Docsity
Docsity

Prepare-se para as provas
Prepare-se para as provas

Estude fácil! Tem muito documento disponível na Docsity


Ganhe pontos para baixar
Ganhe pontos para baixar

Ganhe pontos ajudando outros esrudantes ou compre um plano Premium


Guias e Dicas
Guias e Dicas


Jackson solutions - jackson 1 7 homework solution, Provas de Física

Solução do jackson

Tipologia: Provas

2016

Compartilhado em 28/04/2016

iago_lira
iago_lira 🇧🇷

4

(7)

107 documentos

1 / 3

Toggle sidebar

Esta página não é visível na pré-visualização

Não perca as partes importantes!

bg1
Jackson 1.7 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
Two long, cylindrical conductors of radii a1 and a2 are parallel and separated by a distance d, which is
large compared with either radius. Show that the capacitance per unit length is given approximately by
C≈π ϵ0
(
ln d
a
)
1
where a is the geometrical mean of the two radii. Approximately what gauge wire (state diameter in
millimeters) would be necessary to make a two-wire transmission line with a capacitance of 1.2 x 10-11
F/m if the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm?
SOLUTION:
Place the center of cylinder 1 at the origin and the center of cylinder 2 at x = d. Place a total charge per
unit length -Q/L on cylinder 1 and +Q/L on cylinder 2. We first pretend cylinder 2 is not present and
solve by superposition. Without cylinder 2, the field lines from cylinder 1 are radially outwards. This
symmetry allows us to use Gauss's law in integral form. Draw a cylindrical Gaussian surface centered
on cylinder 1 and enclosing it.
E1
nda=qenc
ϵ0
E1=Q
2πr L ϵ0
r
The second cylinder by itself will create a similar field, only shifted to the right:
E2=Q
2π
xd
i
2Lϵ0
(xd
i)
So that the total field is:
E=Q
2π( x2+y2)Lϵ0
(x
i+y
j)+ Q
2π (( xd)2+y2)Lϵ0
(( xd)
i+y
j)
The potential is then:
Φ= Q
2π(x2+y2)Lϵ0
(x
i+y
j)− Q
2π (( xd)2+y2)Lϵ0
(( xd)
i+y
j)
pf3

Pré-visualização parcial do texto

Baixe Jackson solutions - jackson 1 7 homework solution e outras Provas em PDF para Física, somente na Docsity!

Jackson 1.7 Homework Problem Solution

Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: Two long, cylindrical conductors of radii a 1 and a 2 are parallel and separated by a distance d , which is large compared with either radius. Show that the capacitance per unit length is given approximately by C ≈π ϵ (^0) (ln d a ) − 1 where a is the geometrical mean of the two radii. Approximately what gauge wire (state diameter in millimeters) would be necessary to make a two-wire transmission line with a capacitance of 1.2 x 10- F/m if the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm? SOLUTION: Place the center of cylinder 1 at the origin and the center of cylinder 2 at x = d. Place a total charge per unit length - Q / L on cylinder 1 and + Q / L on cylinder 2. We first pretend cylinder 2 is not present and solve by superposition. Without cylinder 2, the field lines from cylinder 1 are radially outwards. This symmetry allows us to use Gauss's law in integral form. Draw a cylindrical Gaussian surface centered on cylinder 1 and enclosing it. ∮ E 1 ⋅ n^ da =^ qenc ϵ 0 E 1 =

− Q

2 π r L ϵ 0

r The second cylinder by itself will create a similar field, only shifted to the right: E 2 =

Q

2 π∣ x − d ̂ i ∣

2 L ϵ 0 ( xd ̂ i ) So that the total field is: E =

− Q

2 π( x 2

  • y 2 ) L ϵ 0 ( x ̂ i + y ̂ j )+

Q

2 π (( xd ) 2

  • y 2 ) L ϵ 0 (( xd ) ̂ i + y ̂ j ) The potential is then: ∇ Φ=

Q

2 π( x 2

  • y 2 ) L ϵ 0 ( x ̂ i + y ̂ j )−

Q

2 π (( xd ) 2

  • y 2 ) L ϵ 0 (( xd ) ̂ i + y ̂ j )

To calculate conductance, we don't need to know the potential everywhere, only the potential difference between the two conductors. Also, the conductor surfaces are equipotentials, so we only need to know the potential at one point on the surface to know it everywhere. Let us calculate the potential difference between the points ( x = a 1 , y = 0) and ( x = d - a 2 , y = 0). Dot the above equation with ̂ i. d Φ d x

Q

2 π( x 2

  • y 2 ) L ϵ 0 x

Q

2 π(( xd ) 2

  • y 2 ) L ϵ 0 ( xd ) Integrate between our two points: d Φ=

Q

2 π( x 2

  • y 2 ) L ϵ 0 x

Q

2 π(( xd ) 2

  • y 2 ) L ϵ 0 ( xd ) d x Δ V = (^) ∫ a 1 da 2 d Φ= (^) ∫ a 1 da 2

[

Q

2 π x L ϵ 0

Q

2 π( xd ) L ϵ 0 d x

]

Δ V =

Q

2 π ϵ 0 L [^

a 1 da 2 1 x d x − (^) ∫ a 1 da 2 1 xd d x

]

Δ V =

Q

2 π ϵ 0 L [

ln

da 2

a 1 )

−ln

a 2

a 1 − d )]

Δ V =

Q

2 π ϵ 0 L [

ln

da 1 a 2

da 2 a 1

)]

C L =

Q

Δ V

C = 2 π ϵ 0

[

ln

da 1 a 2

da 2 a 1

)]

− 1 For d >> a 1 , a 2 C ≈π ϵ 0

[

ln

d

√ a 1 a 2 )]

− 1

C ≈π ϵ 0 (ln

d

a )

− 1 where a =√ a 1 a 2 ad e −πϵ 0 C