+2 Pay=c -L/2 + L/2 Note: Figure not drawn to scale. 1. A very long, thin, nonconducting cylinder of length L is centered on the y-axis, as shown above. The cylinder has a uniform linear charge density +2.. oint. is located on the y-axis at y = c, where L >> c. A) i. On the figure shown below, draw an arrow to indicate the direction of the electric field at point F due to the long cylinder. The arrow should start on and point away from the dot. P ii. escribe the shape and location of a Gaussian surface that can be used to determine the electric field at point P due to the long cylinder. iii. Use your Gaussian surface to derive an expression for the magnitude of the electric field at point P. E press your answer in terms of 2, c, L, and physical constants, as appropriate.

icon
Related questions
Question
100%
v=x0+at
x = x + x² +
²= ² + 2q₂ (---)
d=
F =
m
P =
J = √F 1 = Ap
P = mv
FS HEN
AE=W-F-dr
9/6
P = F.j
AU-mg h
2₁ ===
:=FxP
=
1-4-4
I=fr dm = Emr²
Σm.x,
V = r@
Far
m
K=
=fxj = Ic
ww
= 0 + at
6-8₂+₂ +²²
9 = 0
Factor
10⁹
Universal gas constant,
Boltzmann's constant,
UNIT
SYMBOLS
1
1.9
10-12
MECHANICS
PREFIXES
10
10³
1-2 centi
milli
micro
+++
nano
pico
ADVANCED PLACEMENT PHYSICS C EQUATIONS
a acceleration
E- ene y
F = _orce
с
J
Ft =.cght
1
-.cy enc
m
n
Jm, ulse
Prefix Symbol
giga
G
mega
M
kilo
k
Р
k
l
t = len ul
Langu. ir...mentu...
m = mass
P=₁0: er
P mo...ent m
i=1
otatio ali etia
- kinetic ee y
spring constant
W
X
1 = li...e
J=, otential ener ty
clas
= elocity or s, cea
Y
agus dista ce
-evioa
30
a
Net ron ... ss, m, = 1.67 x 10
kg
Electron mass, m = 9.11 10³ kg
A og du's umbe, N = 5.02 x 1023 mol-1
e = angle
torque
work done on a > stem
osition
= coeficient of friction
a gulur speed
a gulur acceleration
pase a gle
F=-KAX
U. - -/-k(4x)²
3= max COS(col + 0)
cos(at
o)
1 - 1²/13 - 7
T =
T₁ = 2√
T₂ =2=√1/2
|FG|- Gym
Ug =
R= 8.31 J/(mol-K)
kg = 1.38 x 10-23J/K
1 unified atomic mass unit,
Planck's constant,
Gm,m₂
r
m
meter,
kilogram, kg
second, S
ampere,
kelvin,
kg
mole, mol
hertz, Hz
newton, N
A pascal,
Pa
K joule, J
8-2
.
E-di-
Aneo
E₂ = d
V =
.
0°
0
1
0
ΔΙ' = -
19192
ΔΙ
X
-SE-dir
15%
Are
v-2
ELECTRICITY AND MAGNETISM
A = area
B magnetic field
C = capacitance
U₂ =q'-
لم = =
C=xe04
C₂-C₁
Ξ-ΣΕ
1-de
=
ΔΙ'
1 = R
R-ER
Uc-AV-C(AV)²
R-2
1 = Nev A
P = IAV
ADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATION
LANTS AND CONVER. FORS
27
Froton mass, m, 1.67 x 10
1 9192
Απεργ
Acceleratio i due to gravity
at Earth ssu..ace,
Vacuum permeability.
Ho = 4 x 10 (T-m)/A
Magnetic constant, k = 1/(4x)=1x10-7 (T-m)/A
1 atmosphere ressure,
distance
wall,
coulomb,
volt,
ohm,
henry,
E
E
✔
electric field
emf
-force
W
C
V
Ω
H
1 = current
J= current density
L = inductance
<= length
n = number of loops of wire
per unit length
N-number of charge carriers
per unit volume
P = power
Q = charge
= point c arge
1 u = 1.66 x 10-27 kg = 931 MeV/c²
h = 6.63 x 10-34 J-s = 4.14 x 10-15 eV-s
h: 1.99 x 10-25 J-m = 1.24 x 10³ eV nm
8 = 8.85 x 10-12 C²/(N-m²)
Coulomb's law constant, k = 1/(4m) = 9.0 x 10° (Nm²)/c²
Vac ium permittiv t,,
R- resistance
r-radius or distance
t = time
J = potential or stored energy
V = electric potential
v = velocity or speed
p = resistivity
- flux
x= dielectric constant
F =qvxB
b-dl= Hol
Idix?
= He 1 dbxr
dB
F = [I dè x B
B = onl
Electron charge na gn..ude,
.eron ol, 1
V-1.60 x 10
J
Sd of light,
c = ..00 x 10 m/s
Universal gravitational G=7x10-¹1 (N-m²/kg²
constant,
8 = 9.8 m/s²
8=E-de=
8=
2=-L
U₂ - 2012
150x1¹9
1 atm = 1.0 x 105 N/m² = 1.0 x10³ Pa
farad,
tesla.
dog.
dt
degree Celsius,
electron volt,
F
Т
VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES
60"
90°
3,⁰
1/2
√3/2 4/5 √2/2
.7
3/5 √2/2
0
aine
cose
tane
1
4/5 √3/2
3/5 1/2
4/3 √3
0
√3/3
3/4
1
point charge.
IV. All batteries and meters are ideal unless otherwise stated.
C
RUST
ev
The following assumptions are used in this exam.
T The frame of reference of any problem is inertial unless otherwise
stated.
8
II. The direction of current is the direction in which positive charges
would drift.
III. The electric potential is zero at an infinite distance from an isolated
V. Edge effects for the electric field of a parallel plate capacitor are
negligible unless otherwise stated.
Transcribed Image Text:v=x0+at x = x + x² + ²= ² + 2q₂ (---) d= F = m P = J = √F 1 = Ap P = mv FS HEN AE=W-F-dr 9/6 P = F.j AU-mg h 2₁ === :=FxP = 1-4-4 I=fr dm = Emr² Σm.x, V = r@ Far m K= =fxj = Ic ww = 0 + at 6-8₂+₂ +²² 9 = 0 Factor 10⁹ Universal gas constant, Boltzmann's constant, UNIT SYMBOLS 1 1.9 10-12 MECHANICS PREFIXES 10 10³ 1-2 centi milli micro +++ nano pico ADVANCED PLACEMENT PHYSICS C EQUATIONS a acceleration E- ene y F = _orce с J Ft =.cght 1 -.cy enc m n Jm, ulse Prefix Symbol giga G mega M kilo k Р k l t = len ul Langu. ir...mentu... m = mass P=₁0: er P mo...ent m i=1 otatio ali etia - kinetic ee y spring constant W X 1 = li...e J=, otential ener ty clas = elocity or s, cea Y agus dista ce -evioa 30 a Net ron ... ss, m, = 1.67 x 10 kg Electron mass, m = 9.11 10³ kg A og du's umbe, N = 5.02 x 1023 mol-1 e = angle torque work done on a > stem osition = coeficient of friction a gulur speed a gulur acceleration pase a gle F=-KAX U. - -/-k(4x)² 3= max COS(col + 0) cos(at o) 1 - 1²/13 - 7 T = T₁ = 2√ T₂ =2=√1/2 |FG|- Gym Ug = R= 8.31 J/(mol-K) kg = 1.38 x 10-23J/K 1 unified atomic mass unit, Planck's constant, Gm,m₂ r m meter, kilogram, kg second, S ampere, kelvin, kg mole, mol hertz, Hz newton, N A pascal, Pa K joule, J 8-2 . E-di- Aneo E₂ = d V = . 0° 0 1 0 ΔΙ' = - 19192 ΔΙ X -SE-dir 15% Are v-2 ELECTRICITY AND MAGNETISM A = area B magnetic field C = capacitance U₂ =q'- لم = = C=xe04 C₂-C₁ Ξ-ΣΕ 1-de = ΔΙ' 1 = R R-ER Uc-AV-C(AV)² R-2 1 = Nev A P = IAV ADVANCED PLACEMENT PHYSICS C TABLE OF INFORMATION LANTS AND CONVER. FORS 27 Froton mass, m, 1.67 x 10 1 9192 Απεργ Acceleratio i due to gravity at Earth ssu..ace, Vacuum permeability. Ho = 4 x 10 (T-m)/A Magnetic constant, k = 1/(4x)=1x10-7 (T-m)/A 1 atmosphere ressure, distance wall, coulomb, volt, ohm, henry, E E ✔ electric field emf -force W C V Ω H 1 = current J= current density L = inductance <= length n = number of loops of wire per unit length N-number of charge carriers per unit volume P = power Q = charge = point c arge 1 u = 1.66 x 10-27 kg = 931 MeV/c² h = 6.63 x 10-34 J-s = 4.14 x 10-15 eV-s h: 1.99 x 10-25 J-m = 1.24 x 10³ eV nm 8 = 8.85 x 10-12 C²/(N-m²) Coulomb's law constant, k = 1/(4m) = 9.0 x 10° (Nm²)/c² Vac ium permittiv t,, R- resistance r-radius or distance t = time J = potential or stored energy V = electric potential v = velocity or speed p = resistivity - flux x= dielectric constant F =qvxB b-dl= Hol Idix? = He 1 dbxr dB F = [I dè x B B = onl Electron charge na gn..ude, .eron ol, 1 V-1.60 x 10 J Sd of light, c = ..00 x 10 m/s Universal gravitational G=7x10-¹1 (N-m²/kg² constant, 8 = 9.8 m/s² 8=E-de= 8= 2=-L U₂ - 2012 150x1¹9 1 atm = 1.0 x 105 N/m² = 1.0 x10³ Pa farad, tesla. dog. dt degree Celsius, electron volt, F Т VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES 60" 90° 3,⁰ 1/2 √3/2 4/5 √2/2 .7 3/5 √2/2 0 aine cose tane 1 4/5 √3/2 3/5 1/2 4/3 √3 0 √3/3 3/4 1 point charge. IV. All batteries and meters are ideal unless otherwise stated. C RUST ev The following assumptions are used in this exam. T The frame of reference of any problem is inertial unless otherwise stated. 8 II. The direction of current is the direction in which positive charges would drift. III. The electric potential is zero at an infinite distance from an isolated V. Edge effects for the electric field of a parallel plate capacitor are negligible unless otherwise stated.
L/2+
L/2
Note: Figure not drawn to scale.
1. A very long, thin, nonconducting cylinder of length L is centered on the y-axis, as shown above. The cylinder
has a uniform linear charge density +2.. oint. is located on the y-axis at y = c, where L >> ¢.
a)
Kecling.c
Abh
ri gle
Circle
GEOMETRY AND TRIGONOMETRY
i. On the figure shown below, draw an arrow to indicate the direction of the electric field at point F due
to the long cylinder. The arrow should start on and point away from the dot.
ii. escribe the shape and location of a Gaussian surface that can be used to determine the electric field at
point P due to the long cylinder.
iii. Use your Gaussian surface to derive an expression for the magnitude of the electric field at point P.
E press your answer in terms of 2, c, L, and physical constants, as appropriate.
A==bh
A = r²
C = 277
=r8
Rectangular Solid
Vi h
Cy.inder
Sphere
V = r²t
Sm
V=²
S-427²
sine
Rig Triangle
2+6²=c²
cos@=
tan=
+2
BIL
a
b
C
2x.
Q
b
ADVANCED PLACEMENT PHYSICS C EQUATIONS
A-arca
C-circumference
V-volume
S =sur. ce area
Pay=c
= TLC
h-hei
EleL
W = IUL
r-full
5
anc le
= unic
50
b
P!
90%
G
CALCULUS
df du
chut cx
df
dx
-72²-1
(0²) -
a
1
(In ax) == /
[sin(x)] = cos(x)
[cs(x)] = -ain,
√x²³ dx = n + 1²
Je dx = -0
+1
| = |n|x + a|
x + a
.n-1
(cos(x) dx =sin(x)
—sin
(sin(ax)dx= cos(ax)
VECTOR PRODUCTS
A-B = AB cose
A x B - ABsine
Transcribed Image Text:L/2+ L/2 Note: Figure not drawn to scale. 1. A very long, thin, nonconducting cylinder of length L is centered on the y-axis, as shown above. The cylinder has a uniform linear charge density +2.. oint. is located on the y-axis at y = c, where L >> ¢. a) Kecling.c Abh ri gle Circle GEOMETRY AND TRIGONOMETRY i. On the figure shown below, draw an arrow to indicate the direction of the electric field at point F due to the long cylinder. The arrow should start on and point away from the dot. ii. escribe the shape and location of a Gaussian surface that can be used to determine the electric field at point P due to the long cylinder. iii. Use your Gaussian surface to derive an expression for the magnitude of the electric field at point P. E press your answer in terms of 2, c, L, and physical constants, as appropriate. A==bh A = r² C = 277 =r8 Rectangular Solid Vi h Cy.inder Sphere V = r²t Sm V=² S-427² sine Rig Triangle 2+6²=c² cos@= tan= +2 BIL a b C 2x. Q b ADVANCED PLACEMENT PHYSICS C EQUATIONS A-arca C-circumference V-volume S =sur. ce area Pay=c = TLC h-hei EleL W = IUL r-full 5 anc le = unic 50 b P! 90% G CALCULUS df du chut cx df dx -72²-1 (0²) - a 1 (In ax) == / [sin(x)] = cos(x) [cs(x)] = -ain, √x²³ dx = n + 1² Je dx = -0 +1 | = |n|x + a| x + a .n-1 (cos(x) dx =sin(x) —sin (sin(ax)dx= cos(ax) VECTOR PRODUCTS A-B = AB cose A x B - ABsine
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 5 images

Blurred answer
Similar questions