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H4 Capacitors and Dielectrics (24.1-5)
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H4 Capacitors and Dielectrics (24.1-5)
Due: 11:59pm on Monday, September 28, 2015
You will receive no credit for items you complete after the assignment is due. Grading Policy
Description: Short conceptual problem involving capacitance and dielectric constants of capacitors. (ranking task)
Six parallel-plate capacitors of identical plate separation have different plate areas A, different capacitances C , and
different dielectrics filling the space between the plates. Below is a generic diagram of what each one of these capacitors
might look like.
Part A
Rank the following capacitors on the basis of the dielectric constant of the material between the plates.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Capacitance of a parallel-plate capacitor
The capacitance of an ideal parallel-plate capacitor is described by
C = κ ϵ 0dA ,
where A is the area of each plate, d is the plate separation, ϵ 0 is the permittivity of free space (a constant),
and κ is the dielectric constant of the material between the plates. Since ϵ 0 is a constant of nature and you
are told that d is the same for all of the capacitors, you may say that C ∝ κA . Solve this proportionality
expression for κ and you will be able to rank the capacitors on the basis of the dielectric constant.
All of the capacitors from Part A are now attached to batteries with the same potential difference.
Part B
Rank the capacitors on the basis of the charge stored on the positive plate.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Definition of capacitance
The capacitance of a capacitor is defined as the amount of charge that can be stored on the capacitor per
unit of potential difference ∆V applied to the capacitor:
Q
.
∆V
Note that Q represents the magnitude of the charge stored on either plate of the capacitor. Solve this
equation for Q and remember that ∆V is the same for each of the capacitors in this part, and you will be
able to determine a ranking on the basis of the charge stored on the positive plate for each of the capacitors.
C=
Exercise 24.2
Description: The plates of a parallel-plate capacitor are x apart, and each has an area of A. Each plate carries a
charge of magnitude q. The plates are in vacuum. (a) What is the capacitance? (b) What is the potential difference
between the plates? (c)...
The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of 9.92
charge of magnitude 4.45×10−8 C . The plates are in vacuum.
Part A
What is the capacitance?
C=
Part B
= 2.68
cm 2 . Each plate carries a
What is the potential difference between the plates?
= 16.6
V =
Part C
What is the magnitude of the electric field between the plates?
E=
= 5.07×106
Exercise 24.9
Description: A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is
negatively charged and the outer is positively charged; the magnitude of the charge on each is q. The inner cylinder
has a radius of r_in, the...
A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged
and the outer is positively charged; the magnitude of the charge on each is 20.0 pC . The inner cylinder has a radius of
0.500 mm , the outer one has a radius of 7.40 mm , and the length of each cylinder is 17.0 cm .
Part A
What is the capacitance?
Use 8.854×10−12 F/m for the permittivity of free space.
C=
= 3.51×10−12
F
Part B
What applied potential difference is necessary to produce these charges on the cylinders?
V =
= 5.70
V
Exercise 24.13
Description: A spherical capacitor is formed from two concentric, spherical, conducting shells separated by
vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer
sphere? (b) If the potential difference...
A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere
has radius 15.0 cm and the capacitance is 116 pF .
Part A
What is the radius of the outer sphere?
R = 0.175 m
Part B
If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?
Q = 2.55×10−8 C
± A Simple Network of Capacitors
Description: ± Includes Math Remediation. Given the capacitance of three capacitors connected in a network, and
the final charge on one of the capacitors, calculate the final charge on the other two capacitors and the potential
difference for the network.
In the figure are shown three capacitors with capacitances
C 1 = 6.00 µF, C 2 = 3.00 µF, C 3 = 5.00 µF . The
capacitor network is connected to an applied potential V ab .
After the charges on the capacitors have reached their final
values, the charge Q2 on the second capacitor is 40.0 µC .
Part A
What is the charge Q1 on capacitor C 1 ?
Hint 1. How to approach the problem
Consider only the initial section of the network (from a to d). Develop a relation between the capacitance and
charge for each capacitor in that section. Use the capacitances C 1 and C 2 and the final charge Q2 on C 2
to calculate the charge Q1 on C 1 .
Hint 2. Series or parallel?
Are capacitors C 1 and C 2 connected in series or in parallel?
series
parallel
Since the two capacitors have corresponding plates connected to the same side of the potential
difference in the first section, that is, the left plates are both connected to point a, and the right plates are
both connected to point d, they are connected in parallel.
Hint 3. Calculate the potential difference across the second capacitor
Calculate the potential difference V 2 across the second capacitor.
Hint 1. Equation for capacitance
Q
, where Q is the charge on each plate of
V
is the potential difference across the capacitor (not the difference across the
Recall that for any capacitor one has the relation C
the capacitor and V
=
whole network).
V 2 = 13.3 V
Hint 4. Calculate the potential difference across the first capacitor, V 1
Calculate the potential difference V 1 across the first capacitor.
Hint 1. Parallel capacitors and potential difference
If two capacitors are connected in parallel, how are the potential differences across each capacitor
related? Note that the left plate and right plate are connected to points a and d, respectively, for both
capacitors.
V 1 = 13.3 V
Since C 1 and C 2 are connected in parallel, the potential difference will be the same for both capacitors,
V 1 = V 2 = V ad .
Q1 = 80.0 µC
Part B
What is the charge on capacitor C 3 ?
Hint 1. How to approach the problem
Determine whether the third capacitor, C 3 , is connected in series or in parallel with the first section of the
network (from a to d). Calculate the total charge of the initial section of the network, Qad . Use the result to
find the charge on the third capacitor, Q3 .
Hint 2. Series or parallel?
Is the third capacitor, C 3 , connected in series or in parallel with the first section of the network (from a to d)?
series
parallel
Since the two capacitors in the initial section are connected between points a and d, and the third
capacitor C 3 is connected between points d and b, the initial section and the third capacitor are
connected together in series.
Hint 3. Capacitors in series
For capacitors connected in series, the positive plate of one capacitor is connected to the negative plate of
the other capacitor. Since the charge on the two plates must come from separation of charges in the initially
neutral plates (charge is never created or destroyed), the two plates must have charges of equal magnitude
and opposite signs. Therefore, the charge must be the same for the two capacitors. In this problem, the
network between a and d is in series with capacitor C 3 . Therefore, the total charge on that network must
equal the charge on C 3 .
Hint 4. Calculate the total charge in the initial section
Calculate Qad , the total charge stored in the initial section.
Hint 1. How to find the charge in the initial section
You already know the charge on the first two capacitors, Q1 and Q2 , in the initial section between
points a and d. The total charge must therefore be the sum of the charges in each capacitor.
Q3 = 120 µC
Part C
What is the applied voltage, V ab ?
Hint 1. How to approach the problem
Reduce the network of capacitors to a single equivalent capacitor connecting points a and b directly.
Determine the capacitance C ab and charge Qab on the equivalent capacitor. Then calculate the total
potential difference V ab .
Hint 2. Calculate the equivalent capacitance
Calculate the equivalent capacitance C ab of the network.
Hint 1. Calculate the equivalent capacitance of the initial section
Calculate the equivalent capacitance C ad of the capacitors in the initial section between points a and
d.
Hint 1. Capacitance for capacitors in parallel
The total capacitance for two capacitors with capacitances C α and C β connected in parallel is
given by C total = C α + C β .
Hint 2. Capacitance for capacitors in series
The total capacitance for two capacitors in series, C α and C β , is given by
1
Ctotal
=
1
Cα
+
1
.
Cβ
Remember to solve for C total and not its reciprocal!
C ab = 3.21 µF
Hint 3. Find the total charge
Find the total charge Qab stored in the equivalent capacitor.
Hint 1. Finding the charge of the equivalent capacitor
Remember that since the section between points a and d and the section between points d and b are
in series, the charge on the equivalent capacitor will be the same as the charge stored in each
section. Using the value of Q3 for the third capacitor in the network, as calculated in Part B, you can
show that Qab = Qad = Qdb = Q3 .
Qab = 120 µC
V ab = 37.3 V
Equivalent Capacitance
Description: Find the equivalent capacitance of a network of capacitors with series and parallel connections.
Consider the combination of capacitors shown in the diagram, where C 1 = 3.00 µF , C 2 = 11.0 µF , C 3 = 3.00 µF ,
and C 4 = 5.00 µF .
Part A
Find the equivalent capacitance C A of the network of capacitors.
Hint 1. How to reduce the network of capacitors
To find the equivalent capacitance of the given network of capacitors, it is most convenient to reduce the
network in successive stages. First, replace the capacitors C 2 , C 3 , and C 4 , which are in parallel, with a
single capacitor with an equivalent capacitance. By doing so, you will reduce the network to a series
connection of two capacitors. At this point, you only need to find their equivalent capacitance.
Hint 2. Find the capacitance equivalent to C2 , C3 , and C4
Find the capacitance C 234 equivalent to the parallel connection of the capacitors C 2 , C 3 , and C 4 .
Hint 1. Find the capacitance equivalent to C3 and C4
Find the capacitance C 34 equivalent to the parallel connection of the capacitors C 3 and C 4 .
Hint 1. Two capacitors in parallel
Consider two capacitors of capacitance C a and C b connected in parallel. They are equivalent
to a capacitor with capacitance C eq given by
C eq = C a + C b .
C 34 =
= 8.00
µF
If you replace the capacitors C 3 and C 4 with a capacitor of capacitance C 34 , the resulting
network would be a parallel connection between C 2 and C 34 .
C 234 =
= 19.0
µF
If you replace the capacitors C 2 , C 3 , and C 4 with a capacitor of capacitance C 234 , the resulting
network would be a series connection between C 1 and C 234 . Its equivalent capacitance is also the
equivalent capacitance of the original network.
Hint 3. Two capacitors in series
Consider two capacitors of capacitance C a and C b connected in series. They are equivalent to a capacitor
of capacitance C eq that satisfies the following relation:
1
Ceq
=
1
Ca
+
1
Cb .
CA =
= 2.59
µF
Part B
Two capacitors of capacitance C 5 = 6.00 µF and C 6 = 3.00 µF are added to the network, as shown in the
diagram. Find the equivalent capacitance C B of the new network of capacitors.
Hint 1. How to reduce the extended network of capacitors
To determine the equivalent capacitance of the extended network of capacitors, it is again convenient to
reduce the network in successive stages. First, determine the equivalent capacitance of the series
connection of the capacitors C 2 and C 6 . Then, combine it with the equivalent capacitance of the parallel
connection of C 3 , C 4 , and C 5 , and replace the five capacitors with their equivalent capacitor. The resulting
network will consist of two capacitors in series. At this point, you only need to find their equivalent
capacitance.
Hint 2. Find the equivalent capacitance of C2 , C3 , C4 , C5 , and C6
Find the equivalent capacitance C 2−6 of the combination of capacitors C 2 , C 3 , C 4 , C 5 , and C 6 .
Hint 1. Find the equivalent capacitance of C2 and C6
Find the equivalent capacitance C 26 of the series connection of C 2 and C 6 .
Hint 1. Two capacitors in series
Consider two capacitors of capacitance C a and C b connected in series. They are equivalent
to a capacitor of capacitance C eq that satisfies the following relation:
1
Ceq
=
1
Ca
+
1
.
Cb
C 26 =
= 2.36
µF
Hint 2. Find the equivalent capacitance of
C 345
C3 , C4 , and C5
C3 C 4
C5
Find the equivalent capacitance C 345 of the parallel connection of
C 3 , C 4 , and C 5
Hint 1. Three capacitors in parallel
Consider three capacitors of capacitance C a , C b , and C c connected in parallel. They are
equivalent to a capacitor with capacitance C eq given by
C eq = C a + C b + C c .
C 345 =
= 14.0
µF
C 2−6 =
= 16.4
µF
If you replace the capacitors C 2 , C 3 , C 4 , C 5 , and C 6 with a capacitor of capacitance C 2−6 , the
resulting network would be a series connection between C 1 and C 2−6 . Its equivalent capacitance is
also the equivalent capacitance of the original network.
Hint 3. Two capacitors in series
Consider two capacitors of capacitance C a and C b connected in series. They are equivalent to a capacitor
of capacitance C eq that satisfies the following relation:
1
Ceq
=
1
Ca
+
1
Cb .
CB =
= 2.54
µF
Exercise 24.26
Description: A parallel-plate vacuum capacitor has U of energy stored in it. The separation between the plates is
x_1. If the separation is decreased to x_2, (a) what is the energy now stored if the capacitor was disconnected from
the potential source before the...
A parallel-plate vacuum capacitor has 6.14 J of energy stored in it. The separation between the plates is 3.90 mm . If
the separation is decreased to 1.65 mm ,
Part A
what is the energy now stored if the capacitor was disconnected from the potential source before the separation of
the plates was changed?
U=
= 2.60
J
Part B
What is the energy now stored if the capacitor remained connected to the potential source while the separation of
the plates was changed?
U=
= 14.5
J
± Energy of a Capacitor in the Presence of a Dielectric
Description: ± Includes Math Remediation. Calculating the energy associated with a dielectric-filled capacitor. Can
be used as a follow-up to the problem "A Capacitor as an Energy Storing Device.
A dielectric-filled parallel-plate capacitor has plate area A = 15.0 cm 2 , plate separation d = 9.00 mm and dielectric
constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 7.50 V . Throughout the
2
problem, use ϵ 0 = 8.85×10−12 C /N ⋅ m 2 .
Part A
Find the energy \texttip{U_{\rm 1}}{U_1} of the dielectric-filled capacitor.
Hint 1. Energy of a charged capacitor in terms of its capacitance and voltage
The energy \texttip{U}{U} of a charged capacitor can be expressed, in terms of its capacitance \texttip{C}{C}
and voltage \texttip{V}{V}, as:
\large{U=\frac{1}{2}CV^2}.
You know the voltage, and can use the given quantities to calculate the capacitance.
Hint 2. Capacitance of a dielectric-filled capacitor
The capacitance of a dielectric-filled capacitor is given by:
\large{k\frac{\epsilon_0 A}{d}}
where \texttip{k}{k} is the dielectric constant, \texttip{A}{A} is the plate area, \texttip{d}{d} is the plate
separation and \texttip{\epsilon_0}{epsilon_0} = 8.85×10−12 {\rm C^2/N \cdot m^2} .
Hint 3. Find the capacitance
What is the capacitance \texttip{C}{C} of the dielectric-filled capacitor?
8.85×10−12 {\rm C^2/N \cdot m^2} .
\texttip{C}{C} =
= 5.90×10−12
{\rm C/V}
\texttip{U_{\rm 1}}{U_1} =
= 1.66×10−10
\rm J
Part B
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the
energy \texttip{U_{\rm 2}}{U_2} of the capacitor at the moment when the capacitor is half-filled with the dielectric.
Hint 1. What quantity remains constant?
Since the capacitor remains connected to the battery, its voltage \texttip{V}{V} = 7.50 {\rm V} remains
constant. However, both the capacitance and the charge of the capacitor change.
Hint 2. Modeling the capacitor
The half-filled capacitor can be viewed as a combination of two capacitors: one that is air-filled and the other
dielectric-filled. Each of them has the same voltage \texttip{V}{V} = 7.50 {\rm V} ; that is, these "halfcapacitors" are connected in parallel.
Hint 3. Finding the energy
To find \texttip{U_{\rm 2}}{U_2}, the total energy stored, first calculate the energy of the air-filled and the
dielectric-filled parts separately, then add them.
\texttip{U_{\rm 2}}{U_2} =
Part C
= 1.04×10−10
\rm J
The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out
of the capacitor. Find the new energy of the capacitor, \texttip{U_{\rm 3}}{U_3}.
Hint 1. What quantity remains constant?
Since the capacitor is now disconnected from the battery, each of the plates is isolated, and the charge of the
capacitor remains constant. The capacitance of the capacitor and the voltage across it are the changing
quantities in this process.
Hint 2. Energy of a charged capacitor in terms of its capacitance and charge
The energy \texttip{U}{U} of a charged capacitor can be expressed, in terms of its capacitance \texttip{C}{C}
and charge \texttip{Q}{Q}, as:
\large{U=\frac{1}{2}\frac{Q^2}{C}}.
You need to calculate the charge \texttip{Q}{Q} to use this equation.
Hint 3. Calculate the charge in the capacitor
Consider the half-filled capacitor before it is disconnected. Its energy is \texttip{U_{\rm 2}}{U_2}, which you
found in Part B. This energy can be written in terms of the charge \texttip{Q}{Q} and the voltage \texttip{V}{V}
as
\large{U_2=\frac{1}{2}QV}.
Use this expression to find the charge \texttip{Q}{Q} before the capacitor was disconnected.
\texttip{Q}{Q} =
= 2.77×10−11
\rm C
\texttip{U_{\rm 3}}{U_3} =
= 2.59×10−10
\rm J
Comparing the expressions for \texttip{U_{\rm 2}}{U_2} and \texttip{U_{\rm 3}}{U_3}, one can see that
U_3>U_2; in other words, the energy of the capacitor increases as the plate is being pulled out.
Part D
In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work
\texttip{W}{W} is done by the external agent acting on the dielectric?
Hint 1. Conservation of energy
According to the law of conservation of energy, U_3=U_2+W, since there is no change in the kinetic energy
of the dielectric plate.
\texttip{W}{W} =
= 1.56×10−10
\rm J
Since for every dielectric k>1, the work done in the last process is positive; in other words, an external agent
must apply a force to pull the plate out; the capacitor would exert a net force that would "resist" the pullout.
Energy in Capacitors and Electric Fields
Description: Several questions about the energy of charged capacitors, energy density, and energy of an
electrostatic field. Students are asked to calculate the energy of a charged capacitor by two different methods.
Learning Goal:
To be able to calculate the energy of a charged capacitor and to understand the concept of energy associated with an
electric field.
The energy of a charged capacitor is given by U=QV/2, where \texttip{Q}{Q} is the charge of the capacitor and \texttip{V}
{V} is the potential difference across the capacitor. The energy of a charged capacitor can be described as the energy
associated with the electric field created inside the capacitor.
In this problem, you will derive two more formulas for the energy of a charged capacitor; you will then use a parallel-plate
capacitor as a vehicle for obtaining the formula for the energy density associated with an electric field. It will be useful to
recall the definition of capacitance, C=Q/V, and the formula for the capacitance of a parallel-plate capacitor,
C=\epsilon_0 A/d, where \texttip{A}{A} is the area of each of the plates and \texttip{d}{d} is the plate separation. As usual,
\texttip{\epsilon _{\rm 0}}{epsilon_0} is the permittivity of free space.
First, consider a capacitor of capacitance \texttip{C}{C} that has a charge \texttip{Q}{Q} and potential difference \texttip{V}
{V}.
Part A
Find the energy \texttip{U}{U} of the capacitor in terms of \texttip{C}{C} and \texttip{Q}{Q} by using the definition of
capacitance and the formula for the energy in a capacitor.
\texttip{U}{U} =
Part B
Find the energy \texttip{U}{U} of the capacitor in terms of \texttip{C}{C} and \texttip{V}{V} by using the definition of
capacitance and the formula for the energy in a capacitor.
\texttip{U}{U} =
All three of these formulas are equivalent:
\large{U=\frac{QV}{2}=\frac{Q^2}{2C}=\frac{CV^2}{2}}.
Depending on the problem, one or another may be more convenient to use. However, any one of them would
give you the correct answer. Note that these formulas work for any type of capacitor.
Part C
A parallel-plate capacitor is connected to a battery. The energy of the capacitor is \texttip{U_{\rm 0}}{U_0}. The
capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles.
The new energy of the capacitor is \texttip{U}{U}. Find the ratio U/U_0.
Hint 1. Determine what remains constant
As the plates are being pulled apart slowly, what quantity or quantities remain constant?
capacitance only
voltage only
charge only
both voltage and capacitance
both voltage and charge
Since the geometry of the capacitor is changing, its capacitance changes, too. However, the voltage
remains constant, since it must equal the voltage provided by the battery.
Hint 2. Identify which formula to use
Which formula for energy is most convenient to use in this case?
\large{\frac{QV}{2}}
\large{\frac{Q^2}{2C}}
\large{\frac{CV^2}{2}}
\large{\frac{U}{U_0}} = 0.5
Part D
A parallel-plate capacitor is connected to a battery. The energy of the capacitor is \texttip{U_{\rm 0}}{U_0}. The
capacitor is then disconnected from the battery and the plates are slowly pulled apart until the plate separation
doubles. The new energy of the capacitor is \texttip{U}{U}. Find the ratio U/U_0.
Hint 1. Determine what remains constant
As the plates are being pulled apart, what quantity or quantities remain constant?
capacitance only
voltage only
charge only
both voltage and charge
both voltage and capacitance
The charge remains constant, since the capacitor is disconnected and the charge therefore literally has
nowhere to go.
Hint 2. Identify which formula to use
Which formula for energy is most convenient to use in this case?
\large{\frac{QV}{2}}
\large{\frac{Q^2}{2C}}
\large{\frac{CV^2}{2}}
\large{\frac{U}{U_0}} = 2
In this part of the problem, you will express the energy of various types of capacitors in terms of their geometry and
voltage.
Part E
A parallel-plate capacitor has area \texttip{A}{A} and plate separation \texttip{d}{d}, and it is charged to voltage
\texttip{V}{V}. Use the formulas from the problem introduction to obtain the formula for the energy \texttip{U}{U} of
the capacitor.
Express your answer in terms of \texttip{A}{A}, \texttip{d}{d}, \texttip{V}{V}, and appropriate constants.
\texttip{U}{U} =
Let us now recall that the energy of a capacitor can be thought of as the energy of the electric field inside the capacitor.
The energy of the electric field is usually described in terms of energy density \texttip{u}{u}, the energy per unit volume.
A parallel-plate capacitor is a convenient device for obtaining the formula for the energy density of an electric field, since
the electric field inside it is nearly uniform. The formula for energy density can then be written as
\large{u=\frac{U}{V}},
where \texttip{U}{U} is the energy of the capacitor and \texttip{V}{V} is the volume of the capacitor (not its voltage).
Part F
A parallel-plate capacitor has area \texttip{A}{A} and plate separation \texttip{d}{d}, and it is charged so that the
electric field inside is \texttip{E}{E}. Use the formulas from the problem introduction to find the energy \texttip{U}{U}
of the capacitor.
Express your answer in terms of \texttip{A}{A}, \texttip{d}{d}, \texttip{E}{E}, and appropriate constants.
Hint 1. How to approach the problem
Recall that for the uniform electric field \texttip{E}{E} between the plates of a parallel-plate capacitor, V=Ed,
where \texttip{V}{V} is the potential difference between the plates and \texttip{d}{d} is the distance between
the two plates. You can use this relation to rewrite the equation for energy \large{U=\frac{_1}{^2}CV^2} in
terms of the electric field and the geometry of the capacitor (i.e., the area of the plates and the distance
between them).
\texttip{U}{U} =
As mentioned before, we can think of the energy of the capacitor as the energy of the electric field inside the
capacitor.
Part G
Find the energy density \texttip{u}{u} of the electric field in a parallel-plate capacitor. The magnitude of the electric
field inside the capacitor is \texttip{E}{E}.
Hint 1. How to approach the problem
Since the electric field outside a parallel-plate capacitor is essentially zero, the volume that you are looking
for is the volume of the space between the two plates.
Hint 2. Volume between the plates
Recall that the volume \texttip{V}{V} of a solid with two parallel bases of the same shape and sides
perpendicular to the bases is given by V=Ah, where \texttip{A}{A} is the area of each of the bases and
\texttip{h}{h} is the distance between the bases. Note that the space between the plates of a parallel-plate
capacitor is such a solid.
\texttip{u}{u} =
Note that the answer for \texttip{u}{u} does not contain any reference to the geometry of the capacitor:
\texttip{A}{A} and \texttip{d}{d} do not appear in the formula. In fact, the formula
\large{u=\frac{\epsilon_0 E^2}{2}}
describes the energy density in any electrostatic field, whether created by a capacitor or any other source.
The Capacitor as an Energy-Storing Device
Description: A tutorial containing quantitative questions related to the "work-energy" relationships as applied to
capacitors (including the insertion of a dielectric slab). The problem "Energy of a capacitor in the presence of
dielectric" can be used as a follow-up to this.
Learning Goal:
To understand that the charge stored by capacitors represents energy; to be able to calculate the stored energy and its
changes under different circumstances.
An air-filled parallel-plate capacitor has plate area \texttip{A}{A} and plate separation \texttip{d}{d}. The capacitor is
connected to a battery that creates a constant voltage \texttip{V}{V}.
Part A
Find the energy \texttip{U_{\rm 0}}{U_0} stored in the capacitor.
Express your answer in terms of \texttip{A}{A}, \texttip{d}{d}, \texttip{V}{V}, and \texttip{\epsilon _{\rm 0}}
{epsilon_0}. Remember to enter \texttip{\epsilon _{\rm 0}}{epsilon_0} as epsilon_0.
Hint 1. Formula for the energy of a capacitor
Recall that the equation for the energy of a capacitor is \large{U=\frac{1}{2}CV^2}.
\texttip{U_{\rm 0}}{U_0} =
Part B
The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until
the separation reaches 3d. Find the new energy \texttip{U_{\rm 1}}{U_1} of the capacitor after this process.
Express your answer in terms of \texttip{A}{A}, \texttip{d}{d}, \texttip{V}{V}, and \texttip{\epsilon _{\rm 0}}
{epsilon_0}.
Hint 1. What quantity remains constant?
What characteristic of the capacitor does not change in this process?
charge
voltage between the plates
capacitance
energy
In particular, look carefully at the quantities that do change.
Hint 2. Find the charge on the capacitor
What is the charge \texttip{Q}{Q} that resides on the plates of the capacitor?
Express your answer in terms of some or all of the variables \texttip{V}{V}, \texttip{A}{A}, and
\texttip{d}{d}. Remember to enter \texttip{\epsilon _{\rm 0}}{epsilon_0} as epsilon_0.
Q=
Hint 3. How does the capacitance change?
How does the capacitance change in this process?
remains constant
increases by a factor of 3
decreases by a factor of 3
increases by a factor of 9
decreases by a factor of 9
Hint 4. What is the formula for the energy?
Which of the following formulas is most useful in finding \texttip{U_{\rm 1}}{U_1} for this situation?
\large{\frac{CV^2}{2}}
\large{\frac{QV}{2}}
\large{\frac{Q^2}{2C}}
\texttip{U_{\rm 1}}{U_1} =
The increase in energy of the capacitor comes from the external work that must be done to pull the plates apart.
Keep in mind that the plates have opposite charges and attract each other; some work must be done by an
external agent to pull them apart.
Part C
The capacitor is now reconnected to the battery, and the plate separation is restored to \texttip{d}{d}. A dielectric
plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy
\texttip{U_{\rm 2}}{U_2} of the dielectric-filled capacitor. The capacitor remains connected to the battery. The
dielectric constant is \texttip{K}{K}.
\texttip{\epsilon _{\rm 0}}{epsilon_0}.
\texttip{U_{\rm 2}}{U_2} =
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