The Electric Field 



The electric field is given by E = -VV. In spherical coordinates, when V is independent of f , 

 this becomes 



where N = T^jTj" is the unit vector in the radial direction, and P is the unit vector perpendicular to N 

 in the direction of increasing 9. The partial derivatives are 



^ . 2aVo ( p , 3r! p 5rl p 



||^=2adVo(f2P,'+-^Pi+f>i+ ) 



<^Va.2aVo(^p^ r^ p,^ r^ p,^ ) 



d0 " d2 ^ ' "^1^ d2 5 d-* 5 



where Pi', P3', P5', ... are the derivatives of Pj, P3, P5, ... with respect to 6 (Jahnke and 

 Emde 1945). Thus, the electric field in the two regions is given by 



-P[2adVo(-j5P,'+-^P3+-pP;+ )] volts/cm., r>d (13) 



E2{r,e) = -N[^(P, +^P3+^P5+ )] 



-P[^ (p;+-l5p^+-L^P^+ )] volts/cm., r<d . (U) 



The Current Densit 



1. 



We have seen that the current density J = crE, where a is the conductivity of the nnedium. Thus 

 the equations 



Jj = (tE J annperes /cm. ^ 



J2 = '^^2. ''"ipsres/cm. ^ (15) 



will give the current density at any point in the medium, with Ej and E^ defined by (13) and (14). 



The current density on the perpendicular bisector plane, or the XY-plane, is found by setting 

 6 = T . At this angle the Legendre Polynomials have the values 



Pj =P3 = P5 = .... =0 



Pj' = -1, P3' = 3/2, Ps' = -15/8, P7' = 35/16 



Thus, the current density becomes 



J, = P[2adc5Vo(f3-|f' + |f'-^%•■■)]amp/cm^ r>d (16) 



^2- PL-jl — ( I -"2d2''""8d^ "Ted^"*" ^J amp/cm., r<d (iv) 



for points on the XY-plane. We notice that the current is everywhere in the direction of P, which is 

 a unit vector in the -Z direction when 9 = y . 



The total current is found by integrating this current density over the entire XY-plane. 



1 = if JdA 



19 



