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Field of a Finite Line Charge Kenneth H. Carpenter September 3, 2008 The electric field intensity due to a constant line charge density, ρl , along a straight line segment of length L can be used as a building block for constructing electric field intensity vectors for more complicated charge distributions. This electric field intensity will be derived in a coordinate independent manner in the following. The geometry of the problem ^u_| u^ || α1 ρ l L _ E h α2 β l dl Note the following definitions: • The line segment has total length L. • The point where the electric field intensity E is evaluated is located a perpendicular distance h from the line segment having constant line charge density ρl . • The angles between the perpendicular from the point of E and lines drawn to the ends of the line segment of ρl are α1 and α2 . • A distance l is measured from the point of intersection of the perpendicular from the point of E to the line of ρl , positive when towards the end where α2 is the angle subtended. • The angle β is measured from the perpendicular from the point of E to the line charge, to the line from the point of E to the location of l along the line charge. Thus the relationship, l = h tan β, holds. • Unit vectors ûk and û⊥ are defined as parallel to the line charge and perpendicular to it, respectively. These serve as basis vectors for expressing E. 1 EECE557 Field of a Finite Line Charge - Fall, 2008 2 Calculation of E The incremental charge, dq, contained in incremental length of the line, dl, is dq = ρl dl. In turn, we have dl = h sec2 βdβ. Thus the incremental contribution to E from dq may be written as dE = 1 û⊥ h − ûk h tan β ρl h sec2 βdβ. 3 4π0 [h sec β] Simplifying eq.(1) and then integrating on β gives the desired result for E: Z α2 ρl E = (û⊥ cos β − ûk sin β)dβ 4π0 h −α1 ρl [û⊥ (sin α1 + sin α2 ) + ûk (− cos α1 + cos α2 )]. = 4π0 h (1) (2) Special cases Infinite line charge Suppose L >> h. Then both α1 and α2 approach 90o so that E= ρl û⊥ . 2π0 h (3) This gives a way to judge when a line segment is long enough to be considered infinitely long: when the α angles are close enough to right angles to make the cosine of the angle zero to the desired precision. Half-infinite line charge Another case of interest is when the point of observation of the field is at one end of an otherwise very long line charge. Suppose the end is the one to which α2 is measured. Then α2 = 0 and α1 = 90o . The result for E is ρl E= (û⊥ + ûk ). (4) 4π0 h