Download PES 1120 Spring 2014, Spendier Lecture 18/Page 1 Today

PES 1120 Spring 2014, Spendier
Lecture 18/Page 1
Today:
- Dielectric constant
- more practice!
Last time:
Energy Stored in Capacitors
Q2 1
1
U
 QV  CV 2 units [J] Joules
2C 2
2
Dielectric Constant
What happens when we put a block of dielectric material into an electric field, i.e.
between the plates of a parallel-plate capacitor?
While the entire object remains neutral, there are induced surface charges created on the
opposing surfaces. Those surface charges create an electric field that opposes (points in
the opposite direction to) the electric field in which it is placed.
The net result is a reduction in the electric field between the plates.
Reduced electric field:
 

E  E0  E '

 


E0
E  E0  c E0  (1  c )E0 
k



E
E  (1  c )E0  0
k
(1 c)  1  k  1
PES 1120 Spring 2014, Spendier
Lecture 18/Page 2
Constant κ > 1 is called dielectric constant (no units) and is used to indicate by how
much the electric field E0 inside a dielectric is reduced compared to air.
E
E0
k
(Electric field is reduced)
This reduces the potential across the plates, since the potential is determined by:
 
V   E  dl
f
(potential is reduced, Vab 
i
Vab , vacuum
k
)
The amount that the electric field (and thus, the potential) is reduced. Since the charge on
the capacitor hasn’t changed, the capacitance goes up:
C
Q
Vab
C  kC0
(capacitance increases)
PES 1120 Spring 2014, Spendier
Lecture 18/Page 3
Example 1: A 22.5-μF capacitor is connected to a power supply that keeps a constant
potential difference of 16.0 V across the plates. A piece of material having a dielectric
constant of 5.20 is placed between the plates, completely filling the space between them.
(a) How much energy is stored in the capacitor before and after the dielectric is inserted?
(b) By how much did the energy change during the insertion? Did it increase or decrease?
Why?
PES 1120 Spring 2014, Spendier
Lecture 18/Page 4
Example 2: Figure shows a parallel-plate capacitor with a plate area A = 5.56 cm2 and
separation d = 5.56 mm. The left half of the gap is filled with material of dielectric
constant κ1 = 7.00; the right half is filled with material of dielectric constant κ1 = 12.0.
What is the capacitance?
PES 1120 Spring 2014, Spendier
Lecture 18/Page 5
Example 3: Cell Membranes. Cell membranes (the walled enclosure around a cell) are
typically about 7.5 nm thick. They are partially permeable to allow charged material to
pass in and out, as needed. Equal but opposite charge densities build up on the inside and
outside faces of such a membrane, and these charges prevent additional charges from
passing through the cell wall. We can model a cell membrane as a parallel-plate
capacitor, with the membrane itself containing proteins embedded in an organic material
to give the membrane a dielectric constant of about 10.
(a) What is the capacitance per square centimeter of such a cell wall?
(b) In its normal resting state, a cell has a potential difference of 85 mV across its
membrane. What is the electric field inside this membrane?
PES 1120 Spring 2014, Spendier
Lecture 18/Page 6
For a parallel plate capacitor which is completely filled with a dielectric of dielectric
strength K, the capacitance is now:
C  kC0 
ke0 A
d
The product ke0 is called the permittivity of the dielectric, denoted by ε.
e  ke0
It turn out that we can simply replace e  ke0 by in all our previous equations , where we
assumed that the charges existed in a vacuum. The generalization for dielectric materials:
1 q1 q2 Fvacuum

4 r 2

1 q Evacuum
Electric field: E 

4 r 2

Coulomb’s Law: F 
Gauss’ Law:  
 

 E  r   dA

q enc

or  
 

 E  r   dA

q enc
0
Electric potential of a point charge at origin taking V = 0 at infinity:
1 q Vvacuum ( r )
V (r ) 

4pe r
k