大学物理期末复习

1、General Revision of College Physics,Huang Qiong,Chapter 1. Electrostatic Field,1. Coulomb Law in the Vacuum,(1) Coulomb force,(2) The principle of superposition,The total electrical force on given charge is the vector sum of the electrical forces caused by the other charges, calculated as if each ac

2、ted alone.,Force Due to the System of Point Charges.,(3) Find Electrostatic Force,Force Due to Continuous Charge Distributions,Q,q0,P,(4). Coulomb Law in the Dielectric,2. Electric Field Intensity,Magnitude: the electric field force on unit positive charge,Direction : direct the direction of force e

3、xerted on a positive test charge .,Unit N/C 、V/m,(2) The principle of superposition,(1) Definition,(3). Electric Field Lines,Electric field lines are not real. Field lines are not material objects. They are used only as a pictorial representation to provide a qualitative description of the field.,In

4、 electrostatic field E0, electric field lines begin and end on charges. In Induced electric field Ev, the electric field lines form closed loops, with no beginning and no end.,The electric field lines are denser in the place where the field intensity is stronger, and the electric field lines are spa

5、rser in the place where the field intensity is weaker.,Field lines are never cross. The field at any point has a unique direction.,3. GAUSSS LAW,In vacuum,In dielectric,Electric flux of the Gauss surface is related with charges in Gauss surface and is not related with charges out of Gauss surface.,T

6、he field intensity at a point on the Gauss surface is related with charges in the Gauss surface and is not related with charges out of the Gauss surface.,If the electric flux of a Gauss surface equals zero, there must be not charge in the Gauss surface.,4 . Calculating the Electric Field,Problem-sol

7、ving strategy:,Analysis the distribution of charges.,Solution 1. Applying the superposition principle of the field intensity.,If the charges are countable, the resultant field is the vector sum of the fields due to the individual charges.,when confronted with problems that involve a continuous distr

8、ibution of charge, Element Analysis Method,Solution 2. Applying Gausss law to symmetric charge distribution.,Plane symmetry,Spherical symmetry,The three symmetries:,Cylindrical symmetry,Problem-solving strategy:,Select appropriate gaussian surface.,Select appropriate coordinates, apply Gausss law.,A

9、nalysis the symmetry of the field intensity distribution .,Example 1,Example 2,Example 3,There are two concentric charged spherical shells of radius R1 and R2 . Charge quantities distribute uniformly.,Example 4,Example 5,Example 6,The charge line density of an infinite uniform charged cylindrical su

10、rface of radius R is ,find the electric field intensity.,Example 7,5. Electric potential energy , Electric potential,Electric potential（simply the Potential）,unit：V (volt),The potential at a point equals the work required to bring a unit positive charge from this point to the zero point of electric

11、potential .,Electric potential energy of q0 at a point,The relationship between the electric potential energy and Electric potential,In general, the electric potential is a function of all three spatial coordinates. If V is given in terms of rectangular coordinates, the electric field components Ex

12、,Ey, and Ez can be found from V(x,y,z) as the partial derivatives,For example, if ,then,In a certain region of space, the electric potential is zero everywhere along the x axis. From this we can conclude that the x component of the electric field in this region is (a) zero (b) in the +x direction (c

13、) in the -x direction.,In a certain region of space, the electric field is zero. From this we can conclude that the electric potential in this region is (a) zero (b) constant (c) positive (d) negative.,2020/6/29,35,（1）场强相等的区域，电势处处相等？,（2）场强为零处，电势一定为零？,（3）电势为零处，场强一定为零？,（4）场强大处，电势一定高？,Calculate Potenti

14、al.,Two Method:,Applying element analysis method,Applying the definition of potential,Example 1,Example 2,Find the electric field intensity and the electric potential.,Example 3.,6. Equipotential volumes and surfaces,2、可有,计算电势的方法（2种）,1、微元法,7. SUMMARY,计算场强的方法（3种）,1、点电荷场的场强及叠加原理,2、定义法,（分立）,（连续）,（分立）,（

15、连续）,典型电场的电势,典型电场的场强,均匀带电球面,球面内,球面外,均匀带电无限长直线,均匀带电无限大平面,均匀带电球面,均匀带电无限长直线,均匀带电无限大平面,方向垂直于直线,方向垂直于平面,Example,Solution :,Metallic Conductor and Dielectric in Electrostatic Field,Chapter 2.,1. Electrostatic Equilibrium,(1) The electrostatic equilibrium conditions of conductor,inside the conductor:,The el

16、ectric field just outside the charged conductor is perpendicular to the conductor surface.,When a conductor is in electrostatic equilibrium state, the conductor is a body of equal potential, and its surface is a surface of equal potential.,(2) The distribution of Potential,(3) The distribution of ch

17、arges,Solid Conductor : If the conductor carries charge, the charge resides entirely on its out surface.,Conductor cavity:,Example 1.,r,2. Capacitance,57,EXAMPLE 1,Cylindrical capacitor:,A cylindrical capacitor consists of a cylindrical conductor of radius RA and charge Q coaxial with a larger cylin

18、drical shell of radius RB and charge Q.,Find the capacitance of a cylindrical capacitor if its length is l.,l is large compared with RA and RB.,Suppose the electric quantities per unit length of two poles are respectively and - .,Calculate the potential difference.,Calculate the capacitance.,3 . Ene

19、rgy of Electric field,Chapter 2. Magnetic Field,1. Magnetic field produced by currents.,（1）Biot-savarts law,a finite straight current carrying wire:,Useful Conclusions,I,0,The magnetic induction at a point by an infinitely long, straight current-carrying wire,the half infinitely straight current-car

20、rying wire,at the points of the extended line:,At the centre of the loop:,Magnetic Field Due to a Curved Wire Segment,magnetic moment :,(2) Gausss Law in Magnetism,Find the magnetic flux of YOZ plane.,(3) Amperes circulation theorem (Amperes Law),There are two coaxial cylindrical surfaces as shown i

21、n figure. The current in the inner and the outer metal cylinder is both I, but the direction is opposite. Calculate the magnetic induction at points in the following,EXAMPLE 1.,The magnetic field of an infinite straight solenoid with steady current I is,EXAMPLE 1.,n: the number of turns per unit len

22、gth N： the total number of turns,(4)Ampere force on the current-carrying wire of length L and current I is,the magnetic force on a curved current-carrying wire in a uniform magnetic field is equal to that on a straight wire connecting the end points and carrying the same current.,EXAMPLE,The current

23、 in the long, straight wire CD is .The wire AB carries and its length is L. The wire AB lies in the plane perpendicular to CD. Their distance is d.,Find the net force exerted on the AB by the magnetic field created by the wire CD.,(5) Torque on a Current Loop in a Uniform Magnetic Field (Moment of M

24、agnetic Force ),Chapter 3. Electromagnetic Induction,1. Faradays Law of Electromagnetic Induction,Select the positive direction of the loop arbitrarily (i.e.: the reference direction of emf.).,Determine the normal of the surface bounded by the loop by right hand rule.,Calculate the magnetic flux of

25、the surface at an arbitrary time.,Find the induced emf according to Faraday law of electromagnetic induction .,Find out the induced electromotive force.,(4) MOTIONAL EMF,is the velocity of wire element .,The integral is over the circulation L. When L is closed, we have,The motional EMF could be calculated with the Faradays law of electromagnetic induction.,If,If,2020/6/29,EXAMPLE,A conducting bar of length l rotates with a constant angular speed about a pivot at one end. A uniform magnetic field is directed perpendicular to the plane of rotation as in figure. Find t