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

21-7 Electric Field Calculations for Continuous Charge DistributionsA 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.

The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.

21-8 Field Lines( 場線 )

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.

21-8 Field Lines

Electric dipole( 電偶極 ): two equal charges, opposite in sign:

21-8 Field Lines

The electric field between two closely spaced, oppositely charged parallel plates is constant.

21-8 Field Lines

The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.

21-9 Electric Fields and Conductors

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.

E$$$$$$$$$$$$$$

E$$$$$$$$$$$$$$

F$$$$$$$$$$$$$$

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 10-31 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.

E$$$$$$$$$$$$$$

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.

E

E

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.

$$$$$$$$$$$$$$p

d

dQp

21-11 Electric DipolesAn electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:

21-11 Electric Dipoles

21-11 Electric DipolesThe 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:

21-11 Electric DipolesExample 21-17: Dipole in a field.

The dipole moment of a water molecule is 6.1 x 10-30 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?

Example, Electric Field of a Charged Circular Rod

Chapter 22 Gauss’s Law

Electric flux:

Electric flux through an area is proportional to the total number of field lines crossing the area.

22-1 Electric Flux ( 電通量 )

Flux through a closed surface:

22-1 Electric Flux

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.

22-2 Gauss’s Law

22-2 Gauss’s LawFor a point charge( 點電荷 ),

Therefore,

Solving for E gives the result we expect from Coulomb’s law:

22-2 Gauss’s LawUsing Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives:

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.

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.

22-2 Gauss’s LawConceptual 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?

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?

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).

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.

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.

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.

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.

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?

Chapter 23Electric Potential

The electrostatic force is conservative – potential energy can be defined.

Change in electric potential energy is negative of work done by electric force:

23-1 Electrostatic Potential Energy 靜電位能 and Potential Difference 電位差

Electric potential is defined as potential energy per unit charge:

Unit of electric potential: the volt (V):

1 V = 1 J/C.

23-1 Electrostatic Potential Energy and Potential Difference

Only changes in potential can be measured, allowing free assignment of V = 0:

23-1 Electrostatic Potential Energy and Potential Difference

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?

Analogy between gravitational and electrical potential energy:

23-1 Electrostatic Potential Energy and Potential Difference

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:

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 = +5000 V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration?

23-2 Relation between Electric Potential and Electric Field

The general relationship between a conservative force and potential energy:

Substituting the potential difference and the electric field:

23-2 Relation between Electric Potential and Electric Field

The simplest case is a uniform field:

23-2 Relation between Electric Potential and Electric FieldExample 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 0.050 m, calculate the magnitude of the electric field in the space between the plates.

23-2 Relation between Electric Potential and Electric FieldExample 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.

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:

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.

23-3 Electric Potential Due to Point ChargesTo find the electric potential due to a point charge, we integrate the field along a field line:

23-3 Electric Potential Due to Point ChargesSetting the potential to zero at r = ∞ gives the general form of the potential due to a point charge:

23-3 Electric Potential Due to Point ChargesExample 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 0.500 m from a charge Q = 20.0 µC?

23-3 Electric Potential Due to Point ChargesExample 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.

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

or

23-4 Potential Due to Any Charge DistributionExample 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.

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.

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