大学物理期末复习.ppt
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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
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