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Lecture 3-1
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 3-2
Torque on the Dipole
F
The magnitude of the torque:
  Fx sin   F (d  x) sin 
 Fd sin   qEd sin 
 pE sin 
x

CM
F
The torque points into the screen.
 
  p E

  rF
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
Continuous Charge Distribution 1: Charged Line
dx
E  Ex  k 
 L / 2 ( x  x )2
p
L/2
 k  
xp L / 2
xp L / 2
du
u2
u  x
xp L / 2
1
 k  
 u  xp L / 2
k L
 2
x p  ( L / 2) 2
x p / L    kQ / x p 2
At a point P on axis:
p
 x
xp  L / 2
Q

L
Lecture 3-5
Again: Continuous Charge Distribution 1: Charged Line
At a point P on perpendicular axis:
E  Ey  
L/2
k
 dx
cos
x y
L/2
y
 k 
dx
 L / 2 ( x 2  y 2 ) 3/ 2
L/2
2
2
 x  y tan  
y 2 sec 2 
 k 
d
 ( y 2 tan 2   y 2 ) 3/ 2
k 
k

cos  d  (2sin  )



y
y


L       / 2 and E  2k  / y
x
1   2  
Lecture 3-6
Physics 241 –Warm-up quiz
The rod is uniformly charged with a positive charge
density . 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 3-7
General location of P: Charged Line
x2
 dx
x1
r2
Ex   k
sin   k  
x2
x1
x
dx
2
2 3/ 2
(x  y )
At a point P off axis:
x2


 cos  2 cos 1 
1
 k  2
 k 


2 1/ 2 
(
x

y
)
y
y

 x1


k

 cos 2  cos1 
y
x2
Ey   k
x1
 dx
r
2
cos  k  
x2
x1
y
dx
2
2 3/ 2
(x  y )
k  2
sec 2 
k

d 
sin  2  sin 1 

2
3/ 2


y 1 (tan   1)
y
 x  y tan  
Lecture 3-8
Continuous Charge Distribution 2: Charged Ring
At point P on axis of ring:
ds
Q

2 R
Use symmetry!
E  Ex   k

 ds
x a
2
2
cos
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 3-9
Continuous Charge Distribution 3: Charged Disk
Use the ring result
x
dE x  k 2
dq
2 3/ 2
(x  a )
At a point P on axis:
x
E  E x   dE x   k 2
dq
2 3/ 2
(x  a )
R
x
 k 2
 2 ada
2 3/ 2
0
(x  a )
R
a
 kx 2 
da
2
2
3/
2
0 (x  a )
2 1 / 2 R
 kx 2  ( x  a )
2

0


1
 2 k  1 
  x  0
2

1  ( R / x ) 

R    whole plane and E   / 2 0
dq   2 a da
Superposition of rings!


2

k


 x R

2 0
E

2
k

R
kQ


(x
2
 x 2
x
<= Independent of x
R)
Lecture 3-10
Continuous Charge Distribution 4: Charged Sheets
 ( )   ( )
E=const in each region
Superposition!
Capacitor
geometry
Lecture 3-11
Physics 241 –Quiz 1A
The left half of a rod is uniformly charged with a
positive charge density , whereas the right half is
uniformly charged with a charge density of . 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 zeo.


Lecture 3-12
Physics 241 –Quiz 1b
The upper half of a ring is uniformly charged with a
positive charge density , whereas the lower half is
uniformly charged with a charge density of . What is
the direction of the electric field at a point on the
perpendicular axis through the center of the ring and to
the left of the ring as shown?
a)
b)
c)
d)
e)
to the right
to the left
up
down
E is zeo.


Lecture 3-13
Physics 241 –Quiz 1C
The upper half of a rod is uniformly charged with a
positive charge density , whereas the lower half is
uniformly charged with a charge density of . What
is the direction of the electric field at a point on the
perpendicular bisector and to the right of the rod as
shown?
a)
b)
c)
d)
e)
to the right
to the left
up
down
E is zeo.

