Physics 227: Lecture 8
Dielectrics, and Capacitors
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Lecture 7 review:
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Capacitors come in many sizes and shapes
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Q=CV - charge ±Q on the two plates, voltage V across them
Energy Stored = U = Q2/2C = CV2/2
u = U/Volume = ½ε0E2
In a circuit, capacitors in parallel add, capacitors in series
add inversely
Thursday, September 29, 2011
Capacitive Iclicker Puzzle
Student A says: The plates of my parallel plate
capacitor attract each other, so I have to hold them
to keep them apart.
Student B says: The plates of my capacitor repel each
other, so I have to hold them to keep them together.
Who is right?
A. A.
B. B.
C. Both A and B.
D. Neither.
E. This does not make sense.
Thursday, September 29, 2011
Remember your answer to
this brain teaser. We will
find out the right answer
in the next few slides.
Capacitive Iclicker Puzzle
Student A says: The plates of my parallel plate
capacitor attract each other, so I have to hold them
to keep them apart.
Student B says: The plates of my capacitor repel each
other, so I have to hold them to keep them together.
Who is right?
A. A.
B. B.
C. Both A and B.
D. Neither.
E. This does not make sense.
Thursday, September 29, 2011
Remember your answer to
this brain teaser. We will
find out the right answer
in the next few slides.
Changing Capacitance
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A parallel plate capacitor has Q = CV with capacitance C = ε0A/d.
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It is natural to think that, since the plates have opposite charge,
the forces between them are attractive. Student A is right!?
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How can we change the capacitance? The easy way is to move the
plates closer together or farther apart.
Let’s look in more detail at what happens when the plates come
closer together ...
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Since Q is the same, the electric field is constant, so the energy
density u = ½ε0E2 is constant.
Constant u with decreasing volume → less stored potential
energy, so U is being converted into kinetic energy of the plates.
It all makes sense, student A was right! The force is attractive,
the plates would naturally come together, they have to be held
apart.
Thursday, September 29, 2011
Changing Capacitance - Not so fast
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A parallel plate capacitor has Q = CV with capacitance C = ε0A/d.
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What if the capacitor is in a circuit? Do we get the same answer?
We made an assumption: ``Since Q is the same...’’ We assumed the
capacitor is not connected to a circuit and the charge is constant.
Now the voltage V is fixed. When we move the capacitor plates, V
stays the same.
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Since V = Ed, if we push the plate together, the distance
decreases, so a larger field is needed. If we pull the plates
apart, a smaller field is needed.
Let’s change the distance between the plates a factor of 2. The
field changes a factor of 2, the opposite way, to keep the voltage
fixed. What happens to the energy?
Thursday, September 29, 2011
Changing Capacitance - Not so fast
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A parallel plate capacitor has Q = CV with capacitance C = ε0A/d.
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Increasing d by a factor of 2 decreases C and U by a factor of 2.
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What happens to the energy?
With the voltage V fixed, and U = ½CV2, the energy stored is
proportional to C.
The force between the plates is in the direction to reduce the
potential energy, so the force would increase d - it is repulsive!
Does this make sense considering the energy stored in the electric
field? With constant V and d bigger by a factor of 2, V = Ed tells
us that the field E and the charge Q decrease by a factor of 2.
The energy density u = ½ε0E2 decreases a factor of 4, the volume
Ad increases a factor of 2, so the stored energy is 2 times less,
agreeing with the result above
Student B is right!
Thursday, September 29, 2011
Summary
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A parallel plate capacitor has Q = CV with capacitance C = ε0A/d.
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Student B has a capacitor (in a circuit) with a fixed voltage V, and
finds the force between the plates is repulsive.
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Student A has a capacitor with a fixed charge Q, and finds the
force between the plates is attractive.
They are both right. The physical situations are different, though
they seem the same when crudely described.
Thursday, September 29, 2011
Dielectrics in
Capacitors
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Capacitance of capacitors
is increased by putting a
dielectric material in
between the conductor
plates.
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It also allows a higher
voltage across the
capacitor than vacuum,
and provides mechanical
support.
How does this work?
When you put a material
in an electric field, it can
become polarized...
Thursday, September 29, 2011
Dielectrics
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This leads to a surface charge density on the dielectric that
reduces the electric field inside the dielectric to be smaller
than the applied field.
Thursday, September 29, 2011
Voltage on a Capacitor filled with a Dielectric
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When you fill the volume of a capacitor with a dielectric, the
electric field decreases, so the voltage across the capacitor
decreases - for fixed charge.
Since Q = CV = constant, V decreasing requires that C increases.
Thursday, September 29, 2011
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Dielectric Constant
We can define the dielectric constant as the ratio of the
capacitance of a capacitor with a dielectric to the capacitance
with vacuum between the plates: K = C / C0.
So C0 = ε0A/d → C = Kε0A/d.
For constant charge Q, E = E0/K, V = V0/K
Table from textbook, page 801
Thursday, September 29, 2011
Dielectric iClicker
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We can define the dielectric constant as the ratio of the
capacitance of a capacitor with a dielectric to the capacitance
with vacuum between the plates: K = C / C0.
So C0 = ε0A/d → C = Kε0A/d.
For constant charge Q, E = E0/K, V = V0/K
By what factor does the stored
potential energy change, for
constant charge Q?
U/U0 = ?
A. same - 1.
B. bigger by K
C. smaller by K.
D. bigger by K2.
E. smaller by K2.
Thursday, September 29, 2011
Dielectric iClicker
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•
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We can define the dielectric constant as the ratio of the
capacitance of a capacitor with a dielectric to the capacitance
with vacuum between the plates: K = C / C0.
So C0 = ε0A/d → C = Kε0A/d.
For constant charge Q, E = E0/K, V = V0/K
By what factor does the stored
potential energy change, for
constant charge Q?
U/U0 = ?
Three ways to get the same answer:
U’ = C’V’2/2 = KC(V/K)2/2 = CV2/2K = U/K
U’ = Q’2/2C’ = Q2/2KC = U/K
u’ = εE’2/2 = Kε0(E/K)2/2 = ε0E2/2K = u/K
Thursday, September 29, 2011
A. same - 1.
B. bigger by K
C. smaller by K.
D. bigger by K2.
E. smaller by K2.
Splitting a Capacitor into Two
d
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+
Area A
+
+
+ +
+
+ + +Q
-
-
-
-
-Q
d
+
Area A/2
+
+
+ +Q/2
-
-
-Q/2
+
Area A/2
+
+
+
+Q/2
-
-
-Q/2
Cut in the middle separate the two halves.
A parallel plate capacitor has Q = CV with capacitance C = ε0A/d.
There is the same electric field in all the capacitors.
There is the same voltage across all the capacitors.
The charge on the left capacitor is the sum of the charges on the
right two.
The energy stored in the left capacitor is CV2/2. Each capacitor on
the right has half as much energy, since C is half as large.
This is what we expect from adding capacitors in parallel.
Thursday, September 29, 2011
Now consider Inserting a Dielectric Into a
Capacitor
What is the force on
the dielectric?
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There is constant voltage across the capacitor.
To figure this out, we are going to divide
the capacitor into two halves, a half with
dielectric between the plates and a half
with vacuum between the plates.
Thursday, September 29, 2011
Splitting a Capacitor into Two
dielectric
constant K
d
+
Area A/2
+
+
+ +Q/2
-
-
-Q/2
+
Area A/2
+
+
+
+Q/2
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dielectric
constant 1
-Q/2
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A parallel plate capacitor has Q = CV with capacitance C = ε0A/d.
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Left side: Voltage V, field V/d, energy stored ½CV2 = ½(Kε0A/2d)V2 =
¼Kε0AV2/d. But: charge density σ = ε0V/d comes from the charge
on the capacitor plate plus the surface charge of the dielectric.
Since the field is reduced a factor of K, σcapacitor increases a factor
of K to keep V constant. Thus, σcapacitor = K ε0V/d and Q = Kε0AV/2d.
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Right side: Voltage V, field V/d, energy stored ½CV2 = ½(ε0A/2d)V2 =
¼ε0AV2/d. Also: charge density σ = ε0V/d, charge Q = ε0AV/2d,
The battery supplies charge to keep the voltage constant.
Thursday, September 29, 2011
Inserting a Dielectric Into a Capacitor in a
Circuit
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What happens if there is constant voltage across the capacitor?
What is the force on
the dielectric? What
would happen if it could
move freely?
A. It is pulled in to the capacitor.
B. It is pushed out of the capacitor.
C. There is no force on the dielectric.
D. There is no sideways force, there is a vertical force.
E. Uh ...
Thursday, September 29, 2011
Inserting a Dielectric Into a Capacitor in a
Circuit
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What happens if there is constant voltage across the capacitor?
What is the force on
the dielectric? What
would happen if it could
move freely?
Forces push ``downhill’’, A. It is pulled in to the capacitor.
towards smaller
B. It is pushed out of the capacitor.
potential energy. For
constant V capacitor,
C. There is no force on the dielectric.
that is without the
dielectric. (For constant D. There is no sideways force, there is a vertical force.
Q, it is with the
E. Uh ...
dielectric.)
Thursday, September 29, 2011
Thank you.
Prof. Gilman will be back Monday Oct. 3.
Exam 1 is 9:40 PM Thursday Oct. 6. There
will be optional review sessions instead of
lectures.
You are allowed one formula sheet. Anyone
with conflicts or needing special
arrangments should contact Prof.
Cizewski ASAP.
Thursday, September 29, 2011