Chapter five Capacitance and Capacitors
CHAPTER 5 CAPACITANCE AND CAPACITORS
In Chapter 2 we considered the presence of electric charge. The next step requires that we introduce a device which can hold a reasonable amount of charge. Such a device is called a capacitor, although some people still use the old term condenser.
Early experimenters found that conductors would hold much greater electric charges provided that they were held in close proximity to one another yet kept apart. They also found that the greater the surface area of the conductors then the greater the stored charge.
A simple capacitor can be made from two strips of metal foil sandwiched with two thin layers of insulation. Waxed paper is a suitable insulant; the wax is needed to keep damp out of the paper which otherwise would quickly cease being an insulator. The foil and paper are rolled as shown in Fig. 5.1. Thus we have a device bringing two conductors of large area into very close proximity with one another yet which are insulated, and this would provide a practical capacitor which can be used to hold electric charge.
A capacitor’s ability to hold electric charge is measured in farads. This is a very large unit and most capacitors are rated in microfarads or less. In circuit diagrams there are two common symbols for a capacitor as shown in Fig. 5.2. In subsequent diagrams the international symbol will be used.
A charged capacitor may be regarded as a reservoir of electricity and its action can be demonstrated by connecting a capacitor of, say, 20μF in series with a resistor R, a centre-zero microammeter A and a two-way switch S, as in Fig. 5.3. A voltmeter V is connected across C. If R has a resistance of, say, 1 MΩ, it is found that when switch S is closed on position a, the ammeter A shows a deflection rising immediately to its maximum value and then falling off to zero. This means that initially there has been a significant current due to the inrush of electric charge into the uncharged capacitor, subsequently reducing to zero once the capacitor was fully charged. This change of current is indicated by curve G in Fig. 5.4.
At the same time the voltmeter indicates a rise in voltage across the capacitor C. This rise of voltage is indicated by curve M.
When the switch S is moved over to position b, the ammeter again performs as before except that the indication is in the reverse direction. The reverse deflection is due to the charge rushing out from the capacitor. The current is indicated by curve H.
At the same time the voltmeter indicates a fall in voltage across the capacitor C. This fall in voltage is indicated by curve N.
If the experiment is repeated with a resistance of, say, 2 MΩ, it is found that the initial current, both on charging and on discharging, is halved, but
Capacitors
5.1
Fig. 5.1 Paper-insulated capacitor
Fig. 5.2 Circuit symbols for a capacitor
Fig. 5.3 Capacitor charged and discharged through a resistor
Fig. 5.4 Charging and discharging currents and p.d.s
Fig. 5.5 Hydraulic analogy of a capacitor
it takes about twice as long to rise up and to fall off, as shown by the dotted curves J and K. Curves P and Q represent the corresponding variation of the p.d. across C during charge and discharge respectively.
The shaded area between curve G and the horizontal axis in Fig. 5.4 represents the product of the charging current (in amperes) and the time (in seconds), namely the quantity of electricity (in coulombs) required to charge the capacitor to a p.d. of Vvolts. Similarly the shaded area enclosed by curve H represents the same quantity of electricity obtainable during discharge.
The operation of charging and discharging a capacitor may be more easily understood if we consider the hydraulic analogy given in Fig. 5.5, where P represents a piston operated by a rod R and D is a rubber diaphragm stretched across a cylindrical chamber C. The cylinders are connected by pipes E and are filled with water.
When no force is being exerted on P, the diaphragm is flat, as shown dotted, and the piston is in position A. If P is pushed towards the left, water is with- drawn from G and forced into F and the diaphragm is in consequence dis- tended, as shown by the full line. The greater the force applied to P, the greater is the amount of water displaced. But the rate at which this displacement takes place depends upon the resistance offered by pipes E; thus the smaller the cross-sectional area of the pipes the longer is the time required for the steady state to be reached. The force applied to P is analogous to the e.m.f. of the battery, the quantity of water displaced corresponds to the charge, the rate at which the water passes any point in the pipes corresponds to the current and the cylinder C with its elastic diaphragm is the analogue of the capacitor.
When the force exerted on P is removed, the distended diaphragm forces water out of F back into G; and if the frictional resistance of the water in the pipes exceeds a certain value, it is found that the piston is merely pushed back to its original position A. The strain energy stored in the diaphragm due to its distension is converted into heat by the frictional resistance. The effect is similar to the discharge of the capacitor through a resistor.
No water can pass from F to G through the diaphragm so long as it remains intact; but if it is strained excessively it bursts, just as the insulation in a capacitor is punctured when the p.d. across it becomes excessive.
Hydraulic analogy
CHAPTER 5 CAPACITANCE AND CAPACITORS 95
The experiment described in section 5.1 shows that charge is transferred but it is unsuitable for accurate measurement of the charge. A suitable method is to discharge the capacitor through a ballistic galvanometer G, since the deflection of the latter is proportional to the charge.
Let us charge a capacitor C (Fig. 5.6) to various voltages by means of a slider on a resistor R connected across a battery B, S being at position a; and for each voltage, note the deflection of G when C is discharged through it by moving S over to position b. Thus, if θis the first deflection or ‘throw’ observed when the capacitor, charged to a p.d. of V volts, is discharged through G, and if kis the ballistic constant of G in coulombs per unit of first deflection, then discharge through G is
Q=kθcoulombs
It is found that, for a given capacitor,
[5.1] Charge on C [coulombs]
PD across C [volts] =a constant
Charge and voltage
5.3
Fig. 5.6 Measurement of charge by ballistic galvanometer
The property of a capacitor to store an electric charge when its plates are at different potentials is referred to as its capacitance.
The unit of capacitance is termed the farad (abbreviation F) which may be defined as the capacitance of a capacitor between the plates of which there appears a potential difference of 1 volt when it is charged by 1 coulomb of
electricity.
Capacitance Symbol: C Unit: farad (F)
It follows from expression [5.1] and from the definition of the farad that
or in symbols
∴ Q=CV coulombs [5.2]
Q
V =C
Charge [coulombs]
Applied p.d. [volts] =capacitance [farads]
Capacitance
Capacitors in parallel
Fig. 5.7 Capacitors in parallel
Capacitors in series
In practice, the farad is found to be inconveniently large and the capacit- ance is usually expressed in microfarads(μF) or in picofarads(pF), where
1μF =10−6
F and 1 pF =10−12
F
Example 5.1 A capacitor having a capacitance of 80μF is connected across a 500 V d.c. supply. Calculate the charge.
From equation [5.2],
Q=CV
∴ Charge=(80 ×10−6
) [F] ×500 [V]
=0.04 C =40 mC
Suppose two capacitors, having capacitances C1and C2farads respectively,
to be connected in parallel (Fig. 5.7) across a p.d. of Vvolts. The charge on C1is Q1coulombs and that on C2is Q2coulombs, where
Q1=C1V and Q2=C2V
If we were to replace C1and C2by a single capacitor of such capacitance C
farads that the same total charge of (Q1+Q2) coulombs would be produced
by the same p.d., then Q1+Q2=CV.
Substituting for Q1and Q2, we have
C1V+C2V=CV
C=C1+C2 farads [5.3]
Hence the resultant capacitance of capacitors in parallel is the arithmetic sum of
their respective capacitances.
Suppose C1and C2in Fig. 5.8 to be two capacitors connected in series with
suitable centre-zero ammeters A1and A2, a resistor R and a two-way switch
S. When S is put over to position a, A1and A2are found to indicate exactly
the same charging current, each reading decreasing simultaneously from a maximum to zero, as already shown in Fig. 5.4. Similarly, when S is put over to position b, A1and A2indicate similar discharges. It follows that during
charge the displacement of electrons from the positive plate of C1 to the
negative plate of C2is exactly the same as that from the upper plate (Fig. 5.8)
of C2to the lower plate of C1. In other words the displacement of Qcoulombs
of electricity is the same in every part of the circuit, and the charge on each capacitor is therefore Qcoulombs.
If V1and V2are the corresponding p.d.s across C1and C2respectively,
then from equation [5.2]:
Q=C1V1=C2V2
so that
5.5
CHAPTER 5 CAPACITANCE AND CAPACITORS 97
and [5.4]
If we were to replace C1 and C2 by a single capacitor of capacitance C
farads such that it would have the same charge Qcoulombs with the same p.d. of Vvolts, then
Q=CV or
But it is evident from Fig. 5.8 that V=V1+V2. Substituting for V, V1and
V2, we have
∴ [5.5]
Hence the reciprocal of the resultant capacitance of capacitors connected in series
is the sum of the reciprocals of their respective capacitances.
From expression [5.4]
[5.6] But V1+V2=V
∴ V2=V−V1
Substituting for V2in equation [5.6], we have
∴ [5.7]
and
[5.8]
Example 5.2 Three capacitors have capacitances of 2, 4 and 8μF respectively. Find the total capacitance when they are connected
(a) in parallel; (b) in series. V V C C C 2 1 1 2 = × + V V C C C 1 2 1 2 = × + V V V C C − 1 = 1 1 2 V V C C 2 1 1 2 = 1 1 1 1 2 C =C +C Q C Q C Q C = + 1 2 V Q C = V Q C 2 2 = V Q C 1 1 =
Fig. 5.8 Capacitors in series
Distribution of voltage across capacitors in series
(a) From equation [5.3]:
C=C1+C2+C3
Total capacitance =2 +4 +8 =14μF
(b) If Cis the resultant capacitance in microfarads when the capacitors are in series, then from equation [5.5]:
=0.5 +0.25 +0.125 =0.875
∴ C=1.14μF
Example 5.3 If two capacitors having capacitances of 6μF and 10μF respectively are connected in series across a 200 V supply, find
(a) the p.d. across each capacitor; (b) the charge on each capacitor.
(a) Let V1 and V2 be the p.d.s. across the 6μF and 10μF capacitors
respectively; then, from expression [5.7],
and V2=200 −125 =75 V
(b) Charge on each capacitor
Q=charge on C1
=6 ×10−6×
125 =0.000 75 C =750μC
It follows from expression [5.3] that if two similar capacitors are connected in parallel, the capacitance is double that of one capacitor. But the effect of connecting two similar capacitors in parallel is merely to double the area of each plate. In general, we may therefore say that the capacitance of a capacitor is proportional to the area of the plates.
On the other hand, if two similar capacitors are connected in series, it follows from expression [5.5] that the capacitance is halved. We have, however, doubled the thickness of the insulation between the plates that are connected to the supply. Hence we may say in general that the capacitance of a capacitor is inversely proportional to the distance between the plates; and the above relationships may be summarized thus:
In order to clarify this relationship, we now need to consider the space between the charged plates of a capacitor. In this space, the charges set up electric fields. The study of such electric fields is known as electrostatics.
Capacitance area of plates distance between plates
∝ V1 200 10 6 10 125 = × + = V 1 1 1 1 1 2 1 4 1 8 1 2 3 C =C +C +C = + + Capacitance and the capacitor 5.8
CHAPTER 5 CAPACITANCE AND CAPACITORS 99
The space surrounding a charge can be investigated using a small charged body. This investigation is similar to that applied to the magnetic field surrounding a current-carrying conductor. However, in this case the charged body is either attracted or repelled by the charge under investigation. The space in which this effect can be observed is termed the electric field of the charge and the force on the charged body is the electric force.
The lines of force can be traced out and they appear to have certain properties:
1. In an electric field, each line of force emanates from or terminates in a charge. The conventional direction is from the positive charge to the neg- ative charge.
2. The direction of the line is that of the force experienced by a positive charge placed at a point in the field, assuming that the search charge has no effect on the field which it is being used to investigate.
3. The lines of force never intersect since the resultant force at any point in the field can have only one direction.
The force of attraction or of repulsion acts directly between two adjacent charges. All points on the surface of a conductor may be assumed to be at the same potential, i.e. equipotential, and the lines of force radiate out from equipotential surfaces at right angles. The most simple case is that of the isolated spherical charge shown in Fig. 5.9. However, most electric fields exist between two conductors. The two most important arrangements are those involving parallel plates (as in a simple capacitor) and concentric cylinders (as in a television aerial cable). The resulting fields are shown in Fig. 5.10.
It should not be overlooked that the space between the conductors needs to be filled with an insulator, otherwise the charges would move towards one another and therefore be dissipated. The insulant is called a dielectric.
We can investigate an electric field by observing its effect on a charge. In the SI method of measurement this should be a unit charge, i.e. a coulomb. In practice this is such a large charge that it would disrupt the field being investigated. Therefore our investigation is a matter of pure supposition.
The magnitude of the force experienced by this unit charge at any point in a field is termed the electric field strength at that point. Electric field
Electric fields
5.9
Fig. 5.9 Electric field about an isolated spherical charge
Fig. 5.10 Electric fields between oppositely charged surfaces. (a) Parallel plates; (b) concentric cylinders (cable)
Electric field strength and electric flux density
Fig. 5.11 A parallel-plate capacitor
strength is sometimes also known as electric stress. It can be measured in newtons per unit charge and represented by the symbol E. (Since E
can also represent e.m.f., we use a bold type for E when representing electric field strength and later we will meet D representing electric flux density.)
It should be recalled that 1 J of work is necessary to raise the potential of 1 C of charge through 1 V. When a charge moves through an electric field, the work done against or by the electric field forces is indicated by the change in potential of the charge. Therefore to move a unit charge through a field so that its potential changes by Vvolts requires Vjoules of work.
The most simple field arrangement which we can investigate is that between parallel charged plates as shown in Fig. 5.11. Let us suppose that the plates are very large and that the distance between them is very small. By doing this, we can ignore any fringing effects of the type shown in Fig. 5.10 and assume that all the field exists between the plates. Let us also assume that there is free space between the plates.
There is a potential difference of Vvolts between the plates, therefore the work in transferring 1 C of charge between the plates is Vjoules. But work is the product of force and distance, and in this case the distance is dmetres. Therefore the force experienced by the charge is the electric field strength E
given by
volts per metre [5.9]
The total electric effect of a system as described by the lines of electric force is termed the electric flux linking the system. Flux is measured in the same units as electric charge, hence a flux of Qcoulombs is created by a charge of
Qcoulombs.
The electric flux density is the measure of the electric flux passing at right angles through unit area, i.e. an area of 1 m2. It follows that if the area of the
plates in the capacitor of Fig. 5.11 is Athen the electric flux density Dis given by
coulombs per square metre [5.10] From expressions [5.9] and [5.10]
In electrostatics, the ratio of the electric flux density in a vacuum to the elec- tric field strength is termed the permittivity of free spaceand is represented by
0. Hence,
Electric flux density
Electric field strength = = ÷ = × =
D E Q A V d Q V d A Cd A D = Q A E =V d
CHAPTER 5 CAPACITANCE AND CAPACITORS 101
or farads [5.11]
Permittivity of free space Symbol: 0 Unit: farad per metre (F/m)
The value of 0can be determined experimentally by charging a capacitor,
of known dimensions and with vacuum dielectric, to a p.d. of Vvolts and then discharging it through a ballistic galvanometer having a known ballistic constant kcoulombs per unit deflection. If the deflection is θdivisions,
Q=CV=kθ
∴
From carefully conducted tests it has been found that the value of 0 is