Added NotesDefinitionsPoint ChargesLine of ChargeA Harder ProblemSlide 6Completing the MathSurface ChargeThe GeometrySlide 10Final ResultLook at the “Field Lines”Kinds of continuously distributed chargesThe SphereSummarySlide 16Slide 17Slide 18Slide 19Slide 20Slide 21Added NotesAugust 31, 2005DefinitionsElectric Field = Force per unit charge.Types of charge distributionsPoint ChargesLines of ChargeAreas of ChargeVolumes of ChargeGeneral:unitjeschalljqk rE arg 2rPoint ChargesIn the Figure, the four particles are fixed in place and have charges q1 = q2 = +5e, q3 = +3e, and q4 = -12e. Distance d = 9.0 mm. What is the magnitude of the net electric field at point P due to the particles?Line of ChargeNeed =charge per unit lengthunitrdsk rE2dsdq= dsrA Harder ProblemA line of charge=charge/lengthsetupsetupdxLrxdEdEy2/02/3222/02/322222222)(2)(2)()cos()()cos(LxLxLLxxrdxkrExrdxrkExrrxrdxkE(standard integral)Completing the MathrkLrklELrLLrrkLExx224:line long VERY a oflimit In the4:nintegratio theDoing22221/r dependenceSurface ChargeNeed Surface Charge density = =charge per unit area.unitrdAk rE2dArrunitThe GeometryDefine surface charge density=charge/unit-areadq=dAdA=2rdr(z2+r2)1/2dq= x dA = 2 rdr(z2+r2)1/2 RzzrzrdrzkErzzrzdrrkrzdqkdE02/3222/122222222)cos((z2+r2)1/2Final Result0z2202E,R 12WhenRzzEzLook at the “Field Lines”Kinds of continuously distributed chargesLine of charge or sometimes = the charge per unit length.dq=ds (ds= differential of length along the line)Area = charge per unit areadq=dAdA = dxdy (rectangular coordinates)dA= 2rdr for elemental ring of chargeVolume=charge per unit volumedq=dVdV=dxdydz or 4r2dr or some other expressions we will look at later.The Spheredqrthk=drdq=dV= x surface area x thickness= x 4r2 x drSummary222,222)()()(rrdskrrdAkrrdVkrQkqGeneralrQkqrqQkunitjjjjjunitunitErFEErFErF(Note: I left off the unit vectors in the lastequation set, but be aware that they shouldbe there.)The figure shows two concentric rings, of radii R and R ' = 3.25R, that lie on the same plane. Point P lies on the central z axis, at distance D = 1.90R from the center of the rings. The smaller ring has uniformly distributed charge +Q. What must be the uniformly distributed charge on the larger ring if the net electric field at point P due to the two rings is to be zero?[-5.39] QThe figure shows a plastic ring of radius R = 46.0 cm. Two small charged beads are on the ring: Bead 1 of charge +2.00 µC is fixed in place at the left side; bead 2 of charge +5.60 µC can be moved along the ring. The two beads produce a net electric field of magnitude E at the center of the ring. At what positive and negative values of angle should bead 2 be positioned such that E = 2.00 105 N/C? (Measure the angle from the positive x axis taking counterclockwise to be positive.)In the figure, a thin glass rod forms a semicircle of radius r = 4.00 cm. Charge is uniformly distributed along the rod, with +q = 4.00 pC in the upper half and -q = -4.00 pC in the lower half. (a) What is the magnitude of the electric field at P, the center of the semicircle?[28.6] N/C(b) What is its direction?[-90]° (counterclockwise from the positive x
View Full Document