chapter 8 capacitors - cape...

Chapter 8

Capacitors

You can store energy as potential energy by pulling a bowstring, stretching a spring, com-

pressing a gas, or lifting a book. You can also store energy as potential energy in an electric

field, and a capacitor is a device you can use to do exactly that.

There is a capacitor in a portable battery-operated photoflash unit, for example. It

accumulates charge relatively slowly during the charging process, building up an electric

field as it does so. It holds this field and its energy until the energy is rapidly released

during the flash.

Charging a capacitor

One way to charge a capacitor is to place it in an electric circuit with a battery. An electric

circuit is a path through which charge can flow. A battery is a device that maintains a

certain potential difference between its terminals (points at which charge can enter or leave

the battery) by means of internal electrochemical reactions.

In Fig. 8.1a, a battery B, a switch S, an uncharged capacitor C, and interconnecting

wires form a circuit. The same circuit is shown in the schematic diagram of Fig. 8.1b, in

which the symbols for a battery, switch, and a capacitor represent those devices. The battery

maintains potential difference V between its terminals. The terminal of higher potential is

labelled + and is often called the positive terminal; the terminal of lower potential is labelled

- and is often called the negative terminal.

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55

Figure 8.1: (a) Battery B, switch S, and plates h and l of capacitor C, connected in a circuit.(b) A schematic diagram with the circuit elements represented by their symbols.

The circuit shown in Figs. 8.1a and b is said to be incomplete because switch S is

open; that is, it does not electrically connect the wire attached to it. When the switch is

closed, electrically connecting those wires, the circuit is complete and charge can then flow

through the switch and the wires. The charge that can flow through a conductor, such as

a wire, is that of electrons. When the circuit of Fig.8.1 is completed, electrons are driven

through the wires by an electric field that the battery sets up in the wires. The field drives

electrons from capacitor plate h to the positive terminal of the battery; thus plate h becomes

positively charged. The field drives just as many electrons from the negative terminal of

the battery to capacitor plate l; this plate l, gaining electrons, becomes negatively charged

just as much as plate h becomes positively charged.

The potential difference between the initially uncharged plates is zero. As the plates

56 CHAPTER 8. CAPACITORS

become oppositely charged, that potential difference increases until it equals the potential

difference V between the terminals of the battery. Then plate h and the positive terminal

of the battery are at the same potential, and there is no longer an electric field in the wire

between them. Similarly, plate l and the negative terminal reach the same potential and

there is then no electric fied in the wire between them. Thus, with the field zero, there is

no further drive of electrons. The capacitor is said to be fully charged, with a potential

difference V and a charge Q, which are related by Equ. 8.1.

Q =CV (8.1)

Note: Q is the magnitude of the charge on one plate of the capacitor.

C is the capacitance of a the capacitor and its value depends only on the geometry of the

plates and not on their charge or potential difference. The capacitance is a measure of how

much charge must be put on the plates to produce a certain potential difference between

them: the greater the capacitance, the more charge is required.

The SI unit of capacitance that follows from Equ. 8.1 is the coulomb per volt. This unit

occurs so often that it is goven a special name, the farad (F).

Capacitor Networks

When there is a combination of capacitors in a circuit, we can sometimes replace that

combination with an equivalent capacitor, that is, a single capacitor that has the same

capacitance as the actual combination of capacitors. With such a replacement, we can

simplify the circuit, affording easier solutions for unknown quantities of the circuit. Here

we discuss two basic combinations of capacitors that allow such a replacement.

Capacitors in Parallel

Figure 8.2 shows three capacitors in parallel to a battery. They are connected “in parallel”

because the terminals of the battery are effectively wired directly to the plates of each of

the three capacitors. Because the battery maintains a potential difference V betwen its

terminal, it applies the same potential difference V across each capacitor.

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Figure 8.2: (a) Capacitors in parallel. Each is connected directly to the voltage source justas if it were all alone, and so the total capacitance in parallel is just the sum of the individualcapacitances. (b) The equivalent capacitor has a large plate area and can therefore holdmore charge than the individual capacitors.

For the three capacitors we can write

Q1 =C1V ,

Q2 =C2V ,

Q3 =C3V ,

The total charge on the parallel combination is

Q = Q1 +Q2 +Q3 (8.2)

Q =C1V +C2V +C3V (8.3)


Q = (C1 +C2 +C3)V (8.4)

The equivalent capacitance Ceq, with the same total charge Q and applied potential

difference V as the combination, is then

Ceq =QV

=C1 +C2 +C3 (8.5)

Capacitors in Series

Figure 8.3 shows three capacitors connected in series to a battery, which maintains a po-

tential difference V across the left and right terminals of the series combination. This

arrangement produces potential differences V1, V2, and V3 across capacitors C1, C2, and C3,

respectively, such that V1 +V2 +V3 =V .

We seek the single capacitance Ceq that is equivalent to this series combination and thus

can replace the combination.

When the battery is connected, each capacitor must have the same charge Q. This is

true even though the three capacitors may be of different types and may have different

capacitances.

The potential difference across each capacitor:

V1 =QC1

(8.6)

V2 =QC2

(8.7)

V3 =QC3

(8.8)

The potential difference across the entire series combination is then

V =V1 +V2 +V3 (8.9)

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Figure 8.3: (a) Capacitors connected in series. The magnitude of the charge on each plate isQ. (b) The equivalent capacitor has a larger plate separation d. Series connections producea total capacitance that is less than that of any of the individual capacitors.

V =QC1

+QC2

+QC3

(8.10)

V = Q(

1C1

+1

C2+

1C3

)(8.11)

The equivalent capacitance Ceq, is then

Ceq =QV

=1(

1C1

+ 1C2

+ 1C3

) (8.12)

1Ceq

=1

C1+

1C2

+1

C3(8.13)


Storing Energy in an Electric Field

Once the charging of a capacitor has begun, the addition of electrons to the negative plate

involves doing work against the repulsive forces of the electrons which are already there.

Equally, the removal of electrons from the positive plate requires that work is done against

the attractive forces of the positive charges on that plate. The work which is done is stored

in the form of electrical potential energy.

W =Q2

2C=

12

CV 2 =12

QV (8.14)

where

W = the energy stored by a charged capacitor (J)

Q = the charge on the plates (C)

V = the PD across the plates (V )

C = the capacitance (F)

Figure 8.4: Energy stored in the large capacitor is used to preserve the memory of anelectronic calculator when its batteries are charged.

The Medical Defibrillator

The ability of a capacitor to store potential energy is the basic of defibrillator devices, which

are used by emergency medical teams to stop the fibrillation of heart attack victims. In the

portable version, a battery charges a capacitor to a high potential difference, storing a large

amount of energy in less than a minute. The battery maintains only a modest potential

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difference; an electronic circuit repeatedly uses that potential difference to greatly increase

the potential difference of the capacitor. The power, or rate of energy transfer, during this

process is also modest.

Conducting leads (“paddles”) are placed on the victim’s chest. When a control switch

is closed, the capacitor sends a portion of its stored energy from paddle to paddle through

the victim. As an example, when a 70µF capacitor in a defibrillator is charged to 5000V ,

the energy stored in the capacitor

W =12

CV 2 =12

(70×10−6

)(50002)= 875J (8.15)

About 200J of this energy is sent through the victim during a pulse of about 2.0ms. The

power of the pulse is

P =Wt=

2002.0×10−3 = 100kW (8.16)

which is much greater than the power of the battery itself.

Figure 8.5: Automated external defibrillators are found in many public places. Theseportable units provide verbal instructions for use in the important first few minutes fora person suffering a cardiac attack.


Joining two charged capacitors

Fig. 8.6 shows two charged capacitors whose capacitances are C1 and C2. On closing

S, charge flows until the potential difference across each capacitor is the same. The final

charge, potential difference and energy of each capacitor can be calculated by making use

of:

(i) there is no change in the total amount of charge,

(ii) the two capacitors acquire the same potential difference,

(iii) the capacitors are in parallel and therefore the capacitance, C, of the combination

is given by C =C1 +C2.

Figure 8.6: Joining two charged capacitors.

It turns out that unless the intial potential differences of the capacitors are the same,

the total energy stored by the capacitors decreases when they are joined together. Energy

is dissipated as heat in the connecting wire when charge flows from one capacitor to the

other, and this accounts for the decrease in stored energy.

Charging and Discharging a Capacitor through a resistor

Many electronic circuits involve capacitors charging or discharging through resistors.

When you use a flash camera, it takes a few seconds to charge the capacitor that powers

the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does

charging take longer than discharging?

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Charge

In the circuit of Fig. 8.7, when the switch is in position 1, the capacitor of capacitance C

charges through the resistor of resistance R from a 9V battery. The microammeter records

the charging current I and the voltmeter reads the PD Vc across the capacitor at different

times t.

Figure 8.7: Position 1: Charging, Position 2: Discharging.

If readings of I and Vc are taken at 10 second intervals and graphs of these quantities

plotted against t, Fig. 8.8a and b, they show that

Figure 8.8: (a) I vs t, (b) V vs t.

1. I has its maximum value at the start when the capacitor begins to charge and then

decreases more and more slowly until it becomes zero.


2. Vc rises rapidly from zero and slowly approaches its maximum value (9V) which it

reaches when the capacitor is fully charged and I = 0.

If the readings of the PD VR across the resistor were also taken, a graph of VR against t

would have the same shape as that of I against t (since VR = IR at all times) and is shown

as the dashed graph in Fig. 8.8b. All three graphs are exponential curves; those of I and

VR are decay curves and that of Vc is a growth curve.

A graph of the charge on the capacitor Q against t has the same shape as the Vc against

t graph since Q =VcC.

Vc =V0

(1− e−t/RC

)(8.17)

Qc = Q0

(1− e−t/RC

)(8.18)

I = I0

(e−t/RC

)(8.19)

Also note that during charging the sum of the voltages across the resistor and capacitor

equals the battery voltage V , that is V =Vc +VR.

Initially VC = 0, so V =VR.

Finally, when the capacitor is fully charged, I = 0, therefore VR = 0 and VC =V .

Time constant

If the charging current I remained steady at its starting value, a capacitor would be fully

charged after time T = C× R seconds which, for the circuit of Fig. 8.7, equals 500×10−6×100×103 = 50s. In fact I decreases with time, as Fig. 8.8a shows, and the capacitor

has only 0.63 of its full charge and PD after 50s. Nevertheless, T = CR, called the timeconstant, is a useful measure of how long it takes a capacitor to charge through a resistor.

The greater the values of C and R, the greater is T and the more slowly the PD across it

rises. In general, T = CR is the time for VC (and Q) to rise to 63% of the charging PD at

the start of the time, V .

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Discharge

In Fig. 8.7, when the switch is moved from position 1 to position 2, the capacitor discharges

through the resistor. If the graphs of I, VC and VR are plotted as before, they are again

exponential curves, as shown in Figs 8.9a and b. Note that:

1. The discharge current I, and so also VR, are in the opposite direction to that during

charge.

2. VC and VR are in opposition during discharge.

Figure 8.9: (a) I vs t, (b) V vs t.

Using calculus it can be shown that VC decays according to the equation

VC =V0e−t/CR (8.20)

where V0 is the PD across the capacitor initially and e, the base of natural logarithms,

equal 2.7. Substituting t = T =CR we get

VC =V0

e=

V0

2.7= 0.37V0 (8.21)


In this case the time constant T is the time for VC to fall to 37% of its value at the start

of the discharge.

Also,

QC = Q0e−t/CR (8.22)

CR is known as the time constant of the circuit and is the time taken for the charge on

the capacitor to fall to 1/e (i.e. 36.8%) of its initial value.

Since VC = −VR, i.e. the PD across the capacitor is equal and opposite to that across

the resistor.

VR =−V0e−t/CR (8.23)

I =−(V0/R)e−t/CR =−(Q0/CR)e−t/CR (8.24)

i.e. the current during discharge decreases exponentially. When doing calculations the

minus sign in Equ. 8.24 can be ignored; it merely means that the capacitor’s charge Q is

decreasing.

Figure 8.10: This stop-motion photograph of a hummingbird feeding on a flower was ob-tained with an extremely brief and intense flash of light powered by the discharge of acapacitor through a gas.

Now we can explain why the flash camera in our scenario takes so much longer to

charge than discharge; the resistance while charging is significantly greater than while

discharging. The internal resistance of the battery accounts for most of the resistance while

charging. As the battery ages, the increasing internal resistance makes the charging process

even slower.

chapter 8 capacitors - cape...

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