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UCF PHY 2049C - Week 2 Added Notes - Electric Field

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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, 2005DefinitionsElectric Field = Force per unit charge.Types of charge distributionsPoint ChargesLines of ChargeAreas of ChargeVolumes of ChargeGeneral: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 ChargeNeed  =charge per unit lengthunitrdsk rE2dsdq= dsrA Harder ProblemA line of charge=charge/lengthsetupsetupdxLrxdEdEy2/02/3222/02/322222222)(2)(2)()cos()()cos(LxLxLLxxrdxkrExrdxrkExrrxrdxkE(standard integral)Completing the MathrkLrklELrLLrrkLExx224:line long VERY a oflimit In the4:nintegratio theDoing22221/r dependenceSurface ChargeNeed Surface Charge density = =charge per unit area.unitrdAk rE2dArrunitThe GeometryDefine surface charge density=charge/unit-areadq=dAdA=2rdr(z2+r2)1/2dq= x dA = 2 rdr(z2+r2)1/2     RzzrzrdrzkErzzrzdrrkrzdqkdE02/3222/122222222)cos((z2+r2)1/2Final Result0z2202E,R 12WhenRzzEzLook at the “Field Lines”Kinds of continuously distributed chargesLine of charge or sometimes  = the charge per unit length.dq=ds (ds= differential of length along the line)Area = charge per unit areadq=dAdA = dxdy (rectangular coordinates)dA= 2rdr for elemental ring of chargeVolume=charge per unit volumedq=dVdV=dxdydz or 4r2dr or some other expressions we will look at later.The Spheredqrthk=drdq=dV= x surface area x thickness= x 4r2 x drSummary222,222)()()(rrdskrrdAkrrdVkrQkqGeneralrQkqrqQkunitjjjjjunitunitErFEErFErF(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


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UCF PHY 2049C - Week 2 Added Notes - Electric Field

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