CAPACITORS AND CAPACITIVE CIRCUITS
Any two conductors separated by an insulator can be electrically charged so that one conductor has a positive charge and
the other conductor has an equal amount of negative charge; such an arrangement is called a capacitor. A capacitor can
be made up of two strange-shaped blobs of metal or it can have any number of regular symmetric shapes such as that of
one hollow sphere inside another, or one hollow rod inside another.
The type of capacitor that is of the most practical interest is the parallel plate capacitor. Thus, we will focus
exclusively on the study of the properties of parallel plate capacitors. There are a couple of reasons why you will be
studying parallel plate capacitors. First, the parallel plate capacitor is the easiest to use when making mathematical
calculations or using physical reasoning. Second, it is relatively easy to construct. Third, parallel plate capacitors are
used widely in electronic circuits to do such diverse things as defining the flashing rate of a neon tube, determining what
radio station will be tuned in, and storing electrical energy to run an electronic flash unit. Materials other than
conductors separated by an insulator can be used to make a system that
behaves like a simple capacitor. Although many of the most interesting
properties of capacitors come in the operation of alternating current circuits,
we will limit our present study to the properties of the parallel plate
capacitor and its behavior in direct current circuits like those you have been
constructing in the last couple of units. The circuit symbol for a capacitor
is a pair of lines as shown below.
CAPACITORS

A Capacitor is a device that stores electric charge and electrical potential energy.

Capacitance is a measure of the ability of a device to store charge per unit of voltage applied across the device.
Definition:
The capacitance of a given capacitor is defined mathematically as the ratio of the magnitude of the charge, q, on either
one of the conductors to the voltage, V, applied across the two conductors
Thus, capacitance is defined as a measure of the amount of net or excess charge on either one of the conductors
per unit voltage.
C = Capacitance
(SI: Farad = F = C/V)
Q = Charge stored in the capacitor
(SI: C = Coulomb)
V = Voltage drop across the capacitor
(SI: V = Volt)

The value of this definition is that the capacitance C is a constant that only depends upon the structure of the
capacitor and does not change when a different voltage is applied across the capacitor.
Charging a Capacitor:

The important point to observe is that the voltage across the capacitor is not instantaneously equal to that of the
voltage across the battery when the switch is closed. The voltage on the capacitor builds up as more and more
charge flows onto the capacitor until the "back" voltage produced by the capacitor is equal to the voltage of the
battery.
Energy Storage:

The work it takes to charge up a capacitor is equal to the electrical energy stored in a capacitor.

The work it takes to move a differential charge dq through a voltage V is V dq. Integrating from when there is
no charge on the capacitor until the capacitor is fully charge with a charge Q we find

Alternate forms of the energy stored in a capacitor:
You can draw on some of your experiences with electrostatics to think about what might happen to a parallel
plate capacitor when it is hooked to a battery. This thinking can give you an intuitive feeling for the meaning of
capacitance. For a fixed voltage from a battery, the net charge found on either plate is proportional to the capacitance of
the pair of conductors.
Activity: Predicting Dependence on Area and Separation
a. Consider two identical metal plates of area A, separated by a
nonconducting material that has a thickness d. They are
connected in a circuit with a battery and a switch, as shown
above. When the switch is open, there is no excess charge on
either plate. The switch is then closed. What will happen to
the amount of charge on the metal plate that is attached to the
negative terminal of the battery? What will happen to the
amount of charge on the plate that is connected to the positive
terminal of the battery? Explain.
b. Can excess charges on one plate of a charged parallel plate capacitor interact with excess charges on the other
plate? If so, how? Note: To say that two charges interact is to say that they exert forces on each other from a
distance.
c. Is there any limit to the amount of charge that can be put on a plate? Explain.
d. Use qualitative reasoning to anticipate how the amount of charge a pair of parallel plate conductors can hold will
change as the area of the plates increases. Explain your reasoning.
e. Do you think that the amount of charge a given battery can store on the plates will increase or decrease as the
spacing, d, between the plates of the capacitor increases? Explain.
CAPACITANCE MEASUREMENTS FOR PARALLEL PLATES
The unit of capacitance is the farad, F, named after Michael Faraday. One farad is equal to one coulomb/volt. As you
will demonstrate shortly, one farad is a very large capacitance. Thus, actual capacitances are often expressed in smaller
units with alternate notation as shown below:
Units of capacitance
-6
microfarad: 10 F=1 F=1 UF
picofarad: 10-12 F =1 pF = 1 F = 1 UUF
nanofarad: 10-10 F = 1 nF = 1000 F = 1000 UUF
-6
Note: Sometimes the symbol m is used instead of  or U on capacitors to represent 10 , despite the fact that in other
-3
situations m always represents 10 !
Typically, there are several types of capacitors used in electronic circuits, including disk capacitors, foil
capacitors, electrolytic capacitors, and so on. You might want to examine some typical capacitors. To do this you’ll
need:
• 4 capacitors (assorted collection)
To complete the next few activities you will need to construct a parallel plate capacitor and use a multimeter to
measure capacitance. Thus, you’ll need the following items:
•
•
•
•
•
•
2 pieces aluminum foil, 12 cm  12 cm
1 textbook
1 multimeter (with a capacitance mode)
2 insulated wires (stripped at the ends, approximately 12" long)
1 ruler
1 Vernier caliper
You can make a parallel plate capacitor out of two rectangular sheets of aluminum foil separated by pieces of
paper. A textbook works well as the separator for the foil since you can slip the two foil sheets between any number of
sheets of paper and weight the book down with something heavy and non-conducting like another massive textbook.
You can then use your digital multimeter in its capacitance mode for the measurements. Note: Insert short wires into
the capacitance slots of your multimeter as “probes”
When you measure the capacitance of your “parallel plates,” be sure the aluminum foil pieces are arranged
carefully so they don’t touch each other and “short out.”
Activity: Measuring How Capacitance Depends on Area or on Separation
a. Devise a way to measure how the capacitance depends on either the foil area or on the separation between foil
sheets. If you hold the area constant and vary separation, record the dimensions of the foil so you can calculate
the area. Alternatively, if you hold the distance constant, record its value. Take at least five data points in either
case. Describe your method and then create a data table with proper units and display a graph of the results.
b. Is your graph a straight line? If not, you should make a guess at the functional relationship it represents and
create a model that matches the data. Affix your overlay graph showing both the data and your model in the
space the follows. Be sure to label your graph axes properly.
c. What is the function that best describes the relationship between spacing and capacitance or between area and
capacitance? How do the results compare with your prediction based on physical reasoning?
d. Use the ohmmeter on a piece of paper in your book. What is its resistance? Can current flow through the pages
of your book?
DERIVING A MATHEMATICAL EXPRESSION FOR CAPACITANCE
We can use what we know about charges in
conductors and the relationship between
potential difference, V, and electric field to
derive an expression for the capacitance, C,
of a parallel plate capacitor in terms of the
area, A, and separation, d, of the aluminum
plates. The diagram in at right is useful in
this regard.
Activity: Derivation of Capacitance vs. A and d
a. Given that Q = EA = 0EA where the net positive charge on the bottom plate is denoted by Q. Notes: (1)
We are assuming that since the positive charges on the bottom plate are attracted to the negative charges on
the top plate, the excess charges are on the inside surface of each plate. Remember that in a uniform electric
field the voltage drop over a distance d is given by V = Ed. Plug the definition for charge and voltage into
the definition of charge to write an expression for capacitance as a function of 0dA
C
 0 A
d
where A is the area of the plates and d is their separation.
b. Use one of your actual areas and spacings from the measurements you made to calculate a value of C. Assume
that the dielectric constant, , for paper is about 3.5. How does the calculated value of C compare with the
directly measured value? Note:  (kappa) is a quantity called the dielectric constant and is a property of the
insulating material that separates the two plates. For air,  = 1. It is usually greater than 1 for other materials.
c. Now for an unusual question. If you have two square foil sheets, separated by paper with a dielectric constant of
3.5 that is 1 mm thick, how long (in miles) would each side of the sheets have to be in order to have C = 1 F?
Show your calculations Warning: Miles are not meters.
L=
___________________________
miles
d. Your capacitor would make a mighty large circuit element! How could it be made smaller physically and yet
still have the same value of capacitance? You may want to examine the collection of sample capacitors for some
ideas
. PARALLEL-PLATE CAPACITORS:
o = Permittivity of empty space ( air o)
= 8.854x10-12 C 2/N.m2 = 1/(4k)
 = Permittivity of material between plates
A = Surface Area of one of the plates (SI: m 2)
d = Separation of the plates (SI: m)

Electric Field in a Ideal Parallel Plate Capacitor:

The Electric Field due to a sheet of charged infinite sheet is very simple in that the E-field is constant at any
distance from the sheet.

When two plate of different charge are placed near each other, the two E-fields between the plates add while the
E-field outside the plate cancel.

When the plates are close to each other to form a capacitor, the E-field between the plates is constant through
out the interior of the capacitor as long as one is not near the edges of the plates.

Since the electric field is the negative of the gradient of the potential and the E-field is constant inside a
capacitor, the magnitude of the Electric field has a very simple relation to the voltage between the plates and
their separation d.

This relationship is also true in an electrical wire were V is the voltage across the ends of the wire and d is the
length of the wire.

Using the definition of capacitance we can determine the capacitance C of an ideal capacitor as a function of its
structure.

This equation for the capacitance of a parallel capacitor shows that C is a constant independent of the charge
stored in on the plates or the voltage across the capacitor.

By placing a thin insulating material (a dielectric) between the plates the separation d can be reduced thus
increasing the capacitance of the capacitor and prevent the plates from touching.

It takes more voltage to store the same amount of charge on a capacitor because of the presence of the dielectric.
Typically a dielectric contains polar molecules which partially line up in the presence of the electric field. The
dielectric creates an E-field in the opposite direction which reduces the overall E-field between the plates.

You could calculate the capacitance C of a parallel plate capacitor by replacing the permittivity of empty space
o by the permittivity  of the dielectric material placed between the plates.

Except for simple capacitors (like the parallel plate capacitor) we do have an equation from which we can
calculate the capacitance C. In practice we simply measure the value of a capacitor and assume that it is
constant.
Energy Density of the Electric Field in a Capacitor:

The electrical energy stored in the Electric Field between the plates of an ideal capacitor has a simple form when
expressed as the electrical energy per unit volume, u = U/Vol

This is a general expression that is valid for the energy density of the Electric Field no matter how the electric
field is generated, i.e. it is true at any point in space where there is an electric field E.

For parallel plate capacitors this can easily be derived since the E-field is constant through out the interior of the
capacitor and equal to V/d. Here,
EARTH AS A SPHERICAL CAPACITOR:
Approximate the Earth as a spherical shell with some excess charge –Q surrounded by another spherical shell
with and equal but opposite charge +Q.
Voltage drop between two charged spherical shells:
Spherical Capacitance:
Earth:
Let R1 = 6,378 km, the radius of the Earth.
Let R2 approach infinity, or simply let R2 >> R1, then
DIELECTRICS






Most Capacitors have nonconducting material - a dielectric - between their plates.
Dielectric materials increase the capacitance C of a capacitor; usually by several orders of magnitude compared
to air filled capacitors.
Dielectrics help keep the electrical plates separated without electrical contact. This also allows the separation of
the plates to be reduced and thus increasing the capacitance.
When a dielectric is placed between the plates of a capacitor, the permittivity of the space between the capacitors
change from that of empty space o to that of the dielectric material 
For practical reasons it is more useful to define and use dielectric constant  which is the ratio of o over when
determining the change in capacitance due to the presence of a dielectric.
The Dielectric Strength is the maximum Electric Field that the material can withstand before the material breaks
down as an insulator and permits current to flow through the material.
Material
Dielectric Constant Dielectric Strength
(V/m)
 =  / o
Vacuum
1
Air (1 atm)
1.00054
Air (100 atm)
1.0548
1-3 x10 6
4.9
24x106
4.5-5.5
15x106
Polystyrene (typical)
2.6
25x106
Mylar (typical)
3.5
6-13 x10 6
Paper (typical)
3.5
14x106
Porcelain (typical)
7
4x106
Teflon (typical)
2
60x106
Mineral Oil (typical)
4.5
12x106
Water (20 oC)
80.4
Conductor
Bakelite (typical)
Glass (Pyrex)
Effects of the Presence of Dielectric on Capacitors
Voltage Held Constant
Charge Held Constant
(The Capacitor is connected to an constant emf source and
(The Capacitor is charged up by an emf source and then
then the dielectric is inserted. Equivalently, the capacitor with
disconnected. The dielectric is then inserted when
dielectric are connected to a constant voltage source.)
capacitor is not connected to a voltage source.)
The Voltage remains constant during the process.
The Charge remains constant during the process.
C = Co
C = Co
Q = Qo
Q = Qo
V = Vo
V = Vo /
E = Eo
E = Eo /
U =  Uo
U = Uo /
Parallel Plate Capacitor Problem
A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V
battery. After the capacitor is completely charge up, the battery is removed.
(A) What must be the separation of the plates and the charge density on the plates?
(B) What is the magnitude of the electric field between the plates, and how much energy is stored in the electric field
between the plates?
(C) If a sheet of Mylar is placed between the plates, how will the voltage across the plates and the charge density on
the plates change?
(D) If the sheet of Mylar has an area of 3.00 cm 2 so that it does not cover the whole area of the plates, what would be
the effective capacitance of this arrangement?
(E) If the Mylar is removed and the plates are moved 1.00 mm farther from each other, what will be the voltage
between the plates, and how much work will it take to move the plates apart?
Sketch and Process:
Charge is placed on a capacitor of known area and capacitance.
Relevant Physics:
For parallel plate capacitors:
a. The electric field is constant between the parallel plates and can be determined from,
b. The capacitance can be calculated from either the definition of capacitance or the capacitance of parallel
plates,
c. The energy stored in the electric field between the plates is the same as the work required to charge up a
capacitor.
For parts C, D and E, it is important to keep in mind that no voltage source is connected. Thus, the charge on
the capacitor can not change (conservation of charge) even if a dielectric is inserted or the separation of the
plates is changed. However, both the capacitance and voltage can and do change.
In part D, we can model the partial covered plates as two capacitors in parallel with each other. One (2 cm 2
area) with air between plates, and the other (3 cm 2 area) with Mylar. Once we have calculated each ones
capacitance we simply add the two capacitances since they are in parallel.
(A) Find d and .
Using the equation for the capacity of a parallel plate capacitor,
The surface charge is constant, thus  = Q/A and we will need to find the charge Q stored on one of the plates.
Using the definition of capacitance C = Q/V
Then,
(B) Find E and U.
We could find the E-field between the plates from our knowledge the charge density on the plates,
As a double check,
The energy stored,
This also represents the energy the capacitor discharges if its ends are connected to say a resistor.
(C) Find V and  if Mylar is placed between the plates.
Note this is not the same problem if the Mylar was placed between the plates and then the battery was
connected to charge up the plates. In this case, the charge on the plates would be different from the value we
found in part A. For the problem at hand, there is no were for the charge go so it remains on the plates. See
Effects of the Presence of Dielectric on Capacitors for more details.
In both problems, the permittivity would change from that of free space o to that of  - the permittivity of
Malar.
Looking up the dielectric constant for Mylar, we find it to be 3.5. This means that
Since capacitance depends upon the permittivity, then
Thus, we can that the capacitance has increased,
Since C = Q/V and Q does not change, then the voltage drop.
One advantage of adding a dielectric is that if the capacitor is connected to a battery source, the battery will
supply more charge. This means that the effective capacitance goes up.
(D) Find C when a 3.00 cm2 Mylar sheet is placed between the capacitor's plates.
The key to solving this problem is the realization that any parallel plate capacitor could be broken up into two
capacitors, both in parallel with each other provided we ignore all edge effects. The main requirement is that
the total surface area of the two capacitors is same as the original capacitor. The total or effective capacitance
of two parallel capacitors is just their sum.
For the problem at hand, we envision breaking the capacitor into one air filled capacitor with an area of 2.00
cm 2, and one Mylar filled capacitor with an area of 3.00 cm 2.
Since we know the separation (from part A) and the area of each capacitor, we can calculate their capacitances
separately and then add them together again.
Notice that the capacitance has still increased.
The charge will redistribute itself so that the surface charge density  will be different over the area covered by
the Mylar. To find the charge densities we have use the fact that any two components in parallel with each
other must have the same voltage, i.e. V1 = V2, and that charge is still conserved.
This gives us two equations with two unknowns Q1 and Q2. Solving we find
The charge densities (ignoring edge effects) will be
From this we can see that the charge density is higher where there is Mylar between the plates.
(E) Find V and work W needed, when the plates are moved 1.00 mm farther apart.
Again, the conservation of charge means that the charge on the plates will not change from the value we found
in part A. The new parallel plate capacitance is given by
The capacity has dropped by almost a factor of two. The new voltage will be,
The new energy stored in the E-field of the capacitor is now,
This is more that twice the stored energy stored in part B which was 252 pJ. The energy must come from
somewhere. The difference represents the (minimum) work needed to move the plates farther apart.
Looking back at part C, the voltage across the capacitor was lower with the Mylar present, 3.42 V, than that
the original 12.0 V. Thus no external work was needed to insert the Mylar, i.e. it will be pulled in. External
work would be needed to remove the Mylar. This is similar to the fact that no work is needed to drop a rock
but work is needed to raise the rock back up to its original height.
CAPACITORS IN SERIES AND PARALLEL
You can observe and measure the equivalent capacitance for series and parallel combinations. For this study you can use
two identical capacitors. You’ll need:
• 2 cylindrical capacitors, approx. 0.1 F
• 1 capacitance meter
Activity: Capacitance for a Parallel Arrangement
a. Use direct physical reasoning to predict the equivalent capacitance of a pair of identical capacitors wired in parallel.
Explain your reasoning below. Hint: What is the effective area of two parallel plate capacitors wired in parallel?
Does the effective spacing between plates change?
b. What is the equivalent capacitance when your two cylindrical capacitors are wired in parallel? Summarize your
actual data.
c. Guess a general equation for the equivalent capacitance of a parallel network as a function of the two
capacitances C1 and C2.
Ceq =
Next, consider how capacitors that are wired in series, as shown behave.
Activity: Capacitance for a Series Arrangement
a. Use direct physical reasoning to predict the equivalent capacitance of a pair of capacitors wired in series.
Explain your reasoning. Hint: If you connect two capacitors in series, what will happen to the charges along
the conductor between them? What will the effective separation of the “plates” be? Will the effective area
change?
b. Measure the equivalent capacitance when your two cylindrical capacitors are wired in series. Report your actual
data. Are the results compatible with the expected values?
c. Guess a general equation for the equivalent capacitance of a series network as a function of C1 and C2.
Ceq =
d. How do the mathematical relationships for series capacitors compare to those of resistors? Do series capacitors
combine more like series resistors or parallel resistors? Explain.
PARALLEL AND SERIES CAPACITORS
SERIES CONECTION:



Series connected Capacitors always have the same Charge. They do not the same voltage unless the
capacitors have the same Capacitance C.
The Charge on the equivalent capacitor Ce is the same as the charge on either capacitor.
The Voltage across the equivalent capacitor Ce is the sum of the voltage across both capacitors.
PARALLEL CONNECTION



Parallel connected Capacitors always have the same voltage drop across each of them. They do not have
the same charge unless they have the same capacitance C.
The Charge on the equivalent capacitor Ce is the sum of the charges on both capacitors.
The Voltage on the equivalent capacitor Ce is the same as the voltage across either capacitor.
Reducing Capacitors Circuits to Equivalent Capacitance:
Equivalent or Effective Capacitance

Given some circuit containing only capacitors, the objective is to find the
capacitance of a single capacitor that has the same capacitance as combined
capacitance of all the capacitors in the circuit. The combination of
capacitors can then be replaced by its equivalent capacitance.

The next step is to decide which capacitors are in series and which are in
parallel - Series and Parallel Connections.

By using the rules of parallel and series capacitors you can reduce the
various subcombinations until you are left with a single capacitor.
Reduction for circuit above:
First combine the capacitors that are in series - C1 & C2 and C3 & C4.
This leaves three capacitors in parallel (C34, C5, and C12) that can be combined into one capacitor Ce that has a
capacitance equivalent to the combination of the original five capacitors.
QUANTITY
SERIES
PARALLEL
Voltage:
ADD
SAME
SAME
ADD
ADD
RECOPAL ADD
V
Current:
I
Resistance:
R
Capacitance: RECOPAL ADD
ADD
C
Charge:
Q
Symbolic Representation
SAME
ADD
Resistors:
Capacitors:
Capacitors in a Circuit Problem
Three capacitors are connected to a battery of unknown voltage as shown in the circuit diagram? The voltage drop across
the first capacitor C1 is 11.0 V.
(A) What is the effective capacitance of the three capacitors?
(B) What is voltage of the battery?
(C) Which capacitor has the larges charge stored in it?
(D) How much work did it take the battery to charge up all the capacitors?
Sketch and Process:
A combination of capacitors are connected to each other and to a battery.
Relevant Physics:
Ultimately, the only relation we wave for a capacitor comes from the definition of capacitance.
When there is more than one capacitor in a circuit we must reduce the capacitor combination to one capacitor
that replaces the capacitor combination so that we can use the above equation.
To accomplish this we must examine the circuit given to determine which components are in series and which
components are in parallel. We can then apply the rules for combining parallel and series capacitors.
Looking at the circuit diagram given it is fairly easy to see that capacitors C1 and C2 are in series with each
other
.
To more clearly see other relation it often helpful to redraw the original circuit diagram. Two possible
redrawings are:
Here we can clearly see that capacitor C3, the battery Vb, and the series connected capacitors C1 and C2 in the
top branch are all in parallel with each other.
Important guides to keep in mind are:
i. Series Components:
When two capacitors are in series with each other they have the same charge. Thus C1 and C2 have the same
charge. Moreover, when two capacitors are in series with each other we can be replaced them by one capacitor
with an equivalent capacitance C12,
With this observation we could reduce the circuit to:
ii. Parallel Components:
When two capacitors are in parallel with each other then they have the same voltage drop. This voltage rule is
also true in general for any circuit-branches that are in parallel with each other whether or not the branches
contain capacitors. This means that the voltage drop across C3 is the same as that of the battery Vb. Moreover,
the voltage drop across C12 (the series combination of C1 and C2) is same as voltage of either the battery Vb or
capacitor C3.
Two capacitors in parallel with each other can be replaced by one capacitor with an equivalent capacitance Ce
equal to the sum of the capacitances. For our circuit,
For this arrangement of capacitors some of the relations we could write down are:
(A) Find the effective capacitance of the circuit.
The key to solving this problem is reduce the number of capacitor components by using the rules for combining series
and parallel capacitors.
First notice that capacitors C1 and C2 in circuit given are in series so that their effective capacitance, call it C12,
adds using the their inverses,
Given Circuit
First Reduction
The next important observation is that three remaining components (the battery and the two capacitors C3 and
C12) are all in parallel with each other. Thus these two remaining capacitors are in parallel and their effective
capacitance is just their sum,
Note that we didn’t need to convert the units from nF to F before doing any of the calculations since there were
no other units involved in the equations we used to find the equivalent capacitance for parallel or series
capacitor.
We can now reduce the original circuit to much simpler circuit:
(B) Find the voltage of the battery, Vb.
Since we are given the voltage drop across C1 as 11.0 V, we can start with the branch containing C1 and see what can
determine about this branch. We do this because we know that this branch is in parallel with the battery and thus the total
voltage drop across this top branch must be the same as that of the battery. Redrawing the circuit given,
Since C1 and C2 are in series with each other the voltage drop across them is the sum of voltage drop across
each capacitor.
Because C1 and C2 are in series we also know that both capacitors must both have the same charge so that Q1
= Q2 = Q12. We can find Q1 using the definition of capacitance because we are given the voltage drop across
C1.
Equivalently, knowing that the effective capacitance C12 of both capacitors combined is 12.0 nF from part A
we can determine the voltage drop across the effective capacitance C12 which is also equal to the voltage of the
battery.
Observe that capacitance of the third capacitor C3 has no effect on the voltage of the battery in this particular
problem. Changing the capacitance of C3 will only change the charge stored by C3. The charge Q3 will change
to a value that will make the voltage drop across C3 be the same as that of the battery since they are in parallel
with each other.
An alternate approach is to use the overall effective capacitance Ce. Since C12 and C3 are in parallel the
effective charge is equal to their sum.
This solution looks like it depends upon the capacitance of C3, however, if we substitute the symbolic form for
the effective capacitance Ce we get,
This represents the symbolic solution expressed only in terms of the variables stated in the problem.
(C) Find the capacitor with the largest charge.
From the schematic diagram given, one can see that capacitors C1 and C2 are in series with each other. This means that
both capacitors must have the same charge. Thus if we can find the charge on C1 we will know the charge on C2. Using
the definition of capacitance for C1,
To find the charge on C3 we will need the voltage drop across C3. Since the battery and C3 are in parallel with
each other, the voltage V3 is the same as that of the battery Vb which we found in part B we be 18.333 V.
We could have predicted that C3 would be the largest since it is in parallel with C12. They both have the same
voltage drop and C3 has the largest capacitance.
(D) Determine the work done by the battery to charge up the system of capacitors.
The work needed to charge up the capacitors is equal to the energy stored in each of the capacitors.
Which form of U we use depends upon what information we know most readily about the capacitor in
question. We were given the capacitance of all the capacitors and we determined the voltage drop and charges
across the rest. Thus we could any one of the forms as easy as next.
As a double check let us use one of the other forms.
From the calculation above we can see that capacitor C3 has the most energy stored in it. This is not surprising
since we showed in part C that capacitor C3 also has the largest charge stored in it.
An alternate approach is to use the effective capacitance Ce we found in part A to find the energy stored.
If we were to disconnect the battery and replace it by a resistor then this represent the total amount of energy
that would be dissipated by the resistor as the capacitors discharge.