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2π
2π
2 2
F → e = senθcos ϕ e F → e = senθsenϕ⇒ cos ϕcos ϕ dϕ= cos ϕ sen ϕ d ϕ= 0 .
x rx y ry
∫
0 0
Portanto, só haverá força em z, como já sabíamos pela simetria do problema.
∫
2π
Continuando o cálculo,
{{ }
* * * π
n′+ 1 n+ 1 2iρ
∫ ( EE θ θ EE ϕ ϕ) d ∑ EE n n′ 2
( i) ( i) e [ n′ n′
][ an n bn n an n bn n]
0
+ ϕ= − − − π −τ τ + π − π − τ −
2ρ
n
{() + 1 n′
() + 1 − 2 i ρ
* * * *
i i e [ n n]
⎡an ′ n′ bn′ n′ an′ n′ bn′ n′
⎤}
π −τ
⎣
τ + π − π − τ
⎦
−
n
{( ) + 1 n′
( ) + 1
i i [ n′ n′
][ an n bn n an n bn n]
}
− π +τ π + τ + τ + π −
n
() + 1 n′
( ) + 1 * * * *
{ i −i [ π n +τn]
⎡an ′ π n′ + bn′ τ n′ + an′
τn′ bn′ n′
⎤}
⎣
+ π
⎦
+
n
{ ( ) + 1 n′
( ) + 1 * * * *
2 −i i {[ anτ n + bnπn]
⎡
′ τ ′ + ′ π ′
⎤} + { ⎡
′ π ′ + ′ τ ′
⎤
⎣
an n bn n ⎦ ⎣
an n bn n ⎦[ anπ n+ bnτn]
}}
Rearranjando os termos de
EE
+ EE e chamando o último de I
1
, temos
* *
θ θ ϕ ϕ
2 π n
( ) + 1 n′
( ) + 1 * * *
I1 = ∫ dθsenθcos
θ r ∑ −i i E ′ { ′[ τ τ ′ +π π ′] + ′[ π τ ′ +τ π ′]
+
2
nEn bnbn n n n n anbn n n n n
ρ
[ ] aa [ ]}
* *
n n′ n n′ n n′ n n′ n n′ n n′
ba
τ π +π τ + π π +τ τ =
2 π
* * * *
= ∫ dθsen θcos θr ∑ − i i E E b b + a a τ τ +π π + 2a b π τ +τ π
2
ρ
⎣ ⎦
n+ 1 n′
+ 1
( ) ( ) n n′ { ⎡
n n′ n n′ ⎤[ n n′ n n′ ] n n′ [ n n′ n n′
]}
Já os terceiro e quarto termos ficam como I
2
,
π
2 π *
1 1
2 ∫ sen cos ∑ 2 n n′ { n′ n′
n n n n n n n n
2ρ 0
π
2 π
* n ′
{ ( ) + 1 n
= θ θ θ ( ) + 1
∫ d r sen cos ∑ E ′ − − 2 Re[
π π ′ + τ π ′ + π π
2 nEn i i an n n an n n bn n n′
2ρ
0
n+ n′
+
( ) ( ) 2Re[ ][ ]}
I = dθ r θ θ E E − −i i π +τ a π + b τ + a τ + b π =
π
+ b τπ + a πτ + a ττ + b πτ + b ττ ] =
n n n′ n n n′ n n n′ n n n′ n n n′
{
π
= ∫ dθr θ θ ∑ E E − −i i ⎣a π π +τ τ + a τ π +π τ
2ρ
0
n
( ) + 1 n′
( ) + 1
2 Re ⎡ [ ] [ ]
2 *
sen cos
2 n n′ n n n′ n n′ n n n′ n n′
[ ′ ′] b [ ′ ′] ⎤}
+ bn πnπ n +τnτ n + n τnπ n +πnτ n ⎦ =
233
}