Chap19to21-Efie1d.ppt

+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 to 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 give 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 choose for G.S. as an imaginary spherical shell with A2 r R: R O In the similar way, we can get of spherical shell （带电球壳的电场） P491 . r O A section of an infinitely long cylindrical plastic rod with a uniform +?. Let us find an expression for the magnitude of the ?? at a distance r from the axis of the rod. Our Gaussian surface should match the symmetry of the problem ——axis symmetry 轴对称性 or cylindrical symmetry. Solution: 场强沿垂直轴线的方向；距中心同远处场强相同 Example 20-5 P493 : line of charge The direction of is radially outward from the line of charge