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Lecture 3-1
Electric Field
Define electric field, which is independent of the test charge,
q, and depends only on position in space:
F
E
q
• One is > 0, the other < 0
-q
-
d
q
electric dipole of
dipole moment:
p  qd
+
dipole
Lecture 3-2
Dipole in uniform electric fields
• No net force. The electrostatic
forces on the constituent point
charges are of the same
magnitude but along opposite
directions. So, there is no net
force on the dipole and thus its
center of mass should not
accelerate.
 
  p E

• Net torque! There is clearly a net
torque acting on the dipole with
respect to its center of mass, since
the forces are not aligned.
http://qbx6.ltu.edu/s_schneider/physlets/main/dipole_torque.shtml
Lecture 3-3
Electric Field from Coulomb’s Law
Bunch of Charges
r
+
+
+
-
qi+
-
-
1
i P
+
-
Continuous Charge Distribution
P
r
dq
+
k
qi
E
rˆ

2 i
4 0 i ri
Summation over
discrete charges
http://www.falstad.com/vector3de/
1 dq
E
rˆ   d E
2
4 0 r
  dV

dq   dA
dL

(volume charge)
(surface charge)
(line charge)
Integral over continuous
charge distribution
Lecture 3-4
Reading Quiz 1
Which one of the following statements is incorrect ?
• A) Electric fields leave positive charges and end on negative charges
• B) Electric field lines can intersect at some points in space.
• C) Electric field field lines from a dipole fall off faster than 1/r2.
• D) Electric fields describe a conservative force field.
Lecture 3-5
SUMMARY: FIND THE ELECTRIC FIELD
GIVEN THE CHARGES
1)
GEOMETRY FOR
2)
DISCRETE CHARGES
qi
3)
CONTINUOUS CHARGES
dqi
line charge density
surface charge density
volume charge density
qi or dqi
λ (x)
σ (x.y)
ρ (x,y,z)
Geometry may suggest other coordinate systems, R,θ,Φ or R,θ,Z
Lecture 3-6
Continuous Charge Distribution 1 Charged Line
At a point P on axis:
Ex = k λ / (1/r1–1/r2) = Ex kλ/ (1/x1- 1/x2)
Eqn 22-2a
Ex = k λ {1/( XP + L/2 ) - 1/( XP - L/2)}
Ex = k ( Q/L) L [ XP2 – (L/2)2 ] -1
For XP2 >> (L/2)2
For XP = 0
Ex = k Q / XP2
Ex = 0
xp  L / 2
Q

L
Lecture 3-7
Again: Continuous Charge Distribution 1: Charged Line
At a point P on perpendicular axis:
L/2
x2
 dx

dx
y 
E

E

k
cos
E y   ky 2 cos
  k 2  22 2 3/ 2 dx
L/2
x1
x x1 (yx  y )
r
x2
y
2
k

sec

kdx
  x  y tan  
 k 
2
2
3/
2
   L / 22( x 3/ y2 d)   sin  2  sin 1   x  y tan  
y 1 (tan   1)
y

y 2 sec 2 
 k 
d
 ( y 2 tan 2   y 2 ) 3/ 2
k 
k

 cos d  (2sin  )
2
y
L/2

θ
x
y
L       / 2 and E  2k  / y
1   2  
Lecture 3-8
Physics 241 –Warm-up quiz 2
The rod is uniformly charged with a positive charge
density σ. What is the direction of the electric field at
a point P on a line perpendicular to the rod? Note that
the line and the rod are in the same plane.
a)
b)
c)
d)
e)
to the right
to the left
up
down
lower right
p
Lecture 3-9
Continuous Charge Distribution 2: Charged Ring
At point P on axis of ring:
ds
Q

2 R
Use symmetry!
θ
 ds
EE =Ekx Q x k( x22+ a2 )2-3cos
x
x a
x
k 2
  ds
2 3/ 2
(x  a )
x
k 2
Q
2 3/ 2
(x  a )
kQ / x 2 ( x
a)
(kQ / a 3 ) x ( x
a)
Lecture 3-10
Continuous Charge Distribution 3: Charged Disk
At a point P on axis:
Use the ring with radius a EX value
x
E dE
Exx = k dq
dExx (  k+ 2)
dq
2 3/ 2
(x  a )
R
x dq   2 a da
 k 2
 2 ada
2 3/ 2
0
(x  a )
Integrate rings from 0 to R
R
a
 kx 2 
da
2
2
3/
2
0 (x  a )
x2
a2
-3
Superposition of rings!
2
2 21 / 2 R2 1/2


Exkx=2-2πσkx
 ) – 1/x )

  ( x( 1/(
 a x) + R
2

k


 x R
0

2 0
E 


1
2
k
=
1/4πε
E
=
σ/2ε
k

R
kQ
o
o

 2 k  1 
x  0



( x R)
2 

2
2

1

(
R
/
x
)
x
 x


R    whole plane and E   / 2 0
<= Independent of x
Lecture 3-11
Continuous Charge Distribution 4: Charged Sheets
 ( )   ( )
E=const in each region
Superposition!
Capacitor
geometry
Lecture 3-12
MULTIPLE CHARGE SHEET EXAMPLE
DOCCAM 2
(SKETCH)
Lecture 3-13
Gauss’s Law: Qualitative Statement
 Form any closed surface around charges
 Count the number of electric field lines coming through the
surface, those outward as positive and inward as negative.
 Then the net number of lines is proportional to the net
charges enclosed in the surface.
Lecture 3-14
Electric flux
To state Gauss’s Law in a quantitative form,
we first need to define Electric Flux.
# of field lines N
= density of field lines
x “area”
where “area” = A2 x cos θ
N  E A  E An
Sum over
surface
E
An
General definition of electric flux:
 E   E n dA
S
(must specify sense, i.e., which way)
Lecture 3-15
Electric Flux through
E 

S
 E Closed Surface
E n dA 

S
En dA
• The integral is over a CLOSED surface.
• Since E n is a scalar product, the electric flux is a
SCALAR quantity
• The integration element
n is a vector normal to
the surface and points OUTWARD from the
surface. Out is +, In is -
ΦE proportional to # field lines
coming through outward
Lecture 3-16
Why are we interested in electric flux?
 E is closely related to the charge(s) which cause it.
Consider Point charge Q
E 
E
ndA  kq 
r
ndA
2
r
kQ
Q
2
 2  4 r  
r
0
If we now turn to our previous
discussion and use the analogy to
the number of field lines, then the
flux should be the same even when
the surface is deformed. Thus should
only depend on Q enclosed.
Lecture 3-17
Gauss’s Law: Quantitative Statement
The net electric flux through any closed surface equals the
net charge enclosed by that surface divided by 0.
 E ndA  
E

Qenclosed
0
How do we use this equation??
The above equation is TRUE always but it doesn’t
look easy to use.
BUT - It is very useful in finding E when the physical situation
exhibits a lot of SYMMETRY.
Lecture 3-18
Physics 241 – 10:30 Quiz 3, August 30, 2011
The left half of a rod is uniformly charged with a
positive charge density σ, whereas the right half is
uniformly charged with a charge density of -σ. What
is the direction of the electric field at a point on the
perpendicular bisector and above the rod as shown?
a)
b)
c)
d)
e)
to the right
to the left
up
down
E is zero.
+σ
-σ
Lecture 3-19
Physics 241 – 11:30 Quiz 3, September 1, 2011
The upper half of a ring is uniformly charged with a
positive charge density σ, whereas the lower half is
uniformly charged with a charge density of -σ. What is
the direction of the electric field at a point on the
perpendicular axis and to the left of the ring as shown?
a)
b)
c)
d)
e)
to the right
to the left
up
down
E is zero.
+σ
-σ
Lecture 3-20
Physics 241 – 11:30 Quiz 3, January 18, 2011
The upper half of a ring is uniformly charged with a
negative charge density -σ, whereas the lower half is
uniformly charged with a charge density of +σ. What is
the direction of the electric field at a point on the
perpendicular axis and to the left of the ring as shown?
a)
b)
c)
d)
e)
to the right
to the left
up
down
E is zero.
-σ
+σ