Hw assignment IV: Due Nov 6  12 points 

Note: Submit solutions as a Mathematica notebook (no handwritten work will be accepted !). No late submissions.

Consider the ring with total charge Q=20 C and radius a=0.5 meters, as shown in Fig. 3.11 in the text. The outer conductor is grounded and has a radius that is twice the radius of the ring.

(a)  Calculate the value of the electric field (to four figure accuracy) at the point z=0.75 meters.
(b)  Calculate the value of the electric field (to four figure accuracy) at the point x=2.0 meters.

Using Eq.(3.131) in the text we get

Φ=(Q/(4 π (Subscript[ϵ, 0]))) (∑_(n=0)^∞)  (((((-1))^n)(2n-1)!!)/((2^n) n!)) (a^(2n)) ((1/(z^(2n+1)))-((z^(2n))/(((2a))^(4n+1)))) (Subscript[P, 2n])(0)

where 2a > z > a  is a point on the z-axis.   Using SI units for z we get

the term under the sum is a dimensionless number.

term[z_,n_]= (-1)^(n) (2 n -1)!!/2^n/n!  a^(2n)(1/z^(2n+1) - z^(2n)/(2 a)^(4n+1) )LegendreP[2n,1] /. a->1/2

ξ[z_,max_]:=Sum[term[z,n],{n,0,max}]

Table[ξ[0.75,i],{i,1,25}]

We need to calculate electric field, we know from symmetry that it will be along the z-direction, so all we need
to do is take the derivative of ξ, with respect to z. However we need four figure accuracy so we need to check
convergence properties of this series.

dterm[z_,n_]=D[term[z,n],z]

dξ[z_,max_]:=-Sum[dterm[z,n],{n,0,max}]

dξ[0.75,25]

(Q/(4 π (Subscript[ϵ, 0]) (1 m)^2)) /. {Q->20 C,(Subscript[ϵ, 0])->8.854187817×10^(-12) C/V/m}

Efield= % dξ[0.75,25]

(b) Since there is no charge outside the grounded sphere (V=0) the potential must vanish everywhere outside it, therefore
the elcectric field vanishes (i.e. the grounded sphere shields the charge inside it ).