Lesson 6 – Capacitors and Capacitance
© Lawrence B. Rees 2007. You may make a single copy of this document for personal use without written permission.
6.0 Introduction
In 1745 Pieter van Musschenbroek, while trying to see if electricity is soluble in water,
discovered that electrical charge could be stored. The device came to be known as a Leyden jar.
By being able to store electricity, the Leyden jar turned experimentation with electricity from an
amusing pastime into a power that had to be reckoned with.
The Leyden jar is one type of capacitor. Capacitors are still used when a large amount of
charge needs to be delivered over a short period of time. In this role, they function much like a
water tower – charge is put into the capacitor slowly over a long time so that when there is
demand for a large current, there will be enough charge available. More often, however,
capacitors are used because of their behavior in oscillating circuits. We will discuss these
applications later.
6.1 Defining Capacitors and Capacitance
The general definition of a capacitor (also called a condenser in older books) is any
conductor which can store charge. When we connect a wire between a charged capacitor and
ground, current will flow until the capacitor is left with no net charge. Since any conductor has a
capacity to hold charge, any conductor can be a capacitor.
When we speak of capacitors in this course, we will restrict ourselves to a pair of two
conductors, one with charge +Q and the other with charge –Q. Hence the net charge on a capacitor is always zero. When we use the term “charge on a capacitor,” we always mean the charge
+Q on the positive conductor.
!Q
+Q
Figure 6.1 An example of a capacitor.
A charged capacitor always has an electric field and an electric potential between the
positive and negative conductors. As we learned in Sect. 3.5, the electric potential anywhere on
1
or within each conductor is constant when no current is flowing. If the electric potential varied
from one point to another within the conductor, then current would flow as the electrons would
be free to move to lower energy. (Because electrons are negatively charged, they move to higher
voltage when they move to lower energy, which is just another way of saying that electrons are
attracted by positive charge.) This means that a capacitor can be conveniently characterized in
terms of the voltage difference between the conductors. We will always consider this voltage
difference to be positive and use the letter V to denote this value.
Clearly, different arrangements of conductors will have different abilities to hold charge.
In general, capacitors will be able to hold more charge if there is a greater electrostatic attraction
between the positive charges on the one conductor and the negative charges on the other
conductor. Thus capacitors with large surface areas and the positively and negatively charged
conductors close together hold charge more effectively.
We define a quantity called “capacitance” to measure how much charge a capacitor can
hold. Of course, the amount of charge on a capacitor will depend on how much voltage we apply
to it. If we charge the capacitor by connecting it to a high-voltage battery, more charge will go on
the conductors than if we attach the capacitor to a low-voltage battery. In fact, most capacitors
will hold twice as much charge when we double the voltage. For this reason, we define
capacitance as the charge per volt that a capacitor can hold.
Definition of Capacitance
(6.1)
C=
Q
V
where
C is capacitance in farads (F).
Q is the charge on the capacitor in coulombs (C).
V is the voltage difference between the capacitor plates in volts (V).
One farad is a huge capacitance. Capacitors that are used in most electronic devices are
measured in microfarads ( F) or even picofarads (pF). In practice, picofarads are often called
“micromicrofarads” ( F) or just “puffs.”
To charge a capacitor, all we need to do is connect the positive terminal of a battery to
one conductor and the negative terminal to the other conductor. Charge will continue to flow
onto the conductors until the voltage across the capacitor is equal in magnitude and opposite in
direction to the voltage across the battery. That means that an electron in a wire between the
battery and the capacitor is pushed one direction by the battery with the same force that it is
pushed the other direction by the capacitor, so no current will flow in the circuit.
2
Things to remember:
• Capacitors have charge +Q one conductor and charge –Q on a second conductor.
• Capacitance is defined by the relation Q = CV.
• In steady state, the voltage across a capacitor is equal in magnitude to the applied voltage.
• In steady state, no current flows in the branch of a circuit containing a capacitor.
6.2. Parallel-Plate Capacitors
The simplest sort of capacitor is the parallel-plate capacitor. Such a capacitor is made of
two identical, parallel, conducting plates separated by a vacuum. (In practice, air is very similar
to a vacuum in this application.) The plates can have any shape, but we do demand that the
separation distance between the plates be small compared to the length and width of the plates. If
this is true, we can ignore edge effects, the irregularities in the electric field near the edges of the
capacitor. Let’s then connect the capacitor to a battery and charge the plates to values of +Q and
–Q. We now wish to find the electric field and the voltage of the capacitor. First, let’s consider
the electric field lines of an infinite sheet of charge with a uniform positive charge density. By
symmetry, the direction of the field must be perpendicular outward from the sheet of charge as
there is nothing in space to distinguish between right and left, or up and down. Furthermore, the
spacing between the lines must be the same everywhere as well, because the electric field
+
+
+
+
+
+
+
Figure 6.2 The electric field of an infinite sheet of positive charge.
cannot be larger in one area than in a nearby area. From this, we can see that the electric field
cannot become weaker as we move away from the plane. (Why do we know that? Hint: Look at
the field lines!) The direction of the electric field is to the left on the left side of the plane, and to
the right on the right side of the plane. The magnitude of the electric field is the same
everywhere outside of the plane itself. The electric field of a negative plane of charge is similar
to that of a positive plane of charge, except that the field lines point toward the plane rather than
away from it.
3
A parallel-plate capacitor is constructed from two identical conducting planes. The planes
are not infinite, but if they are close together the fields are nearly the same as for an infinite sheet
of charge, except at the edges. In Fig. 6.3, the electric field outside of the capacitor on the right
will be the sum of the field from the positive plate pointing right and the field from the negative
plate pointing left. Since these two fields have the same magnitude but opposite directions, they
cancel out completely. The fields between the plates, however, are both to the right, producing a
uniform electric field (ignoring the edges) from the positive to the negative plate.
r
E=0
−
−
−
−
−
−
−
+
+
+
+
+
+
+
r
E=0
Figure 6.3. The electric field lines of a parallel-plate capacitor.
Note that the positive charge and the negative charge will move to the inside surfaces of each
conducting plate. (Why?)
+
+
+
+
+
+
+
Figure 6.4 The electric field of the positive plate alone.
4
Another thing we should note is that the charge on the negative plate cannot create nor
destroy any of the field lines that would be present if the positive plate were alone in space. The
negative charges pull the positive charges to the inside of the plate, and pull all the electric field
lines so that they point toward the negative plate, but they cannot destroy any of the field lines.
Note that in both Fig.6.3 and Fig. 6.4 there are ten field lines coming from the positive charges.
In fact, if we were to place all the charges on the plate onto a small sphere, the electric field lines
would point radially outward from the center of the sphere, but there would still be ten of them.
Now we are in a position to actually determine the electric field inside the capacitor. We
know that for a point charge, the strength of the electric field is related to the number of field
lines by the relations found n Sect. 3.4. Letting be a constant and N be the number of field lines
between the plates (ten in our drawing):
α
N
=α E.
A
(6.2)
We don’t know but we can determine its value because we know what E would be if the entire
charge on the positive plate of the capacitor were to be on one point charge. For a point charge,
the perpendicular surface we would use for counting N in Eq. (6.2) would be a sphere of radius r.
This then would give us:
α
N
1 Q
=α
2
4π ε 0 r 2
4π r
⇒α =
Using this value of
α
ε0N
Q
in Eq. (6.2), we can find the electric field in the capacitor.
E=
N
N Q
=
αA A ε 0 N
E=
(6.3 Electric field in a parallel plate capacitor)
Q
ε0 A
where
E is the electric field in a parallel-plate capacitor in volts (V).
Q is the charge on the capacitor in coulombs (C).
ε 0 is the permittivity of free space, 8.85×10–12 C/Nm2.
A is the area of one capacitor plate in square meters (m2)
What we have just proven is a special case of Gauss’s law of electricity. We will study this in
more detail in Lesson 7.
5
Now that we know the electric field in a parallel plate capacitor, we can easily determine
the voltage across the capacitor. Since the field is uniform, we have E = − ∆V / ∆x or E = V / d
where d is the distance between the plates. (The minus sign just gives direction, which we
ignore here.) This gives us:
V =
(6.4 Voltage across a parallel plate capacitor)
Qd
ε0 A
Note that the voltage between the plates is larger when the plates are separated farther from each
other. Sometimes this seems counterintuitive, as the voltage seems like it should be larger when
the plates are close together. Remember, though, that the electric field in the capacitor is
uniform, and the force between the plates is uniform (as long as d is small). Because the plates
are attracted to each other, it takes work to separate the plates to a larger distance, so the
potential energy increases as we separate the plates. This is similar in concept to the fact that the
gravitational potential energy of a ball is larger when the ball is raised further from the ground.
Finally, we can rearrange Eq. (6.4) to obtain an expression for the capacitance, C=Q/V.
C=
(6.5 Capacitance of a parallel plate capacitor)
ε0 A
d
where
C is the capacitance of a parallel-plate capacitor in farads (F).
d is the distance between the capacitor plates in meters (m).
ε 0 is the permeability of fee space, 8.85×10–12 C/Nm2.
A is the area of one capacitor plate in square meters (m2)
Things to remember:
• The electric field inside a capacitor is uniform so E=V/d.
ε A
• The capacitance of a parallel-plate capacitor is C = 0 .
d
• Memorize the capacitance expression, and you can deduce V and E from it.
6.3. Energy in a Parallel-Plate Capacitor
Since one of the principal uses of a capacitor is to store energy, it is important to know
just how much energy is stored in a charged capacitor. To do this, we calculate how much work
it takes to put the charge on the two plates of the capacitor. Although this sounds like a
formidable task, we can use a clever approach to make the problem easy. We begin with an
uncharged parallel-plate capacitor. Rather than charging the plates in the traditional way, by
connecting the plates to a battery, we pull electrons off the left plate with a pair of tweezers, and
6
move them over to the right plate. The left plate gradually becomes positive and the right plate
becomes negative. As this happens, the electrons we remove are attracted by the positive plate
and repelled by the negative plate so that we have to do work against the electrostatic force to
move additional charge from plate to plate. The small amount of work it takes to move a little bit
of charge Q from one plate to the other is:
∆W = Fd = ∆QEd = V∆Q
All we have to do now is add up all the little bits of work necessary to bring the capacitor to a
final charge Q.
Q
Q
Q2
Q
W = ∫ dW = ∫ VdQ = ∫ dQ =
C
2C
0
0
The work done to move the charge Q from one plate to the other is just the potential energy U
stored in the capacitor.
Energy Stored in a Capacitor
U=
(6.6 Potential energy of a capacitor)
1 Q2 1
= CV 2
2 C
2
When physicists thought of electric fields as actual entities existing in space, they thought
the energy in a capacitor was stored in the electric field itself. Although we no longer think of the
electric field quite so literally, it is still useful to talk about the energy “stored in the field.” For a
parallel-plate capacitor, we can obtain a convenient expression by simply substituting for V and
C in Eq. (6.6).
U=
1
1 ε0A
CV 2 =
(Ed )2 = 1 ε 0 E 2 Ad
2
2 d
2
Since Ad is just the volume of the capacitor, we can rewrite this as:
u=
U
1
= ε0E2
Volume 2
where u is called the “energy density” of the electric field. Although this expression was
specifically derived for the special case of a parallel-plate capacitor, it turns out to be generally
true. We will use it later when we wish to find the energy carried by electromagnetic radiation.
7
Energy Density of the Electric Field
(6.7)
u=
1
ε0E2
2
Things to remember:
• The energy it takes to charge a capacitor becomes available as the capacitor is discharged.
1
• The energy stored in a capacitor is U = CV 2 .
2
• We can think a capacitor’s energy as being stored in the electric field. The energy density
1
(energy per unit volume) of an electric field is u = ε 0 E 2 .
2
6.4. Dielectrics
As we mentioned earlier, the first capacitor was developed in 1745 by Pieter van
Musschenbroek, a professor at the University of Leyden in the Netherlands. Van Musschenbroek
wanted to know if electricity could be stored by dissolving it in water, so he took a jar, put a
conductor through a cork in the top of it, and dangled a chain from that conductor into a jar full
of water.
Figure 6.5 A Leyden jar.
His assistant held the jar in one hand and with the other he happened to touch the
conductor. As he did this, he received a violent shock. Van Musschenbroek traded roles with his
assistant and tried the experiment again. The experiment was a near-fatal success. In inventing
the Leyden jar, Van Musschenbroek stumbled onto the fact that putting water near the
capacitor’s conductors greatly increases the capacitor’s ability to hold charge.
8
Think About It
What constituted the two conductors in the first Leyden jar? – Soon afterward, Leyden
jars were made with metal foil lining the inside and outside surfaces of an empty jar. (If you’re
interested in making your own Leyden jar, you can find directions on the internet.)
To understand how water makes a capacitor more powerful, we need to think about the
structure of a water molecule. A water molecule, of course, consists of two hydrogen atoms
attached to one oxygen atom. The hydrogen atoms are not opposite each other, but are separated
by an angle of about 105°. The electrons of the hydrogen atom tend to remain between the
hydrogen nucleus – a proton – and the oxygen atom. The net result is that the two protons make
one side of the molecule positively charged while the other side of the molecule has an excess of
negative charge. Such a molecule that has zero net charge, but has one side that is positively
charged and the opposite side negatively charged, is called a “polar molecule” or a ‘dipole.”
Figure 6.6 A water molecule.
If a dipole is placed in a uniform electric field, there is no net force on it, because it is neutral.
However, the molecule will tend to line up in the electric field, as shown in Fig.6.7. A slab of
insulating dipoles placed between the plates of a capacitor is called a “dielectric.”
9
Figure 6.7 Dipoles between capacitor plates.
Note that the total charge in the middle of the dielectric – that is, in any region such as
that bounded by the dashed rectangle in Fig. 6.7 – is zero. However, near each of the plates of
the capacitor, there is an effective sheet of charge on the surface of the dielectric. We can view
the effect of the dielectric in two different ways:
(1) If we connect a battery to the capacitor, the negative charge on the dielectric near the
positive plate of the capacitor attracts positive charge, and the positive charge near the negative
plate of the capacitor attracts negative charge. This means that more charge will flow onto the
plates with the dielectric in place than without it. Because the battery fixes the voltages across
the capacitor, the voltages with and without the dielectric are the same. The capacitor holds more
charge with the same voltage, so the dielectric causes the capacitance to increase.
(2) If we charge the capacitor, then remove the wires so the charge is fixed, the effective
charge on the surfaces of the dipole next to the capacitor plates produces an electric field which
opposes the electric field of the plates. The electric field with the dielectric is smaller than
without the dielectric. Hence, the voltage with the dielectric is also smaller. Since the same
charge is held with a smaller voltage, the dielectric causes the capacitance to increase.
10
E of plates
E of dielectric
Figure 6.8 The electric field inside a capacitor with a dielectric.
If the dielectric were perfect, the field produced by the dielectric would be equal to the
field produced by the plates and the capacitor voltage would be zero. However, in practice,
dielectrics never approach perfection. We characterize the effectiveness of a dielectric by a
number (kappa) called the “dielectric constant.”
κ
Definition of the Dielectric Constant
 1
E dielectric = 1 −  E plates
 κ
(6.6)
The minimum value of is 1, which corresponds to no dielectric field at all. As gets large, the
magnitude of the dielectric field approaches the magnitude of the plates’ field. Some typical
values of the dielectric constant are given in Table 3.1.
κ
κ
11
Table 3.1 Dielectric constants of Materials
Material
Dielectric Constant
Air
1.00059
Water
90
Glass
5.6
Paper
3.7
Nylon
3.4
κ
Now let us see how the dielectric affects the other characteristics of the system: charge,
voltage, and capacitance. As before, the distance between the plates is d and the area of each
plate is A. Keeping in mind that the electric field on the plates alone (with or without a dielectric)
Q
, and that V = Etotal d , we may
is given by the parallel-plate capacitor equation, E plates =
ε0A
conclude:
 1
E dielectric = 1 −  E plates
 κ
1
1

E total = E plates − E dielectric = 1 − 1 +  E plates = E plates
κ
κ

V 1 Q
E total = =
d κ ε0A
C=
κε A
ε A
Q
=Q 0 =κ 0
V
Qd
d
Finally, we can express this in terms of C 0 the capacitance with no dielectric present:
Capacitance with a Dielectric
(6.7)
C = κC 0
Now let us see what this means in terms of charge and voltage. First, let us do a fixed
voltage experiment. That is, we keep a battery connected to the plates of a capacitor, and then
insert a slab of dielectric to completely fill the space between the plates.
12
Q Q0
=
C C0
V = V0 ⇒
Q=
Q0 C
C0
⇒ Q = κQ0
In this case the capacitance and charge both increase by a factor in order to keep the voltage
constant.
κ
As a second case, let us do a fixed charge experiment. We first charge the capacitor with
an initial voltage V and then disconnect the battery. This time the voltage will change while the
charge remains fixed.
Q0 = Q ⇒ C 0V0 = CV
C 0V0
C
1
⇒ V = V0
V=
κ
So this time, the capacitance increases by a factor while the voltage drops by a factor to keep
the charge constant.
κ
κ
Our derivation of the energy stored in a capacitor still holds, so we can compare the
energy with a dielectric to without a dielectric:
Fixed charge: Q = Q0, V =
V0
κ
,
U=
V2 1
1
1
CV 2 = κ C 0 02 = U 0
2
2
κ
κ
If the surface charge of a dielectric is opposite the charge of the nearby plates, there should be an
attractive force pulling the dielectric into the capacitor. If the force is attractive, the total
potential energy must be smaller as the dielectric is pulled in.
Fixed voltage: V = V0 , Q = κQ0 ,
U=
1
1
CV 2 = κ C 0V02 = κU 0
2
2
This time the potential energy of the capacitor increases. But shouldn’t the dielectric be pulled
into a capacitor with fixed charge as well? In this case we have to remember that there is a
battery attached to the capacitor, so we have to consider the larger system. As we insert the
dielectric between the capacitor plates, we allow charge to flow from the battery onto the
capacitor plates. This reduces the potential energy of the battery. The net result is that the energy
of the battery-capacitor system reduces as well.
13
You might ask what happens to a dielectric once a capacitor is discharged. The dipoles
will not usually remain aligned, as thermal vibration will randomize their orientation quite
quickly. There are a few materials, however, that have strong enough attractions between
adjacent dipoles that, once aligned, they remain aligned even after the external electric field is
turned off. These materials are the electric equivalent of permanent magnets, and hence are
called “electrets.”
Things to remember:
• A dielectric increases the capacitance by a factor 6. 6$1.
• If the charge on a capacitor is fixed, a dielectric reduces the voltage by a factor 6. It does this
by reducing the electric field in the capacitor.
• If the voltage on a capacitor is fixed, a dielectric increases the charge on a capacitor by a factor
6. It does this by pulling charge onto the plates from the battery.
6.5. Capacitors in DC Circuits
In DC circuits, current only flows through capacitors when they charge or discharge. We
will look at those processes in the next section. In steady-state (not changing in time), a capacitor
simply holds a certain charge at a certain voltage. Current can not actually flow through a
capacitor because the plates are separated. However, when a capacitor is in the process of
charging or discharging, current can flow through the branch of a circuit containing a capacitor.
We are usually a little lax about our terminology and just say “current flows through a capacitor
when it charges or discharges.” In steady state, however, no current flows “through a capacitor.”
5
20 F
1
12 V
+
2
Figure 6.9 A DC circuit with a capacitor.
To see how this works, Let’s take the simple circuit shown in Fig. 6.9. (Note that
capacitors are schematically represented by two parallel lines in circuit diagrams.) We wish to
find the currents, voltages, and the charge on the capacitor. Once the capacitor is charged, no
current can flow through it, so the voltage across the 5 resistor is zero. The remaining circuit is
just a battery in series with two resistors. The total resistance is 1 + 2 = 3 . The current
through this circuit is I = V / R = 4 A. The voltages across the 1 and 2 resistors are 4 V and 8
V, respectively. The voltage across the capacitor is 4 V, the same as that across the 1 resistor.
(Why?) The charge on the capacitor is Q = C V = 80 C.
14
C1,Q1,V1
A
B
C2,Q2,V2
C,Q,V
B
= A
Figure 6.10 Two capacitors in parallel.
In the same way that we can make networks of resistors, we can also make networks of
capacitors. We would like to combine capacitors in series and parallel to make equivalent
capacitors in the same way we did with resistors. In order to do that, let’s look at how charge and
voltage behave in simple series and parallel combinations of capacitors.
When two capacitors are placed in parallel, their voltages must be the same. If we charge
the capacitors and then connect the wire from point A to point B in Fig. 6.10 to discharge the
capacitor, the total charge that flows from A to B is the sum of the charges on the two capacitors.
Hence, we have:
V = V1 = V2
Q = Q1 + Q2
C=
Q Q1 + Q2 Q1 Q2
=
=
+
= C1 + C 2
V
V
V1 V2
(6.8 Adding capacitors in parallel)
C = C1 + C 2
15
C1,Q1,V1
C2,Q2,V2
A
B
C,Q,V
B
= A
Figure 6.11 Two capacitors in series
When two capacitors are placed in series, it is clear that the voltages are additive, just as
with resistors. However, charge is a little more complicated. Let us assume that the two
capacitors in Fig. 6.11 are initially discharged. Then we attach points A and B to a battery. The
left plate of capacitor 1 then has a charge +Q1. Charge leaves the right plate of capacitor 1 until
its charge becomes –Q1. However, the right plate of capacitor 1 and the left plate of capacitor 2
(enclosed by a dashed box in the figure) have no connection to the outside world, so all the
charge that leaves the right plate of capacitor 1 must go to the left plate of capacitor 2. Hence, the
charge on the two capacitors must be equal. This is further verified by the fact that if we connect
points A and B in either the two capacitors or the equivalent capacitor, the charge that flows
from A to B is Q1. Therefore, we may conclude:
V = V1 + V2
Q = Q1 = Q2
1 V V1 + V2 V1 V2
1
1
= =
=
+
=
+
C Q
Q
Q1 Q2 C1 C 2
(6.9 Adding capacitors in series)
1
1
1
=
+
C C1 C 2
Note that the formulas for adding capacitance in series and parallel are “opposite” the
equivalent formulas for adding resistance in series and parallel, as noted in the table below.
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Table 3.2 Resistors and Capacitors in Series and Parallel
Resistors
Capacitors
Series
V = V1 + V2
Parallel
V = V1 = V2
I = I1 = I 2
I = I1 + I 2
V = IR
R = R1 + R2
V = IR
V = V1 + V2
V = V1 = V2
Q = Q1 = Q2
Q = Q1 + Q2
1
C
1
1
1
=
+
C C1 C 2
V =Q
V =Q
1
1
1
=
+
R R1 R2
1
C
C = C1 + C 2
We note that the equations are identical if we replace I with Q and R with 1 / C in the equations
for resistors and capacitors. This is a result of the fact that the voltage across a resistor is V=IR ,
whereas the voltage across a capacitor is V=Q/C.
Things to remember:
• The equations for adding capacitors in series and parallel are the reverse of those for resistors.
• Capacitors in parallel have the same voltage.
• Capacitors in series have the same charge.
6.6. Charging and Discharging Capacitors: RC Circuits
One last thing we would like to know about capacitors in circuits is exactly how
capacitors charge and discharge. That is, we want to know the charge on a capacitor and the
current flowing through a circuit as a function of time.
Let us begin with the very simple circuit shown in Fig. 6.12(a). In this circuit the
capacitor is originally connected to a battery of voltage V0 and charged to an initial charge
Q0 = C V0. At time t = 0, the battery is connected to the resistor, as shown in the figure. Of
course, current will flow from the positive plate of the capacitor to the negative plate of the
capacitor. The resistor serves to limit the amount of current that can flow. The larger the
capacitance and the larger the resistance, the longer time it will take the capacitor to discharge.
The charge as a function of time is shown in Fig. 6.12(b).
17
+
−
I
(a)
Q0
Q(t)
(b)
t
Figure 6.12 (a) A simple circuit to discharge a capacitor.
(b) Charge on the capacitor as a function of time.
Now let analyze this same circuit quantitatively. To do this, we first make a voltage
diagram of the circuit.
Voltage
−
+
VC =
I
+
−
Q
C
VR = IR
Figure 6.13 A voltage diagram for the circuit in Fig. 6.12.
Taking I to be positive, we can then write an equation for the voltage in the diagram:
18
(6.10)
Q
− IR = 0.
C
Since the current comes from the charge on the capacitor flowing through the circuit, we know
that Q and I must be related. Current is the amount of charge q flowing past a given point in the
circuit during a small time t. The change in charge on the capacitor during this same time is
Q = – q. The minus sign is a consequence of the fact that when current is positive, the charge
on the capacitor is getting smaller so Q is negative. This then means:
∆Q
dQ
=−
∆t → 0 ∆ t
dt
I = − lim
This equation tells us that the current in the circuit, I, comes from discharging the capacitor.
Since Q is the charge on the capacitor and Q is decreasing, I is a positive quantity.
We can now substitute this expression into Eq. (6.10) to obtain:
dQ
Q
+R
=0
dt
C
dQ
1
Q
=−
dt
RC
This is a differential equation (an equation involving derivatives) that can then be solved for Q(t)
subject to the initial condition that Q ( 0 ) = Q 0 = CV0 . Normally, we will just give you the
solutions to differential equations; however, this equation is simple enough we can solve it with
elementary calculus:
dQ(t )
1
Q(t )
=−
dt
RC
dQ
1
dt
=−
Q
RC
dQ
1
=−
∫
∫ dt
Q
RC
1
t + ln K
ln Q = −
RC
We have written the constant of integration as ln K for convenience.
19
ln Q − ln K = −
1
t
RC
Q
1
=−
t
K
RC
Q = Ke −t / RC
ln
Q(0) = Ke 0 = K = CV0
⇒ Q(t ) = CV0 e −t / RC
This solution tells us that the charge is initially CV0 and that the charge decays
exponentially. Since the argument of an exponential function must be dimensionless, RC must
have units of time. In SI units, therefore, 1 × 1 F = 1 s. We use the Greek letter (tau) to
represent this quantity, = RC, and call this the “time constant” of the exponential decay.
τ
It is useful to look at the meaning of the time constant. Clearly if is large, the exponent
is small, and it takes a long time for the capacitor to discharge. Note that this is consistent with
our original analysis of the circuit, wherein we suggested that if the capacitance is large and the
resistance is large, it would take a long time to discharge the capacitor. More quantitatively we
can calculate the time it takes for the capacitor to discharge to various fractions of its original
charge. Let f be the ratio of the charge at time t to the initial charge.
τ
Q
= f = e − t /τ
Q0
1
f
1
t = τ ln
f
Q
If
. , t = τ ln 10 = 2.303τ
= 10% = 01
Q0
e t /τ =
If
Q 1
1
= ≈ 0.3679 ≈ , t = τ ln e = τ
Q0 e
3
If
Q
10
τ
.
= 90% = 0.9, t = τ ln = 01054
Q0
9
These results are shown graphically in Fig. 6.14. Fig. 6.15 shows the discharge curve of
the same capacitor with values of resistance adjusted so that = 1 s, 2 s, and 3 s. Note that the
time constant is the time required for the charge on the capacitor to drop to 1 / e = 0.3679 of its
original value.
τ
20
Figure 6.14 The exponential decay of a discharging capacitor.
Figure 6.15 Q(t) for discharging RC circuits with
time constants of J = 1 s, 2 s, and 3 s.
21
If we wish to find the current as a function of time, we can differentiate the charge as a
function of time:
I (t ) = −
CV0 −t / τ CV0 −t / τ V0 −t / τ
dQ(t )
d
e
e
= − CV0 e −t / τ =
=
= e
dt
dt
τ
RC
R
Note that the current is initially what it would be without the capacitor, but that it decays to zero
with the same time constant as that for the capacitor to discharge.
C
V0
+
R
Figure 6.16 A circuit to charge a capacitor.
The last circuit we wish to consider is one in which a capacitor is charged, as shown in
Fig. 6.16. Note how a switch is represented in the circuit diagram. In this case, we take the initial
charge on the capacitor to be zero. After a long period of time the capacitor completely charges
and current ceases to flow in the circuit. When the capacitor is fully charged, the voltage on the
capacitor becomes V0. The final charge on the capacitor is then Qf = C V0.
Let us draw a voltage diagram of this circuit at some time during the charging process.
+
I
+
−
+
−
Voltage
VR = IR
V0
VC =
Q
C
Figure 6.17 A voltage diagram for the circuit in Fig. 6.16.
22
We proceed much as we did in the previous case, except that this time we recognize that the
current flowing in the circuit causes the capacitor to charge. Since the capacitor is charging
dQ / dt > 0, and
I=
dQ
dQ
=+
dt
dt
The voltage diagram then gives:
Q
=0
C
1
dQ(t ) V0
Q(t )
=
−
dt
R RC
V0 − IR −
This differential equation is not quite so simple. We will just give the solution for the initial
conditions described above.
Q(t ) = CV0 (1 − e − t / τ ), τ = RC
i (t ) =
dQ CV0 −t / τ V0 −t / τ
e
=
= e
τ
dt
R
The current is an exponentially decreasing function with a time constant = RC. This is the same
time constant we found for the charging capacitor. The charge, however, increases toward its
final value with the same time constant. That is, the time constant is the time required for the
current to reach 1 – 1 / e = 0.6321 of its final value.
τ
23
Figure 6.18 Q(t) for charging RC circuits with time constants of 1 s, 2 s, and 3 s.
Things to remember:
• Capacitors charge and discharge with a time constant τ = RC .
• The time constant is the time it takes a decaying exponential function to fall to 1 / e of the
function’s initial value if decreasing or to rise to 1 – 1 / e of its final value if increasing.
• In circuits where capacitors are charged and discharged, the charges and currents are of one of
two forms:
f (t ) = f (0) e −t / τ
f (t ) = f (∞) (1 − e −t / τ )
24