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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 cas
2、e, the 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 221 21230021()44q qaq q RFNRRvvv9 120108.851036 Where is permittivity in free spaceIn free space+2RF8001310 m/
3、sc Though Coulombs law is based on experimental evidence, it is in fact also a postulate. There are two stipulations:cmR981010=1691010Rr 0The total charge enclosed is zero. Thusb) The total charge enclosed is sRQ24=204RQr E204RQEarvv240Rr E0RREa Evv2 04rQE=On the bottom face,Example 2.2.3 Determine
4、the electric field intensity of an infinite planar charge 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
5、 face,Figure 2.2.3Applying Gauss law to an dsEdszEzsdEzz=vv.sQA= 02,s zAE A =00, 02, 02s zs zzEzzzEzz = = br 0dsdEEvv rErThe total outward E flux isThe total charge enclosed within the Gaussian surface isdsrsdErEr,=vrArSErdsEsdE24=vv.343 00rdvdvQ VVv=Figure 2.2.4Applying Gauss law to a spherical ele
6、ctron cloud.We construct a spherical Gaussian surface Sowith rb outside the electron cloud. The surface integral of the E flux is and the total charge enclosed is34Qb= rEr24Therefore the E field inside the electron cloud isbrrrE=0,300 vConsequently,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=,32 03 0 vHomeworkP- 7.1P- 7.2P- 7.3
