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1、General Revision of College PhysicsChapter 1. Electrostatic Field1. Coulomb Law in the Vacuum123021124rrqqf(1) Coulomb force(2) The principle of superpositionThe total electrical force on given charge is the vector sum of the electrical forces caused by the other charges, calculated as if each acted
2、 alone.Force Due to the System of Point Charges. (3) Find Electrostatic ForceForce Due to Continuous Charge DistributionsQ dVdsdldq 3004rrdqqFd 3004rrdqqF Qdqr+q0P(4). Coulomb Law in the Dielectric123r021124rrqqf2. Electric Field Intensity0qFE Magnitude: the electric field force on unit positive cha
3、rge Direction : direct the direction of force exerted on a positive test charge .Unit N/C 、V/m(2) The principle of superposition(1) Definition niinEEEEE121(3). Electric Field Lines Electric field lines are not real. Field lines are not material objects. They are used only as a pictorial representati
4、on to provide a qualitative description of the field.In 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 inten
5、sity is stronger, and the electric field lines are sparser in the place where the field intensity is weaker. istaticqSdE0s10svortexSdEField lines are never cross. The field at any point has a unique direction.3. GAUSSS LAWSeSdE iQ01 In vacuum,In dielectric,0QSdDS Electric flux of the Gauss surface i
6、s related with charges in Gauss surface and is not related with charges out of Gauss surface. The 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
7、, there must be not charge in the Gauss surface. If the electric flux of a Gauss surface equals zero, then field intensity at every point on the Gauss surface is zero. Gauss theorem is tenable only to the electrostatic field whose distribution is symmetrical in space.4 . Calculating the Electric Fie
8、ld Problem-solving 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 conti
9、nuous distribution of charge, Element Analysis MethodSolution 2. Applying Gausss law to symmetric charge distribution.Plane symmetrySpherical symmetryThe three symmetries:Cylindrical symmetryProblem-solving strategy:Select appropriate gaussian surface.Select appropriate coordinates, apply Gausss law
10、.Analysis the symmetry of the field intensity distribution . Example 1Example 2Example 3 , ,RR ,2 211 ErRErERrThere are two concentric charged spherical shells of radius R1 and R2 . Charge quantities distribute uniformly.R1， Q1R2， Q2Example 4Example 5Example 6The charge line density of an infinite u
11、niform charged cylindrical surface of radius R is ,find the electric field intensity. rEExample 71R2R5. Electric potential energy , Electric potential00aababawwwAqE dl“ ”“”rPl dEQWU0 Electric potential（simply the Potential）unit：V (volt)J/C 1 1VThe potential at a point equals the work required to bri
12、ng a unit positive charge from this point to the zero point of electric potential .Electric potential energy of q0 at a pointThe relationship between the electric potential energy and Electric potentialxExUyEyUzEzUIn general, the electric potential is a function of all three spatial coordinates. If
13、V is given in terms of rectangular coordinates, the electric field components Ex ,Ey, and Ez can be found from V(x,y,z) as the partial derivatives For example, if ,then223Vx yyyz6xExy In a certain region of space, the electric potential is zero everywhere along the x axis. From this we can conclude
14、that the x component of the electric field in this region is (a) zero (b) in the +x direction (c) 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.2023-
15、3-135（1 1）场强相等的区域，）场强相等的区域，电势处处相等？电势处处相等？（2 2）场强为零处，电）场强为零处，电势一定为零？势一定为零？（3 3）电势为零处，场）电势为零处，场强一定为零？强一定为零？（4 4）场强大处，电）场强大处，电势一定高？势一定高？QQEU势势0 0rrdEURaP0 Calculate Potential.Two Method:QrdqU040rUE dl Applying element analysis method Applying the definition of potentialExample 1Example 2ABOq1R2R3RQFind
16、 the electric field intensity and the electric potential. Example 3.6. Equipotential volumes and surfaces2、可有、可有EU计算电势的方法（计算电势的方法（2种）种）1、微元法、微元法QrdQU0 04 4 iiirQU0 04 4 0rrdEU7. SUMMARY计算场强的方法（计算场强的方法（3种）种）1、点电荷场的场强及叠加、点电荷场的场强及叠加原理原理iiirrQE3 30 04 4 QrdQrE3 30 04 4 2、定义法、定义法UE （分立）（分立）（连续）（连续）（分立）（分立）（连续）（连续）xExUEU典型电场的电势典型电场的电势典型电场的场强典型电场的场强均匀带电均匀带电球面球面0E304rrqE球面内球面内球面外球面外均匀带电无均匀带电无限长直线限长直线rE02均匀带电无均匀带电无限大平面限大平面02E均匀带均匀带电球面电球面rqU04RqU04均匀带电无均匀带电无限长直线限长直线02lnraU均匀带电无均匀带电无限大平面限大平面02dEdU方向垂直于直线方向
