Longitudinal line charge density and the transverse

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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: 3 2 3 3 2 m C 7 . 12 ) m 00 . 2 ( ) m 10 0 . 25 ( C 10 0 . 50 d d - - - = × × = = = π π ρ l r q q q v v , 1 3 m mC 25 m 00 . 2 C 10 0 . 50 - - = × = = l q ρ λ and 2 2 3 3 2 m C 4 . 25 ) m 10 0 . 25 ( C 10 0 . 50 - - - = × × = = = π π σ r q q a a . while the secondary densities are
8 3 2 1 2 3 1 3 2 m C 7 . 12 m m C 7 . 12 ) m 10 0 . 25 ( m C 10 0 . 25 - - - - - - = × × = = π π λ λ ρ ρ r a . and 3 1 2 2 m C 7 . 12 m m C 7 . 12 m 00 . 2 m C 4 . 25 - - - - = = l a σ . The two secondary densities are identical in value and can simply be found by noting that l = = a a σ ρ λ ρ 2.5.4 Density classification enhanced by transverse space multipliers If we introduce a concept of transverse space multiplier s i for each scalar charge array such that the full space is a product s i and the longitudinal length. Thus for the elemental arrays these are unity 1 d d 1 = l l s , [m m -1 m 0 ] (2.28a) length = l l d d d d t 2 a s [m 2 m -1 m 1 ] (2.28b) and surface area = a d d d d 3 l v s . [m 3 m -1 m 2 ] (2.28c) Then λ will have two secondary interpretations to match those for σ and for ρ . See table 2. Table 2: Electric scalar charge densities classified by transverse space multipliers Primary densities Secondary densities Full space 1 st order partial space 2 nd order partial space Longitudinal line Transverse size Transverse size Longitudinal line l d d q = λ λ λ λ λ = l l d d l l d 1 d d P λ λ = q λ λ λ = 1 λ = l d d P q a d d q = σ = l l d d d d t σ σ λ σ a l l d d d d σ σ λ = a l σ λ σ = l d d σ λ = l d d l v d d q = ρ = a d d d d ρ ρ λ ρ l v l d d d d ρ ρ σ = a a v ρ λ ρ = a d d ρ σ = l d d a Example 2 : Find the electric field and potential at a perpendicular distance ρ from the mid point of a straight line segment of length l 2 and line charge density λ (see figure 2.5). Solution : Let z be position along the line segment varying from l - = z to l + = z , with its centre as the origin O of coordinates. Then within the ρ z –plane, the line elemental electric scalar charge z q d d λ = is at point ( ) 0 , P z , and its electric field E r q d is at point ( ) ρ , 0 P , with the displacement R R ˆ of P from P having magnitude
9 ( ) 2 / 1 2 2 ρ + = z R and inclined at angle θ given by Figure 2.5: Electric field due to a line charge distribution R z = θ cos and R ρ θ = sin The elemental field E r d at P has two field components: ( ) ( ) ( ) 2 / 3 2 2 0 2 / 3 2 2 0 d ˆ 4 1 d ˆ 4 1 d sin ˆ cos ˆ d ˆ d ˆ d z z z z z E E E z + + + = + = + = ρ λρ πε ρ λ πε θ θ ρ ρ z ρ z ρ z E r as 2 0 d 4 1 d R z E λ πε = .

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