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1、Lecture 7 Coulombs Law and Electric Field IntensityIntensityOutlineThe Experimental law of CoulombElectric field intensity Gausss Law and ApplicationsElectrostaticsIn the static electric field(electrostatic),a charge can be either concentrated at a point or distributed in some fashion.In any case,th

2、e charge is assumed to be at rest and constant in time.The electric fields due to the charge do not change with time.St.Elmos firelighting7.1 Coulombs Lawyxzo1rr1q2rr12Rr12Fr2q12211 22121230021()44q qaq q RFNRRvvv9120108.851036Where is permittivity in free spaceIn free space+2RF8001310 m/sc Though C

3、oulombs law is based on experimental evidence,it is in fact also a postulate.There are two stipulations:cmR981010=1691010validated for to a degree of accuracy of The charged bodies be very small in comparison with the distance of separation The force be inversely proportional to the square of the di

4、stance.Since 12040(,)(,)limtqRrrtqF x y zE x y zaqvvvvrrRrravvvvWhere is the unit vector from the source point to the field point .Rarv rvwe have zyP rv1q2qrvRvIf the charge is not located at the origin of a chosen coordinate system,then1q7.2 Electric field intensityFigure 2.1.2 Point charge not at

5、the origin.130()4(/)q rrrrEVmvvvvv1 22024()q qRRFq EaNvvThis is the mathematical form Coulombs law.xy 014qR When a point charge is placed in the field point P,the electric force experienced by due to is2q2q1q0Er7.2.1 Electric Field due to a System of Discrete Charges For the electric field created b

6、y a group of n discrete point charges q1,q2,qn,located at the principle of superposition(vector sum)applies:)/(41)(1)(03mVrEnkrrrrqkkk=vvvvvv,21nkrrrrrLrLrrExample 7.2.1 The electric field of an electric dipole.Figure 2.1.3 Electric field of a dipole.Evdvrv2/drvv+2/drvv+=33022224drdrdrdrqEvvvvvvvvvS

7、olutionThe first term:+=232/3232/3222/3323114222drrrdrrddrrdrdrdrvvvvvvvvvvvv)(rd Rr 0The total charge enclosed is zero.Thusb)The total charge enclosed is sRQ24=204RQr E204RQEarvv240Rr E0RREa Evv204rQE=On the bottom face,Example 2.2.3 Determine the electric field intensity of an infinite planar char

8、ge with a uniform surface charge densitySolution We choose as the Gaussian surface a rectangular box with top and bottom faces of an arbitrary area A equidistant from the charge.If the charged sheet coincides with the xy-plane,then on the top face,Figure 2.2.3Applying Gauss law to an dsEdszEzsdEzz=v

9、v.sQA=02,szAE A=00,02,02szszzEzzzEzz=ESince there is no contribution from the side faces,we haveThe total charge enclosed in the box is Therefore,from which we obtainFigure 2.2.3Applying Gauss law to an infinite planar charge.dsEdszEzsdEzz=)()(vvAEdsEsdEzAzS22=vvEorirExample 2.2.4 Determine the E fi

10、eld caused by a spherical cloud of electrons with a volume charge density for and Solution First we recognize that the given source condition has spherical symmetry.The Gaussian surfaces must be concentric spherical surfaces.a)A hypothetical spherical Gaussian surface Siwith r br 0dsdEEvvrErThe tota

11、l outward E flux isThe total charge enclosed within the Gaussian surface isdsrsdErEr,=vrArSErdsEsdE24=vv.34300rdvdvQVVv=Figure 2.2.4Applying Gauss law to a spherical electron cloud.We construct a spherical Gaussian surface Sowith rb outside the electron cloud.The surface integral of the E flux is an

12、d the total charge enclosed is34Qb=rEr24Therefore the E field inside the electron cloud isbrrrE=0,300vConsequently,which follows the inverse square law.We observe that outside the charged cloud the E field is the same as though the total charge is concentrated on a single point charge at the center.This is true,in general,for a spherically symmetrical charged region even though is a function of r.0.3Qb=vbrrbrE=,32030vHomeworkP-7.1P-7.2P-7.3