Hw assignment IV: Due October 30.  12 points 

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

A circular ring, that has electric charge uniformly distributed on it, of radius a is located at a distance b from the xy plane. The z coordinate axis forms the axis of symmetry for the ring of charge. See Fig. (3.4) in the text. Consider now an additional ring with the same charge located at z=-b.

(a)  Find an expression for the potential Φ at any point along the z-axis. Find the E field on that axis.

On the z axis the potential due to the first ring on the z-axis is

(Subscript[ϕ, 1]) =(1/(4π (Subscript[ϵ, 0]))) (Q/(Sqrt[(a^2)+(((z-b))^2)])) =(Q/(4π (Subscript[ϵ, 0]))) (∑_(L=0)) ((Subscript[r, <]^L)/(Subscript[r, >]^(L+1)))(Subscript[P, L])(Cosθ) (Subscript[P, L])(Cosα)

where (Subscript[r, <]), (Subscript[r, >]) are less, greater than for r,c, and Tanα=a/b and θ=0 is the polar angle on the z-axis 


Like wise for the ring located at z=-b

(Subscript[ϕ, 2]) =(1/(4π (Subscript[ϵ, 0]))) (Q/(Sqrt[(a^2)+(((z+b))^2)])) = (1/(4π (Subscript[ϵ, 0]))) (Q/(Sqrt[(a^2)+(b^2)+(z^2)+2 z b ]))=(1/(4π (Subscript[ϵ, 0]))) (Q/(Sqrt[(c^2)+(z^2)+2 c z Cosα ]))=
(1/(4π (Subscript[ϵ, 0]))) (Q/(Sqrt[(c^2)+(z^2)-2 c z  Cosβ ]))    where β=α+π .

and so ϕ=(Subscript[ϕ, 1])+(Subscript[ϕ, 2])
The E field is simply the minus the gradient along the z-direction (we know the other components of E vanish due to symmetry

FullSimplify[Grad[(1/(4π (Subscript[ϵ, 0]))) (Q/(Sqrt[(a^2)+(((z-b))^2)]))+(1/(4π (Subscript[ϵ, 0]))) (Q/(Sqrt[(a^2)+(((z+b))^2)])),{z}]]

(b) Suppose b=a=0.5 meters, Q, the total charge of each ring, is 100 Coulomb). Find the potential
at the point x=1.0 meters on the x-axis. Give your answer  to four significant figures.

Using the expressions,for arbitrary θ,for(Subscript[ϕ, 1])and(Subscript[ϕ, 2])we get

ϕ=(Q/(4π (Subscript[ϵ, 0])))  (∑_(L=0)) ((Subscript[r, <]^L)/(Subscript[r, >]^(L+1)))(Subscript[P, L])(Cosθ)  ((Subscript[P, L])(Cosα) + (Subscript[P, L])(Cosβ)  ) 

but  ((Subscript[P, L])(Cosα) + (Subscript[P, L])(Cosβ)  ) = (Subscript[P, L])(Cosα) + (Subscript[P, L])(-Cosα)   = 2(Subscript[P, L])(Cosα) for even L only

and so

ϕ=(Q/(2π (Subscript[ϵ, 0])))  (∑_(L=0,2..)) ((Subscript[r, <]^L)/(Subscript[r, >]^(L+1)))(Subscript[P, L])(Cosθ)  (Subscript[P, L])(Cosα)   (sum only over even L)

Since c= (Sqrt[2])a= (Sqrt[2])/2<r we get θ=π/2 for the x-axis

ϕ=(Q/(2π (Subscript[ϵ, 0])))  (∑_(L=0,2..)) ((c^L)/(r^(L+1)))(Subscript[P, L])(0)  (Subscript[P, L])(b/c)   = (Q/(2π (Subscript[ϵ, 0])r)) (∑_(L=0,2..)) ((c^L)/(r^L))(Subscript[P, L])(0)  (Subscript[P, L])(1/(Sqrt[2]))

So we need to calculate the sum (∑_(L=0,2..)) ((((Sqrt[2])/2) )^L)(Subscript[P, L])(0)  (Subscript[P, L])(1/(Sqrt[2]))      (since c/r=(Sqrt[2])/2)  to four significant
figures

sum[max_]:=Sum[(Sqrt[2]/2)^L LegendreP[L,0] LegendreP[L,1/Sqrt[2]],{L,0,max,2}]

Table[N[sum[i]],{i,0,20,2}]

So we get  (Q/(2r π (Subscript[ϵ, 0]))) 0.9088  where r = 1 meter

0.9088  Q/(2 π r ϵ0) /. {Q->100 C, r->1 m,ϵ0-> 8.8541878 10^(-12) C/V/m}

(c) Repeat (b) for the case x=0.25 meters.

In this case c > r and so we need the expansion

ϕ=(Q/(2π (Subscript[ϵ, 0])c))  (∑_(L=0,2..)) ((r^L)/(c^L))(Subscript[P, L])(0)  (Subscript[P, L])(b/c)     and the ratio  r/c -> 1/4/(Sqrt[2])/2 = (1/(2 (Sqrt[2])))

sum[max_]:=Sum[(1/2/Sqrt[2])^L LegendreP[L,0] LegendreP[L,1/Sqrt[2]],{L,0,max,2}]

Table[N[sum[i]],{i,0,20,2}]

and so we have

ϕ =(Q/(2π (Subscript[ϵ, 0])c))0.982104   c=(Sqrt[2])/2

0.982104  Q/(2 π c ϵ0) /. {Q->100 C, c->1/(Sqrt[2]) m,ϵ0-> 8.8541878 10^(-12) C/V/m}