Definitions
• Dielectric—an insulating material placed between plates of a
capacitor to increase capacitance.
• Dielectric constant—a dimensionless factor  that determines
how much the capacitance is increased by a dielectric. It is a
property of the dielectric and varies from one material to another.
• Breakdown potential—maximum potential difference before
sparking
• Dielectric strength—maximum E field before dielectric breaks
down and acts as a conductor between the plates (sparks)
Most capacitors have a non-conductive material (dielectric) between the conducting
plates. That is used to increase the capacitance and potential across the plates.
Dielectrics have no free charges and they do not conduct electricity
Faraday first established this
behavior
Capacitors with Dielectrics
•
Advantages of a dielectric include:
1. Increase capacitance
2. Increase in the maximum operating voltage. Since dielectric
strength for a dielectric is greater than the dielectric strength
for air
Emax di  Emax air  Vmax di  Vmax air
•
3. Possible mechanical support between the plates which
decreases d and increases C.
To get the expression for anything in the presence of a
dielectric you replace o with o
k 0 S
C
;
d
V  Ed  E decreases: E  E0 / k
Field inside the capacitor became smaller – why?
We know what happens to the conductor in the electric field
Field inside the conductor E=0
outside field did not change
Potential difference (which is the
integral of field) is, however, smaller.

V  ( d  b)
o
C
0 A
d [1  b / d ]
There are polarization (induced)
charges
– Dielectrics get polarized
Properties of Dielectrics
Redistribution of charge – called polarization
K
C
dielectric constant of a material
C0
We assume that the induced charge is directly
proportional to the E-field in the material
E
E0
K
when Q is kept constant
V
V0
K
In dielectrics, induced charges do not exactly
compensate charges on the capacitance plates
E0 

;
0
  K 0

E

1
u   E2
2
E
  i
0


1

 i   1  
K
Induced charge density
Permittivity of the dielectric material
E-field, expressed through charge density  on the conductor plates
(not the density of induced charges) and permittivity of the dielectric
 (effect of induced charges is included here)
Electric field density in the dielectric
Example: A capacitor with and without dielectric
Area A=2000 cm2
d=1 cm; V0 = 3kV;
After dielectric is inserted, voltage V=1kV
Find; a) original C0 ; b) Q0 ; c) C d) K e) E-field
Dielectric Breakdown
Plexiglas breakdown
Dielectric strength is the maximum electric field
the insulator can sustain before breaking down
Applications
In 1995, the microprocessor unit on the microwave imager on the DMSP
F13 spacecraft locked up - occurred ~5 s after spacecraft began to charge
up in the auroral zone in an auroral arc. Attributed to high-level charging of
spacecraft surface and subsequent discharge.
Spacecraft surfaces are generally covered with thermal blankets - outer
layer some dielectric material - typically Kapton or Teflon. Deposition of
charge on surface of spacecraft known as surface charging. Incident
electrons below about 100 keV penetrate the material to a depth of a few
microns, where they form a space charge layer - builds up until breakdown
occurs accompanied by material vaporization and ionization. A discharge is
initiated - propagates across surface or through the material, removing part
of bound charge. Typically occur in holes, seams, cracks, or edges - have
been know to seriously damage spacecraft components.
Thermal blankets composed of layers of
dieletric material with vapor
deposited aluminum (VDA)
between each layer. On
DMSP, VDA between layers
(22) not grounded - serve as
plates of a set of 22 parallelplate capacitors - top plate
consists of electrons buried
in top few microns of Teflon.
1 22 1

C i1 c i
C/A=7.310-9 F/m2
 of a parallel plate capacitor to some voltage with
Time to charge outer surface
respect to spacecraft frame is:
t
CV
i(1   )A
Laboratory measurements for Teflon:
-discharge at 3 kV in a 20 keV electron beam
-secondary electron yield () at 20 keV ~ 0.2
Given measured incident precipitating current density of i = 4.8 A/m2, the time
to reach breakdown voltage for conditions experienced by DMSP F13 is 5.7 s this is the time after the spacecraft began to charge up that the lockup
occurred.
If the VDA layer on the bottom side of the outer Teflon layer were grounded to
the spacecraft frame, the capacitance would have been 3.510-7 F/m2 and the
charging time would have been 132 s - no discharge would have occurred.
Molecular Model of Induced Charge
Electronic polarization of nonpolar molecules
Total charge Q   qi  0
i
But dipole moment d   qi ri may be nonvanishi ng
i
For nonpolar molecules d  0 in the absence
of the applied electric field E but they
acquire finite dipole moment in the field :
d   0 E
( is the polarizabi lity of a molecule/a tom)
Electronic polarization of polar molecules
In the electric field more molecular dipoles are oriented
along the field
Polarizability of an Atom
- separation of proton and electron cloud in the applied
electric field
P- dipole moment per unit volume, N – concentration of atoms
When per unit volume, this dipole moment is called
polarization vector P  Nqδ   0E

Property of the material: Dielectric susceptibility   N
Polarization charges induced on the surface:  ind  Pn  P  n
For small displacements: P~E; P= 0 E
The field inside the dielectric is reduced :
   ind
E 
E  free
 0  free
0
K 0K
K  1   ;  ind  (
K 1
) free
K
Gauss’s Law in Dielectrics
EA 
(   i ) A


0
KEA 


 KE d A
Q free
0
1

 i   1  
K
A
0
Gauss’s Law inDielectrics
Forces Acting on Dielectrics
We can either compute force directly
(which is quite cumbersome), or use
relationship between force and energy
F  U
CV 2
Considering parallel-plate capacitor U 
2
Force acting on the capacitor, is pointed inside,
hence, E-field work done is positive and U - decreases
U V 2 C
Fx  

x
2 x
x – insertion length
Two capacitors in parallel
C  C1  C2 
0
d
w( L  x) 
V 2  0w
Fx 
( K  1)
2 d
K 0
wx
d
w – width of the plates
More charge here
constant force