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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 capa-citor is always zero. When we use the term “charge on a capacitor,” we always mean the charge +Q on the positive conductor.

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

+Q

!Q

2

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) V

QC =

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.

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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.

+++++++

4

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.

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.

+

+

+

+

+

+

+

−−−−−−−

+++++++

0=Er

0=Er

5

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 α be a constant and N be the number of field lines between the plates (ten in our drawing):

(6.2) EA

N α= .

We don’t know α 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:

Q

N

r

Q

r

N

0

20

2 4

1

4

εα

επα

π

=⇒

=

Using this value of α in Eq. (6.2), we can find the electric field in the capacitor.

N

Q

A

N

A

NE

0εα==

(6.3 Electric field in a parallel plate capacitor) A

QE

0ε=

where E is the electric field in a parallel-plate capacitor in volts (V). Q is the charge on the capacitor in coulombs (C). 0ε 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.

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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 xVE ∆∆−= / or dVE /=

where d is the distance between the plates. (The minus sign just gives direction, which we ignore here.) This gives us:

(6.4 Voltage across a parallel plate capacitor) A

QdV

0ε=

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.

(6.5 Capacitance of a parallel plate capacitor) d

AC 0ε

=

where C is the capacitance of a parallel-plate capacitor in farads (F). d is the distance between the capacitor plates in meters (m). 0ε 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.

• The capacitance of a parallel-plate capacitor is d

AC 0ε

= .

• 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

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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:

QVQEdFdW ∆=∆==∆ 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.

C

QdQ

C

QVdQdWW

QQ

2

2

00

==== ∫∫∫

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

(6.6 Potential energy of a capacitor) 22

2

1

2

1CV

C

QU ==

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).

( ) AdEEdd

ACVU 2

0202

2

1

2

1

2

1 εε===

Since Ad is just the volume of the capacitor, we can rewrite this as:

202

1E

Volume

Uu ε==

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.

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Energy Density of the Electric Field

(6.7) 202

1Eu ε=

Things to remember: • The energy it takes to charge a capacitor becomes available as the capacitor is discharged.

• The energy stored in a capacitor is .2

1 2CVU =

• We can think a capacitor’s energy as being stored in the electric field. The energy density

(energy per unit volume) of an electric field is 202

1Eu ε= .

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.

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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.”

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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.

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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 κ (kappa) called the “dielectric constant.”

Definition of the Dielectric Constant

(6.6) platesdielectric EE

−=κ1

1

The minimum value of κ is 1, which corresponds to no dielectric field at all. As κ 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.

E of plates

E of dielectric

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Table 3.1 Dielectric constants of Materials

Material Dielectric Constant κ

Air 1.00059

Water 90

Glass 5.6

Paper 3.7

Nylon 3.4

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)

is given by the parallel-plate capacitor equation, A

QEplates

0ε= , and that dEV total= , we may

conclude:

d

A

Qd

AQ

V

QC

A

Q

d

VE

EEEEE

EE

total

platesplatesdielectricplatestotal

platesdielectric

00

0

1

1111

11

εκκεεκ

κκ

κ

===

==

=

+−=−=

−=

Finally, we can express this in terms of 0C the capacitance with no dielectric present:

Capacitance with a Dielectric

(6.7) 0CC κ=

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.

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0

0

0

0

00

QQ

C

CQQ

C

Q

C

QVV

κ=⇒

=

=⇒=

In this case the capacitance and charge both increase by a factor κ in order to keep the voltage constant. 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.

0

00

000

1VV

C

VCV

CVVCQQ

κ=⇒

=

=⇒=

So this time, the capacitance increases by a factor κ while the voltage drops by a factor κ to keep the charge constant. 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: 02

20

020

,0

1

2

1

2

1, U

VCCVU

VVQQ

κκκ

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: 02

002

00 2

1

2

1,, UVCCVUQQVV κκκ =====

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.

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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.”

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.

20 F

12 V

5

+

1

2

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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:

212

2

1

121

21

21

CCV

Q

V

Q

V

QQ

V

QC

QQQ

VVV

+=+=+==

+===

(6.8 Adding capacitors in parallel) 21 CCC +=

C,Q,V

C2,Q2,V2

= A

B A

C1,Q1,V1

B

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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:

212

2

1

121

21

21

111

CCQ

V

Q

V

Q

VV

Q

V

C

QQQ

VVV

+=+=+==

==+=

(6.9 Adding capacitors in series) 21

111

CCC+=

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.

C,Q,V

C2,Q2,V2

= A

B A

C1,Q1,V1

B

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Table 3.2 Resistors and Capacitors in Series and Parallel

Series Parallel

Resistors

21

21

21

RRR

IRV

III

VVV

+==

==+=

21

21

21

111

RRR

IRV

III

VVV

+=

=+===

Capacitors

21

21

21

111

1

CCC

CQV

QQQ

VVV

+=

=

==+=

21

21

21

1

CCCC

QV

QQQ

VVV

+=

=

+===

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).

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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.

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:

I

Vol

tage

+− + −

C

QVC = IRVR =

t

Q(t)

Q0

(b)

(a)

+−

I

19

(6.10) .0=− IRC

Q

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:

dt

dQ

t

QI

t−=

∆∆−=

→∆ 0lim

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:

QRCdt

dQdt

dQR

C

Q

1

0

−=

=+

This is a differential equation (an equation involving derivatives) that can then be solved for Q(t) subject to the initial condition that Q Q CV( ) .0 0 0= = Normally, we will just give you the

solutions to differential equations; however, this equation is simple enough we can solve it with elementary calculus:

KtRC

Q

dtRCQ

dQ

dtRCQ

dQ

tQRCdt

tdQ

ln1

ln

1

1

)(1)(

+−=

∫−=∫

−=

−=

We have written the constant of integration as ln K for convenience.

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RCt

RCt

eCVtQ

CVKKeQ

KeQ

tRCK

Q

tRC

KQ

/0

00

/

)(

)0(

1ln

1lnln

−

−

=⇒

===

=

−=

−=−

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, τ = RC, and call this the “time constant” of the exponential decay.

It is useful to look at the meaning of the time constant. Clearly if τ 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.

Q

Qf e

ef

tf

Q

Qt

Q

Q et e

Q

Qt

t

t

0

0

0

0

1

1

10% 01 10 2 303

10 3679

1

3

90% 0 910

901054

= =

=

=

= = = =

= ≈ ≈ = =

= = = =

− /

/

ln

. , ln .

. , ln

. , ln .

τ

τ

τ

τ τ

τ τ

τ τ

If

If

If

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 τ = 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.

21

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.

22

If we wish to find the current as a function of time, we can differentiate the charge as a function of time:

ττττ

τ/0/0/0/

0

)()( tttt e

R

Ve

RC

CVe

CVeCV

dt

d

dt

tdQtI −−−− ===−=−=

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.

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.

Figure 6.17 A voltage diagram for the circuit in Fig. 6.16.

I

Vol

tage

+ −+ −

C

QVC =

IRVR =

+

0V

V0

C

R +

23

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

dt

dQ

dt

dQI +==

The voltage diagram then gives:

)(1)(

0

0

0

tQRCR

V

dt

tdQC

QIRV

−=

=−−

This differential equation is not quite so simple. We will just give the solution for the initial conditions described above.

ττ

τ

τ

τ

/0/0

/0

)(

),1()(

tt

t

eR

Ve

CV

dt

dQti

RCeCVtQ

−−

−

===

=−=

The current is an exponentially decreasing function with a time constant τ = 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.

24

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 .RC=τ • 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:

)1()()(

)0()(/

/

τ

τ

t

t

eftf

eftf−

−

−∞=

=