1. 今日課程內容
CH21 電荷與電場 電場
電偶極
CH22 高斯定律
CH23 電位

2. 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.

3. 21-8 Field Lines(場線)
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.

4. 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.

5. 21-8 Field Lines
Electric dipole(電偶極): two equal charges, opposite in sign:

6. 21-8 Field Lines
The electric field between two closely spaced, oppositely charged parallel plates is constant.

7. 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┴.

8. 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.

9. 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.

10. 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.

11. 21-11 Electric Dipoles(電偶極)
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.

12. 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.

13. 21-11 Electric Dipoles

14. 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.

15. 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.

16. Example, Electric Field of a
Charged Circular Rod

17. 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.

18. 22-1 Electric Flux (電通量)
Electric flux:
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.
Electric flux through an area is proportional to the total number of field lines crossing the area.

19. 22-1 Electric Flux
Flux through a closed surface:

20. 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.

21. 22-2 Gauss’s Law For a point charge(點電荷), Therefore,
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).
Solving for E gives the result we expect from Coulomb’s law:

22. 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任意 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.

23. 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.

24. 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.

25. 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.

26. 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).

27. 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.

28. 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.
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.

29. 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.

30. 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.

31. 22-3 Applications of Gauss’s Law
Conceptual Example 22-9: 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.

32. Chapter 23 Electric Potential
Chapter 23 opener. We are used to voltage in our lives—a 12-volt car battery, 110 V or 220 V at home, 1.5 volt flashlight batteries, and so on. Here we see a Van de Graaff generator, whose voltage may reach 50,000 V or more. Voltage is the same as electric potential difference between two points. Electric potential is defined as the potential energy per unit charge. The children here, whose hair stands on end because each hair has received the same sign of charge, are not harmed by the voltage because the Van de Graaff cannot provide much current before the voltage drops. (It is current through the body that is harmful, as we will see later.)

33. 23-1 Electrostatic Potential Energy靜電位能 and Potential Difference電位差
The electrostatic force is conservative – potential energy can be defined.
Change in electric potential energy is negative of work done by electric force:
Figure Work is done by the electric field in moving the positive charge q from position a to position b.

34. 23-1 Electrostatic Potential Energy and Potential Difference
Electric potential is defined as potential energy per unit charge:
Unit of electric potential: the volt (V):
1 V = 1 J/C.

35. 23-1 Electrostatic Potential Energy and Potential Difference
Only changes in potential can be measured, allowing free assignment of V = 0:

36. 23-1 Electrostatic Potential Energy and Potential Difference
Conceptual Example 23-1: A negative charge.
Suppose a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change?
Figure Central part of Fig. 23–1, showing a negative point charge near the negative plate, where its potential energy (PE) is high. Example 23–1.
Solution: The electron will move towards the positive plate if released, thereby increasing its kinetic energy. Its potential energy must therefore decrease. However, it is moving to a region of higher potential V; the potential is determined only by the existing charge distribution and not by the point charge. U and V have different signs due to the negative charge.

37. 23-1 Electrostatic Potential Energy and Potential Difference
Analogy between gravitational and electrical potential energy:
Figure (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.

38. 23-1 Electrostatic Potential Energy and Potential Difference
Electrical sources such as batteries and generators supply a constant potential difference. Here are some typical potential differences, both natural and manufactured:

39. 23-1 Electrostatic Potential Energy and Potential Difference
Example 23-2: Electron in CRT.
Suppose an electron in a cathode陰極 ray tube is accelerated from rest through a potential difference Vb – Va = Vba = V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × kg) as a result of this acceleration?
Solution: a. The change in potential energy is qV = -8.0 x J.
b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

40. 23-2 Relation between Electric Potential and Electric Field
The general relationship between a conservative force and potential energy:
Figure To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.
Substituting the potential difference and the electric field:

41. 23-2 Relation between Electric Potential and Electric Field
The simplest case is a uniform field:

42. 23-2 Relation between Electric Potential and Electric Field
Example 23-3: Electric field obtained from voltage.
Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is m, calculate the magnitude of the electric field in the space between the plates.
Solution: E = V/d = 1000 V/m.

43. 23-2 Relation between Electric Potential and Electric Field
Example 23-4: Charged conducting sphere.
Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q.
Solution: The electric field outside a conducting sphere is Q/(4πε0r2). Integrating to find the potential, and choosing V = 0 at r = ∞:
a. V = Q/4πε0r.
b. V = Q/4πε0r0.
c. V = Q/4πε0r0 (the potential is constant, as there is no field inside the sphere).

44. 23-2 Relation between Electric Potential and Electric Field
The previous example gives the electric potential as a function of distance from the surface of a charged conducting sphere, which is plotted here, and compared with the electric field:
Figure (a) E versus r, and (b) V versus r, for a uniformly charged solid conducting sphere of radius r0 (the charge distributes itself on the surface); r is the distance from the center of the sphere.

45. 23-2 Relation between Electric Potential and Electric Field
Example 23-5: Breakdown voltage.
In many kinds of equipment, very high voltages are used. A problem with high voltage is that the air can become ionized due to the high electric fields: free electrons in the air (produced by cosmic rays, for example) can be accelerated by such high fields to speeds sufficient to ionize O2 and N2 molecules by collision, knocking out one or more of their electrons. The air then becomes conducting and the high voltage cannot be maintained as charge flows. The breakdown of air occurs for electric fields of about 3.0 × 106 V/m. (a) Show that the breakdown voltage for a spherical conductor in air is proportional to the radius of the sphere, and (b) estimate the breakdown voltage in air for a sphere of diameter 1.0 cm.
Solution: a. Combining the equations for the field and the potential gives V = r0E.
b. Substituting gives V = 15,000 V.

46. 23-3 Electric Potential Due to Point Charges
To find the electric potential due to a point charge, we integrate the field along a field line:
Figure We integrate Eq. 23–4a along the straight line (shown in black) from point a to point b. The line ab is parallel to a field line.

47. 23-3 Electric Potential Due to Point Charges
Setting the potential to zero at r = ∞ gives the general form of the potential due to a point charge:
Figure Potential V as a function of distance r from a single point charge Q when the charge is positive.
Figure Potential V as a function of distance r from a single point charge Q when the charge is negative.

48. 23-3 Electric Potential Due to Point Charges
Example 23-6: Work required to bring two positive charges close together.
What minimum work must be done by an external force to bring a charge q = 3.00 μC from a great distance away (take r = ∞) to a point m from a charge Q = 20.0 µC?
Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

49. 23-3 Electric Potential Due to Point Charges
Example 23-7: Potential above two charges.
Calculate the electric potential (a) at point A in the figure due to the two charges shown, and (b) at point B.
Solution: The total potential is the sum of the potential due to each charge; potential is a scalar, so there is no direction involved, but we do have to keep track of the signs.
a. V = 7.5 x 105 V
b. V = 0 (true at any point along the perpendicular bisector)

50. 23-4 Potential Due to Any Charge Distribution
The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous): or

51. 23-4 Potential Due to Any Charge Distribution
Example 23-8: Potential due to a ring of charge.
A thin circular ring of radius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center.
Solution: Each point on the ring is the same distance from point P, so the potential is just that of a charge Q a distance (R2 + x2)1/2 from point P.

52. 23-4 Potential Due to Any Charge Distribution
Example 23-9: Potential due to a charged disk.
A thin flat disk, of radius R0, has a uniformly distributed charge Q. Determine the potential at a point P on the axis of the disk, a distance x from its center.
Solution: Consider the disk to be made up of infinitely thin rings, each at a radius R with a thickness dR. Each ring then carries a charge dq = 2QR dR/R02. Integrating to find V then gives the solution in the text.

53. 習題
Ch21: 10, 31, 35, 48, 58, 65
Ch22: 8, 18, 24, 34,
Ch23: 11, 15, 27, 33, 35,