Unformatted text preview: Electric Dipole Field ! Orient the dipole along the xaxis, with q at <s/2, 0> and q at <s/2,0>, defining s . ! We will define the dipole moment vector as p = qs, 0 . Note that this vector points from negative to positive charge, i.e., opposite to the direction of the electric field between the two charges (for reasons that become apparent when we consider the potential energy of a dipole in an external field). ! Now let us calculate the electric field at some distant point from the origin, r . We ! will denote the angle between r and the xaxis as ! . ! ! ! Let r1 denote the vector from charge q to the observation point r . Likewise, r2 ! denotes the vector from charge q to the observation point r . ! ! s ! ! s ! ! Then + r1 = r and ! + r2 = r . 2 2 It follows that ! kq kq ^ ^ E ( r ) = ! 2 r1 ! ! 2 r2 . r1 r2 We ! r2 ! r ! ! r1 ! p can rewrite the above expression as: ! ! ! ! ! E 1 1 = ! 3 (r ! s / 2) ! ! 3 ( r + s / 2 ) kq ! s ! s r! r+ 2 2 This expression is exact. Now we want to work towards an approximate expression ! that takes advantage of the statement that r " s . ! 2 !! s $ , s2 s 2 2) # r & = r + rs cos' ( r +1 cos' . * r " 2% 4
! ! ! ! ! r 3E r !s /2 r +s /2 = ! 3/2 3/2 kq # s & # s & %1 ! r cos" ( %1+ r cos" ( $ ' $ ' Now Taylor expand each denominator keeping only terms first order in p/r: ! # 3s & ! ! # 3s & ! ! r 3E = ( r ! s / 2 ) %1+ cos" ( ! ( r + s / 2 ) %1 ! cos" ( $ 2r ' $ 2r ' kq ! ! ! ! r 3E r ! ^ = 3s cos! " s = 3s # r " s kq r ! " ! ! " ^ 3( p ! r)r " p Result: E ( r )dipole = k r3
In this form, it is immediately apparent that the field along the dipole direction has a magnitude twice that perpendicular to the dipole direction (at same distance, r, from the dipole. ! Finally, note that if we measure the electric field strength at some location, r , we gain knowledge only of p, not the magnitude of q and s separately. ...
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 Spring '07
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 Vector Space, Charge, SEPTA Regional Rail, Electric charge, kq

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