Physics for Scientists and Engineers II , Summer Semester 2009
Lecture 2: May 20th 2009
Physics for Scientists and Engineers II
Physics for Scientists and Engineers II , Summer Semester 2009
Electric Field due to a Continuous Charge Distribution
• We can model a system of charges as being continuous (instead of discrete) if the distance between the charges is much smaller than the distance to the point where the electric field is calculated.
• Procedure: - Divide charge distribution into small charge elements q. - Add contributions to E from all charge elements.
r̂
PE r
q
ii
i
e rr
qkE ˆ
2
rr
dqkr
r
qkE e
ii
i
i
qe
i
ˆˆ22
0lim
rr
qkE e ˆ
2
Physics for Scientists and Engineers II , Summer Semester 2009
Charge Density (a useful concept when calculating E from charge distribution)
dldql
Q
dAdqA
Q
dVdqV
Q
:l)length of line aon ddistributeuniformly is Q (ifdensity chargeLinear
:A) area of surface aon ddistributeuniformly is Q (ifdensity charge Surface
:V) volumeat throughouddistributeuniformly is Q (ifdensity charge Volume
Physics for Scientists and Engineers II , Summer Semester 2009
Example: Electric Field due to a Uniformly Charged Rod
l
x
y
P
a
x dx
E
dq = dx
22 :dq from on toContributi
x
dxk
x
dqkdEE ee
)11
1 :rod) (entire dq all from
22
ala
Qk
alal
Qk
xk
x
dxk
x
dxkEE
ee
al
ae
al
a
e
al
a
e
charge.point a of field the, 0 lFor :otice2a
QkEN e
Physics for Scientists and Engineers II , Summer Semester 2009
Example: Electric Field due to a Uniformly Charged Rod…..this is harder….
l
x
y
P
a
x
Ed dq = dx
j
xa
dxaki
xa
dxxk
jr
dxaki
r
dxxk
jr
a
r
dxki
r
x
r
dxk
jr
dxki
r
dxkEdE
ee
ee
ee
ee
ˆˆ
ˆˆ
ˆˆ
ˆcosˆsin :dq from on toContributi
23
2223
22
33
22
22
r
Physics for Scientists and Engineers II , Summer Semester 2009
Example: Electric Field due to a Uniformly Charged Rod…..this is harder….
l
x
y
P
a
x
Ed dq = dx
j
xa
dxaki
xa
dxxkEdE ee
ˆˆ :dq from on toContributi2
3222
322
r
dx
xaakEdx
xa
xkE
jxa
dxaki
xa
dxxkE
l
ey
l
ex
l
ee
0 23
220 2
322
0 23
2223
22
1
ˆˆ :Ppoint at field electric Total
Physics for Scientists and Engineers II , Summer Semester 2009
….solving the integral for Ex
dx
xa
xkE
l
ex
0 2
322
2
122
21
22 :onSubstitutixa
dxxduxau
22
22
22
20 2
122
22
11
1112222
laa
ala
l
Qk
laak
ukdu
ukdx
xa
x
xakE
ee
la
ae
la
a
e
l
ex
Physics for Scientists and Engineers II , Summer Semester 2009
….solving the integral for Ey
dx
xa
akE
l
ey
0 2
322
that)know you toexpect t wouldn'(Icos
tan :onSubstituti2
da
dxax
2222
max0
002
23
2
02
23
20
22
3222
sinsin
coscos
1
cos1
1
cos
1
tan1
1
costan
max
maxmax
maxmax
ala
Qk
al
l
la
Qk
la
Qk
a
k
da
kd
a
k
da
kd
a
aa
akE
ee
ee
ee
eey
a
l
22 al max
22maxsinal
l
Physics for Scientists and Engineers II , Summer Semester 2009
….and the final result
22 ala
QkE e
y
22
22
laa
ala
l
QkE ex
again)chargepointaof(fieldand0:
:rodshort very aoflimit theIn
20
lim a
QkEE e
yxl
akE
akE eyex
l
and:
:rod longvery aoflimit theIn
lim
Physics for Scientists and Engineers II , Summer Semester 2009
Visualizing Electric Fields with Electric Field Lines
• The electric field vector is always tangent to the electric field line.
• The electric field line has a direction (indicated by an arrow). The direction is the same as that of the electric field (same direction as force on a positive test charge).
• The number of lines per unit area through a normal plane (perpendicular to field lines) is proportional to the magnitude of the electric field in that region.
Example: Electric field lines of a point charge
+
N field lines
Surface density of field lines at an imagined sphere of radius r is
Electric field strength is proportional to
24 r
N
2
1
r
Physics for Scientists and Engineers II , Summer Semester 2009
Visualizing Electric Fields with Electric Field Lines
• For a single positive point charge: Electric field lines go from the positive charge to infinity.• For a single negative point charge: Electric field lines go come from infinity and end at the negative point charge.• For multiple point charges: Lines can start at the positive charges and end at the negative charges.• Electric field lines can never cross (think about why that is so).• For two unequal point charges of opposite sign with charges Q1 and Q2 , the number N1 of field lines terminating at Q1 and the
number N2 of field lines terminating at Q2 are related by the equation
1
2
1
2
Q
Q
N
N
Physics for Scientists and Engineers II , Summer Semester 2009
Motion of a Charged Particle in a Uniform Electric Field
• Assume particle has charge q, mass m.• Particle experiences a force
• The force results in an acceleration (according to Newton’s second law):
• For positive charges: Acceleration is in the same direction as electric field.• For negative charges: Acceleration is in a direction opposite to the electric field.• A uniform electric field will cause a constant acceleration of the particle.
You can use equations of motion for constant acceleration.
• Work is done on the particle by the electric force as the particle moves.
EqF e
m
Eq
m
Fa
e
xFW e
Physics for Scientists and Engineers II , Summer Semester 2009
Example (similar to Ex. 23.10 in book)
- - - - - - - - - -
+ + + + + + + + + +
E-?iv
Electron: m = 9.11x10-31 kg ; q = 1.60x10-19 CElectric Field: E = 800 N/C
L = 0.100 m
The electron leaves the electric field at an angle of = 65 degrees.Q1: What was the initial velocity of the electron?Q2: What is the final velocity of the electron (magnitude)?Q3: How low would the electric field have to be so that the net force on the electron is zero?Q4: Were we justified in neglecting the gravitational force in Q1 and Q2?
Physics for Scientists and Engineers II , Summer Semester 2009
tan;
:1Question
i
fy
i
eyfy v
v
v
Ltt
m
Eqt
m
Ftav
s
m
kg
mCN
CL
m
Eqv
v
L
m
Eqv
v
L
m
Eqv
i
ii
ify
631
19
105.2deg65tan1010.9
100.08001060.1
tan
tan
s
msmv
v if
6
6
101.665cos
105.2
cos
:2Question
Physics for Scientists and Engineers II , Summer Semester 2009
force. nalgravitatio n thelarger thamuch is
electronan on force electric thefields, electric smallextremely for except Yes,
:4Question
106.5106.1
1010.98.9
:3Question
1119
312
C
N
C
kgs
m
q
gmEg
m
qE
Physics for Scientists and Engineers II , Summer Semester 2009
Gauss’s Law – An alternative procedure to calculate electric fields of highly symmetric charge distributions
The concept of “Electric Flux”:
Area = A
E
area. lar toperpendicu being E and Econstant for
:flux Electric E EA
Physics for Scientists and Engineers II , Summer Semester 2009
E
AE is lar toperpendicu Area
AE is lar toperpendicunot Area
cosAA
The electric flux through the two surfaces is the same
cosAEAEE
Normal to green surface
Physics for Scientists and Engineers II , Summer Semester 2009
The electric flux through the two surfaces is the same
cosAEAEE
Normal to green surface
To calculate the flux through a randomly oriented area you need to know the angle between the electric field and the normal to the area.
Physics for Scientists and Engineers II , Summer Semester 2009
surface
E
i
iiE
iiiiiE
AdE
AE
AEAE
:segments surface smallmally infinitesi oflimit in the ....and
:surface entiregh flux throu Electric
cos:element surfacegh flux throu Electric
How to treat situations where the electric field is not constant over the area?
• Divide area into small areas over which E is constant.• Calculate flux for each small area.• Add fluxes up.
Area vector:magnitude = areadirection = perpendicular to area
iE
iiA
“surface integral”
Physics for Scientists and Engineers II , Summer Semester 2009
dAEAdE nE : surface closedgh Flux throu
Flux through a closed surface:
•Convention: Area vectors always point outwards.Field lines that cross from the inside to the outside of the surface : (positive flux because cos is positive)Field lines that cross from the outside to the inside of the surface: (negative flux because cos is negative)
90
18090
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