Transcript PHYS_3342_092911

The graded exams will be returned next Tuesday, Oct 4. You will have until the next
class on Thursday, Oct 6 to rework the problems you got wrong and receive 50% added
credit. Make sure you are in class as you will no have another opportunity to rework the
exam. I will be going over the answers in class on Thursday. This will also be your only
opportunity to ask for corrections/clarifications on any grading mistakes.
The homework assignment will be on line this afternoon but will not be due until
Tuesday, Oct 11. This will give you the opportunity to start work on the problems so that
you will not be overloaded with homework and the exam rework next week.
Energy Storage in Capacitors
Electric Field Energy
Electric potential energy stored = amount of work done to charge the capacitor
i.e. to separate charges and place them onto the opposite plates
V
Q
C
To transfer charge dq between
conductors, work dW=Vdq
Q
Q
q
Q2
Total work W   V ( q)dq   dq 
C
2C
0
0
Q2 1
CV 2
Stored energy U 
 QV 
2C 2
2
Charged capacitor – analog to stretched/compressed spring
Capacitor has the ability to hold both charge and energy
CV 2 ( 0 A /d)(Ed)  0 E 2
uE 


2Ad
2Ad
2

Density of energy (energy/volume)
Energy is conserved in the E-field
In real life we want to store more charge at lower voltage,
hence large capacitances are needed
Increased area, decreased separations, “stronger”
insulators
Electronic circuits – like a shock absorber in the car, capacitor smoothes power
fluctuations
Response on a particular frequency – radio and TV broadcast and receiving
Undesirable properties – they limit high-frequency operation
Example: Transferring Charge and Energy Between Capacitors
Switch S is initially open
1) What is the initial charge Q0?
2) What is the energy stored in C1?
3) After the switch is closed
what is the voltage across each
capacitor? What is the charge on each? What is the total energy?
a)
Q0  C1V0
b)
1
U i  Q0V0
2
c) when switch is closed, conservation of charge
Q1  Q2  Q0
Capacitors become connected in parallel V 
C1V0
C1  C2
1
1
d) U f  Q1V  Q2V  Ui Where had the difference gone?
2
2
It was converted into the other forms of energy (EM radiation)
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
C
K0 A
; V  Ed  E decreases : E  E 0 /K
d
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
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