大学物理双语版奥本汉姆课件ChaptoEfied.pptVIP

* Karl Friedrich Gauss (1777—1835), German mathematician and physicist. He made a lot of contributions in the fields of experimental physics, theoretical physics and mathematics. He made major contributions to the theory of electromagnetism. 3. Gauss’ law (高斯定理) (P489) To find out the relation between the electric field and its source, consider a very simple case of a positive point charge q at the center of a spherical shell: The same result for any shaped-closed surface A2 due to the continuity of electric field line. +q If q is outside of the area, inward=outward, ?=0 —— Gauss’law for vacuum. The relation of the source of electric filed and the field 反映场和源 的关系. The flux of the electric field through a closed surface of any shape equals 1/?0 times of the algebraic sum of charges enclosed within the surface. 真空中静电场内通过任意闭合曲面的电通量等于该曲面所包围的电量的代数和的1/?0倍. (a) Where is the algebraic sum of all interior charges enclosed in the Gaussian surface. ? is not related to the way of distribution and outside charges. (b) The quantity ?on the left side of above equation is the electric field resulting from all charges, both those inside and those outside the Gaussian surface. Descriptions: 4. Applying Gauss’ law——calculating (P491) (d) It gives a simple way to calculate the distribution of electric field for a given charge distribution with sufficient symmetry. (c) Gauss’s Law and Coulomb’s law are equivalent (see P496). However, Gauss’s Law is hold for the produced by moving charges; Coulomb’s law is only true for electrostatic field. How do we choose the surface for calculating the electric flux or ? A uniformly charged spherical volume （带电球体）, radius R and total charge Q. Find the at any point of inside and outside region. Example 20-4 (P492): R ? Question R Solution: To calculate E outside of the sphere, we choose for Gauss’ surface (G. S.) as an imaginary spherical shell A1 with r R: using Gauss’law To calculate E inside of the sphere, we