+ Q free on inner surface. + + + + + + + + + + + +.       .  q bound. Symmetry – fields must be uniform – field lines perpendicular to plates. + q bound. + + + + + +.
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+Qfreeon inner surface
+ + + + + + + + + + + +
      
qbound
Symmetry
– fields must be uniform
– field lines perpendicular to plates
+qbound
+ + + + + +
           
Qfreeon inner surface
Interior points electric
field must be zero
+ + + + + + + + + + + +
+Qfreeon inner surface
plate separationd
Qfreeon inner surface
           
       
+ + + + + + + +
           
Electric field
Electric displacement
Polarization
single turn of wire with current I
B
around integration loop B
dr = 0 and Br = 0
outside loop Bz = 0
.
.
.
.
.
.
.
3
X
X
X
X
X
X
2
Current i
out of page
Circulation loop:
square of length L
1
4
Current i
into page
Crosssection through electromagnet
Induced dipole moment – helium atom
+2e
+2e
e
e
e
e
Zero electric field – helium atom symmetric zero dipole moment
A
B
effectively charge +2e at A and 2e at B
dipole moment p = 2ed
Induced dipole moment – sulfur atom
+16e
+16e
8e
8e
8e
8e
Zero electric field – helium atom symmetric zero dipole moment
A
B
effectively charge +16e at A and 16e at B
dipole moment p = 16ed
q
+q
Width of ring r d
Radius of ring r sin
+
+
+
Area of the shaded ring
between and + d
surface S
d
r

Pcos


Lowest energy state
U = 0
U = + p E
highest energy state



= 90o
+
+
+
= 0
= 180o
r  1
1/T
Po
T
conducting
sphere q
air
a
r
nonconducting
liquid
Symmetry field lines must be radial
sphere q
Eairt
air
Eliquidt
nonconducting
liquid
Symmetry Eairt = Eliquidt Eair = Eliquid = E
+
+
+
+
+
+
+
+
+
+
greater concentration of charge
on surface bounded by liquid
field lines of E
+
+
+


+
+
shift in atoms
due to ionic nature of bond
induced dipoles due
to shift in electron cloud
rotation
orientation of polar molecules
Bar magnet bought near
unmagnetized piece of iron
N
N
N
Bar magnet will attract the iron that was initially unmagnetized
north pole attracts
south pole
M applying Ampere’s Lawiron
PERMANENT MAGNET
Hiron
B
B, Hgap
Mgap = 0
B = Bgap = Biron
ELECTROMAGNET
Miron
Hiron
B
Mgap = 0
B, Hgap
B = Bgap = Biron
Y applying Ampere’s Law
thickness t
area A = wt
width w
X
current in
X direction
Z
magnetic field in Z direction
Schematic diagram of a Hall Probe
width applying Ampere’s Law
w
Y
+ + + + +
    
_
VH
+
VH
    
+ + + + +
X
charge carriers electrons ()
eg wire, Ntype semiconductor
charge carriers positive (+)
eg holes in Ptype semiconductor
I
.
Z direction
out of page
e applying Ampere’s Law

e
length L
area
A
resistance R
resistivity
conductivity
number densityn
+
I
_
_
v
electron
V
+Z applying Ampere’s Law
Electron at A moving parallel to +Yaxis
vy
Electron acted upon by the radial component of the magnetic field force on electron in +X direction +Xcomponent to the velocity
Fx
due to Bz
By
axis for the motion of the electron beam
+Y
Bz
radial component
of magnetic field
+X
+Z applying Ampere’s Law
Electron at B has a velocity component in the +X direction
Fy
due to Bz
Electron acted upon by the axial component of the magnetic field By force on electron in Z direction i.e. towards to axis focusing action
Fz
due to By
vx
By
axis for the motion of the electron beam
+Y
Bz
radial component
of magnetic field
+X
external magnetic field applying Ampere’s Law
Electrostatic capacitor applying Ampere’s Law
Electrolytic capacitor
Electrochemical double layer capacitor applying Ampere’s Law
d
conductive
electrode
conductive
electrode
separator
activated carbon
+ applying Ampere’s Law
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+



























+ applying Ampere’s Law
Zero applied stress

Compressive stress
Induces a voltage
Applied voltage produces
An expansion
 applying Ampere’s Law























+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Ferroelectric material
Antiferroelectric material
Interior points electric applying Ampere’s Law
field must be zero
+0.2Q on outer surface
+
+
+ + + + + + + + + +
+Q on inner surface
         
Q on inner surface
Interior points electric
field must be zero
Symmetry
– electric field must be uniform
– electric field lines perpendicular to conductive plates
Symmetry applying Ampere’s Law
– fields must be uniform
– field lines perpendicular to plates
Interior points electric
field must be zero
+ + + + + + + + + +
+Q on inner surface
         
Q on outer surface
+ + + + + + + + + +
+Q on outer surface
Q on inner surface
         
Interior points electric
field must be zero
... applying Ampere’s Law
...
...
V
Capacitors in series (charge on each plate)
Capacitors in parallel (voltage across each capacitor is the same)
Series
branch
Capacitors in parallel applying Ampere’s Law
Q1
+Q1
C1
Q =Q1+Q2
+Q2
Q2
Ceq = C1+C2
C2
V
V
Capacitors in series
Q
+Q
Q
1/Ceq = 1/C1+1/C2
Q
+Q
C1
C2
V
V
E applying Ampere’s Law = 0
+Q applying Ampere’s Law/2
+Q/2
C1 = Q / 2V1
C1
C1
V1
Q = 2 C1V1
 Q/2
 Q/2
qA = C2V2 = r C1 V2
qB = C1 V2
Q = qA + qB
= C1 V2 (r + 1)
= 2 C1 V1
V2 = 2 V1 / (r + 1)
qA = 2 C1 V1 r / (r + 1)
qB = 2 C1 V1 / (r + 1)
+qA
+qB
V2
C1
C2
 qA
 qB
+ applying Ampere’s LawQf
Qf
Homogenous dielectric – uniformly polarized applying Ampere’s Law
Dielectric is neutral
+
+
+
+
+
+
+







The electrical field is reduced in the dielectric material
+ applying Ampere’s Law
+
+
+
+
+
+
+
+
+
+
Thin long rod L = 0
Zero polarization
Sphere L = 1/3
Concentration of charges
At surface given by











Flat plate L = 1
Max polarization
+ applying Ampere’s LawQ
 Q
R applying Ampere’s Law1
R2