb’s law:
Figure A single point charge Q at the center of an imaginary sphere of radius r (our “gaussian surface”—that is, the closed surface we choose to use for applying Gauss’s law in this case). 55

56. 22-2 Gauss’s Law
Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives:
Figure A single point charge surrounded by a spherical surface, A1, and an irregular surface,A2.
Looking at the arbitrarily shaped surface A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface. 56

57. 22-2 Gauss’s Law
Finally, if a gaussian surface encloses several point charges, the superposition principle shows that:
Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.

58. 22-2 Gauss’s Law Conceptual Example 22-2: Flux from Gauss’s law.
Consider the two gaussian surfaces, A1 and A2, as shown. The only charge present is the charge Q at the center of surface A1. What is the net flux through each surface, A1 and A2?
Solution: The net flux through A1 is Q/ε0, as it encloses charge Q. The net flux through surface A2 is zero, even though the field is not zero on the surface. 58

59. 22-3 Applications of Gauss’s Law
Example 22-3: Spherical conductor.
A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere?
Figure Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3.
Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2).
b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero.
c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.
Spherical 59

60. 22-3 Applications of Gauss’s Law
Example 22-4: Solid sphere of charge.
An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).
Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2).
b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03).
Spherical/radial 60

61. 22-3 Applications of Gauss’s Law
Example 22-5: Nonuniformly charged solid sphere.
Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.
Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05.
b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05. 61

62. 22-3 Applications of Gauss’s Law
Example 22-6: Long uniform line of charge.
A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends.
Linear,
Cylindrical
Solution: If the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere. Therefore, a cylindrical gaussian surface will allow the easiest calculation of the field. The field is parallel to the ends and constant over the curved surface; integrating over the curved surface gives E = λ/2πε0R. 62

63. 22-3 Applications of Gauss’s Law
Example 22-7: Infinite plane of charge.
Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane.
Solution: We expect E to be perpendicular to the plane, and choose a cylindrical gaussian surface with its flat sides parallel to the plane. The field is parallel to the curved side; integrating over the flat sides gives E = σ/2ε0.
Planar & cylindrical 63

64. 22-3 Applications of Gauss’s Law
Example 22-8: Electric field near any conducting surface.
Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by
E = σ/ε0
where σ is the surface charge density on the conductor’s surface at that point.
Solution: Again we choose a cylindrical gaussian surface. Now, however, the field inside the conductor is zero, so we only have a nonzero integral over one surface of the cylinder. Integrating gives E = σ/ε0. 64

65. 22-3 Applications of Gauss’s Law
The difference between the electric field outside a conducting plane of charge and outside a nonconducting plane of charge can be thought of in two ways:
The field inside the conductor is zero, so the flux is all through one end of the cylinder.
The nonconducting plane has a total charge density σ, whereas the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.

66. 22-3 Applications of Gauss’s Law
Conceptual Example : Conductor with charge inside a cavity.
Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a point charge +q. What can you say about the charges on the inner and outer surfaces of the conductor?
Solution: The field must be zero within the conductor, so the inner surface of the cavity must have an induced charge totaling –q (so that a gaussian surface just around the cavity encloses no charge). The charge +Q resides on the outer surface of the conductor. 66

67. 22-3 Applications of Gauss’s Law
Procedure for Gauss’s law problems:
Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field).
Draw the surface.
Use the symmetry to find the direction of E.
Evaluate the flux by integrating.
Calculate the enclosed charge.
Solve for the field.

68. 22-4 Experimental Basis of Gauss’s and Coulomb’s Laws
In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero.
Figure (a) A charged conductor (metal ball) is lowered into an insulated metal can (a good conductor) carrying zero net charge. (b) The charged ball is touched to the can and all of its charge quickly flows to the outer surface of the can. (c) When the ball is then removed, it is found to carry zero net charge. 68

69. Summary of Chapter 22 Electric flux: Gauss’s law:
Gauss’s law can be used to calculate the field in situations with a high degree of symmetry.
Gauss’s law applies in all situations, and therefore is more general than Coulomb’s law.

70. Homework Asignment # 2
Chapter 21 – 40, 58, 62
Chapter 22 – 10, 22, 34
Homework Asignment # 3
Chapter 23 – 4, 14, 30, 36, 52, 62