Download PES 1120 Spring 2014, Spendier Lecture 17/Page 1 Today

PES 1120 Spring 2014, Spendier
Lecture 17/Page 1
Today:
- One more example on capacitor combinations in a circuit
- Storing Energy in a capacitor
- Dielectrics
- HW 4 given out - due next week Wednesday
- http://phet.colorado.edu/en/simulation/capacitor-lab
Last Lecture:
n
Capacitors in parallel: Ceq   Ci
i 1
i.e for two: V1 = V2 = V and Q = Q1 +Q2
Capacitors in series:
n
1
1

Ceq
i 1 Ci
i.e for two: V1 + V2 = V and Q = Q1 = Q2
http://phet.colorado.edu/en/simulation/capacitor-lab
Example 1: In the Figure, the battery has a potential difference of V = 10.0 V and the
three capacitors each have a capacitance of 10.0μF. What is the
(a) charge on capacitor 1 and
(b) charge on capacitor 2?
PES 1120 Spring 2014, Spendier
Lecture 17/Page 2
Storing Energy
Let me ask you a question. Let’s say that today you have a lot of spare energy, but that
there is a high likelihood that tomorrow you would need more than you will have. What
would you do to prepare for tomorrow?
One possibility is to take something at ground level, use the spare energy you have today
to do work on that something and raise it to a higher level. Then, tomorrow, you could
use that gravitational potential energy to do work… This is how a pumped-storage plant
works.
PES 1120 Spring 2014, Spendier
Lecture 17/Page 3
Energy Stored in Capacitors
Let’s look at a parallel plate capacitor with some initial charge on it. The electric
potential energy stored in a charged capacitor is just equal to the amount of work required
to charge it – that is, to separate opposite charges and place them on different conductors.
When the capacitor is discharged, this stored energy is recovered as work done by
electrical forces.
For a parallel-plate capacitor, how much work does it take to put more charge on it?
As more charge is added to the plates, the battery has to do more work
Q - the final amount of charge on the plates
V - the final potential difference
We know:
W
q
Work done by battery adding charge dq:
dW = v dq
(v is potential value when charging, v < V ).
V  Vb Va  
dW  vdq 
q
dq
C
Total Work done by Battery:
Q
W 
0
q
Q2
dq 
 U
C
2C
If we define the potential energy of an uncharged capacitor to be zero, i.e. Uf = 0, then
U  U f U i   U i  U 
Q2
2C
and work done by a battery is saved as potential energy U:
PES 1120 Spring 2014, Spendier
U
Lecture 17/Page 4
Q2 1
1
 QV  CV 2 units [J] Joules
2C 2
2
Example 2: An air-spaced parallel plate capacitor has a capacitance of 1 μF and a plate
separation of 10-3 m. The charge on the capacitor is 6 nC. What is the force between the
plates in N? [Hint: use energy conservation.]
We can think of the capacitor as an electric analog of a stretched spring:
For capacitor: U 
For spring: U el 
Q 2 1  1  2
  Q
2C 2  C 
1 2
kx
2
With 1/C analogous to the spring constant, and Q, the charge analogous to the extent the
spring is compressed or stretched.
Note that the smaller the capacitance, the larger the “spring constant”, and the
harder it is to put charge on it.
PES 1120 Spring 2014, Spendier
Lecture 17/Page 5
Application of Energy storage by capacitors
Most practical applications of capacitors take advantage of their ability to store and
release energy. In electronic flash units used by photographers, the energy stored in a
capacitor is released by depressing the camera’s shutter button. This provides a
conducting path from one capacitor plate to the other through the flash tube. Once this
path is established, the stored energy is rapidly converted into a brief but intense flash of
light. An extreme example of the same principle is the Z machine at Sandia National
Laboratories in New Mexico, which is used in experiments in controlled nuclear fusion.
A bank of charged capacitors releases more than a million joules of energy in just a few
billionths of a second (109). For that brief space of time, the power (Energy per Time)
output of the Z machine is 2.9 x 1014 W, of about 80 times the electric power output of all
the electric power plants on earth combined!
Energy Stored in Electric Fields: Electric Energy Density in Vacuum
Let me play around a little with this result.
U
Q2 1
1
 QV  CV 2
2C 2
2
And now, let me look at the potential energy in the capacitor per volume (electric energy
density u) of a parallel-plate capacitor:
e0 A 2
V
U
1 CV 2 1 d
1 e0V 2 1
u


 e0 E 2
2
volume
2 Ad
2 Ad
2 d
2
This result turns out to be independent of the geometry of the capacitor, and is completely
general:
Energy is stored in all electric fields, and has a density:
1
u  e0 E 2
2
How can you compute U from u?
U   u dV
PES 1120 Spring 2014, Spendier
Lecture 17/Page 6
Example 3: Suppose you want to store 1.0 J of electric potential energy in a volume of
1.00 m3 in vacuum. What is the magnitude of the required electric field?
The value of E found in this example is sizable, corresponding to a potential difference of
nearly a half million volts over a distance of 1 meter.
Dielectrics: Why Dielectrics
There are three reasons why dielectrics are used in capacitors:
1) They act to give a solid separation between plates of a capacitor (which are usually
large, thin sheets of metal separated by a small distance).
2) They increase the dielectric breakdown of the capacitor by increasing the voltage that
can be applied before a spark occurs.
3) They increase the capacitance!
The first of these is self evident, but let’s explore the other two.
Dielectric Breakdown
•An insulator has no free electrons to allow charges to move, but they do have electrons.
•These electrons are bound in the atomic or molecular potentials.
•If a sufficiently large electric force acts on them, they can be pulled free from their
atoms or molecules, transforming the insulator into a conductor. This process is called
"ionization"
PES 1120 Spring 2014, Spendier
Lecture 17/Page 7
Examples:
- Lightning :The strong electric field causes the air around the cloud to "break down,"
allowing current to flow in an attempt to neutralize the charge separation. Simply stated,
the air breakdown creates a path that short-circuits the cloud/earth as if there were a long
metal rod
- Van de Graaff generator
Dry air has a dielectric strength of ~3x106V/m.
By choosing a material with a higher electric field of breakdown (higher dielectric
strength), once can avoid dielectric breakdown.
Dry air has a dielectric strength of ~3x106 V/m.
Polystyrene has a dielectric strength of ~24x106 V/m.
How can dielectrics increase capacitance?
Q
C
Vab
- increase charge
- decrease Vab (decrease the electric field)
Dielectric Constant
•Remember what happens when we put a dipole in an electric field?
•Even if the molecule has no inherent dipole moment, an electric field can cause one:
PES 1120 Spring 2014, Spendier
Lecture 17/Page 8
Now, let’s look at what happens when we put a block of dielectric material into an
electric field.
•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.
Hence we indeed end up with a reduced electric field between the plates!
 

Reduced electric field: E  E0  E '