longitudinal line charge density and the transverse sectional surface density, as well as the transverse sectional density of the latter two primary densities. Solution:Apply the definitions given above. Note that here the transverse section is a circular surface and the full space is a cylinder. Thus: 32332mC7.12)m00.2()m100.25(C100.50dd---=××==≡=ππρlrqqqvv, 13mmC25m00.2C100.50--=×==lqρλand 22332mC4.25)m100.25(C100.50---⊥=××===⊥ππσrqqaa. while the secondary densities are
832123132mC7.12mmC7.12)m100.25(mC100.25------⊥≡=××==ππλλρρra. and 3122mC7.12mmC7.12m00.2mC4.25----≡==⊥laσ. The two secondary densities are identical in value and can simply be found by noting that l⊥==⊥aaσρλρ2.5.4 Density classification enhanced by transverse space multipliersIf we introduce a concept of transverse space multiplier sifor each scalar charge array such that the full space is a product siand the longitudinal length. Thus for the elemental arrays these are unity 1dd1=≡lls, [m m-1≡m0] (2.28a) length ⊥=≡llddddt2as[m2m-1≡m1] (2.28b) and surface area ⊥=≡adddd3lvs. [m3m-1≡m2] (2.28c) Then λwill have two secondary interpretations to match those for σand for ρ. See table 2. Table 2: Electric scalar charge densities classified by transverse space multipliers Primary densities Secondary densities Full space 1storder partial space 2ndorder partial space Longitudinal lineTransverse size Transverse size Longitudinal linelddq=λλλλλ=≡llddlld1ddPλλ=≡⊥qλλλ=1λ=⊥lddPqaddq=σ⊥=≡llddddtσσλσallddddσσλ=≡⊥⊥alσλσ=⊥lddσλ=⊥lddlvddq=ρ⊥=≡addddρρλρlvlddddρρσ=≡⊥⊥aavρλρ=⊥addρσ=⊥lddaExample 2: Find the electric field and potential at a perpendicular distance ρfrom the mid point of a straight line segment of length l2and line charge density λ(see figure 2.5). Solution: Let zbe position along the line segment varying from l-=zto l+=z, with its centre as the origin O of coordinates. Then within the ρz–plane, the line elemental electric scalar charge zqddλ=is at point ()0,Pz′, and its electric field Erqdis at point ()ρ,0P, with the displacement RRˆof Pfrom P′having magnitude
9()2/122ρ+=zRand inclined at angle θgiven by Figure 2.5:Electric field due to a line charge distribution Rz=θcosand Rρθ=sinThe elemental field Erdat P has two field components: ()()()2/32202/3220dˆ41dˆ41dsinˆcosˆdˆdˆdzzzzzEEEz+++=+=+=ρλρπερλπεθθρρzρzρzEras 20d41dRzEλπε=.