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DipolesThe total charge is not the only thing
that mattersThe distribution of charge also matters
Let us define a simple (primary) dipole as consisting of two opposite charges q separated by distance d
Total charge is 0But dipole is not electrically inert!Dipole moment p
Vector direction is from negative to positive charges
qdp =|| r
q
-q
pd
∫= dVxxp )(ρrr
Moment of inertiaDipole moment p
Vector direction is from negative to positive charges
(This equation is only meant to remind you of I from your mechanics class! We won’t actually use it)
Like moment of inertia
qdp =|| r
q
-q
pd
∫= dVxxp )(ρrrm
m
Id
Torque on DipolesEven though neutral, the dipole is
sensitive to electric fieldsLet’s put a third charge Q (positive)
along bisector+q feels outward push, -q feels inward
Net force is smallApproaches 0 as Q becomes far away compared to d
The tangential components, though, result in substantial torque!
Does not disappear as Q goes further away
Dipole would like to spin
q
-q
pQ
Torque on DipolesCompare dipole direction p to
local field direction E(xp)If field is constant over
small region near dipoleIf perpendicular to field
Torque as we just sawIf aligned with field
No total forceIf anti-aligned with field
No total force, but unstable
)( pxEprrr ×=τ
+-
Total force on dipoleIf field is constant
No net force on dipolePositive, negative charges pulled in opposite directions
But held together
+ -
+ -
If field not constantOne end of dipole may feel
stronger force than the other endDipole can be sucked into or
pushed out of region
Total Force on DipoleHow to measure change
of field across ends?Derivative
+ -
+ -
[ ]
xEp
xExq
xExxq
xExxxExEq
xExEqF
∂∂
=∂∂
∆=∂∂
−=
⎥⎦⎤
⎢⎣⎡
⎭⎬⎫
⎩⎨⎧
∂∂
−+−=
−=∆
−+
+−++
−+
)()(
)()()(
)()(
xx EpF ∇⋅=rr
Don’t worry about the details of this formula. It’s for cultural purposes only. The important point is to realize that the dipole is pulled towards regions of larger field
Induced DipolesA charge can induce a dipole in an
originally neutral, dipole-less objectAtoms
Negative charge can repel electrons, attract proton
Electrons re-arrange themselves around nucleus
0)( == ∫ dVxxp ρrr
-
∫= dVxxp )(ρrr
Result: net dipole momentField has gradient
Field of single charge Neutral atom attracted to single charge!
Will also work if single charge is neutral dipoleOrigin of van der Waals forces
(reminder formula, don’t memorize!)
Dipole ForceSuppose a molecule is “polarizable”
It develops dipole moment p that is proportional to E-field, with coefficient ε
What is the force between a single fixed charge Q and this molecule, as function of separation R?
+-
Q p=εE
R
Dielectric BreakdownIf field not constant
One end of dipole may feel stronger force than the other endDipole can be sucked into or pushed out of region
If field very strong and very rapidly varying
Tidal force may result Forces stronger than atomic bond between the charges“Dielectric breakdown”
aka “sparking”~10,000 V / cm in air
+ -
Potential EnergyIn constant E-field of
strength EWhat is potential energy
of dipole with strength p and angle φ, shown at right?
Careful with signss+
- φ
θτ sinpE=θτrr ddU ⋅= =⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=−= ∫
φ
θθ0
sin dpEWU
)cos1( φ−pEp x E into page
θ vector out of page (RH rule)
DielectricsInsert molecule between
capacitor platesConstant E-field
If the molecule highly polarAlign themselves to field
If molecule highly polarizableInduced dipole moments throughout materialAtomic “cloud” of electrons distortedBut atoms themselves stay in place
Recall simple hydrogen atom in E-field of charge
∫= dVxxp )(ρrr
+-
)( pxEprrr ×=τ
MaterialsMaterials consist of many
moleculesLiquids: randomly arranged and oriented
Solids: arranged in arrays
+ -+-
+-
+-
+-
+-
+-
+-+-+
-
+-+-+
-
+-
+-
Dielectric MaterialsConsider now arrays of
molecules after aligning themselves
Rotatable dipoles (liquid)Centers fixed (solid)
Two things to noteDipole moments (induced or aligned) all point along field directionDipoles line up head-to-tail
+-+-+-+-+-
+-+-+-+-+-
+-+-+-+-+-
Head-to-Tail Dipole ChainsHead of one cancels tail of nextUltimately left only with head on
one end, tail on otherIf dipole has length d
Each dipole is qdOne head, one tail separated by ndDipole moment qndSame as adding up all the dipole moments n(qd)
Dipole moments arranged head to tail simply sum upActs simply as if charges are at
the ends of the dipole chainsSometimes the end charges are called “bound charge”
qd+-
qd+-
qd+-
qd+-
qd+-
qnd+-
Dielectric ConstantLarge dipole now induced in the original
field E0Induced electric field Einduced opposes the original, imposed field!
(E-field of dipole points opposite the dipole vector)Induced field is smaller than original, imposed field
Strength of induced dipole depends on material
Can in principle be calculated in quantum mechanics
Field E’ in internal region sum of Original field E0Induced field Eind
+-+-+-+-+-
+-+-+-+-+-
+-+-+-+-+-
KEE 0'r
r=
E0
Dielectric constant
+++++
-----E0
E0
Einduced
Dielectric Constant, Capacitance
Insert dielectric material (K>1) between capacitor plates
Induce dipole momentEquivalent to induced charge on faces of sheet
Head-to-tail chainsReduced E-field
E’=E/KReduced Voltage (work / charge)
∆V’=E’d=Ed/K=∆V/KSame charge Q
C’=Q’/∆V’=Q/(∆V/K)=KQ/∆V=KCCapacitance increased by factor K!
All actual capacitors built with a dielectric layerDefine permittivity Kε0=ε
+Q -QK=2
dA
dAKC εε == 0
Making ComputersA capacitor stores chargeYou could imagine
A capacitor with charge on it = stored value of 1A capacitor without charge = stored value of 0
This is basically how RAM is madeRecall Q=C∆V
For fixed ∆V, larger capacitance means the stored charge is larger
This means the stored signal is larger (and lasts longer)
For computer memory, would like large CC=Kε0A/d
Also want small ACompromise
Use small A, large K!Can find special materials with K>>1000
Lets you shrink your capacitor by 1000
Voltage, Potential, Work, Capacitance, and Fields
E-Field(N/Coulomb or Volts/m)
Electric Potential
(Volts; J/Coulomb)
Potential Energy(Joules or N.m or eV)
Force(Newtons)
Charges(Coulomb)
Integrate E.x over path
Gradient
Integrate F.x over path
GradientEqFrr
=VqU ∆=∆
iesch i
i rr
qE ˆ4arg
20
∑=πε
r
∑esch i
i
rq
arg 04πεGauss’ Law
Voltage(Volts; J / C)
BAABV Φ−Φ=
Q=C∆V
Coulomb’s Law
∫ =⋅=Φ0
)(εenclQadEflux rr
Dipole(Coulomb.m)
qdp =
Φ
U F
E
p
∆V q