ABSTRACT
At the end of this chapter you should be able to:
• understand field plotting by curvilinear squares
• show that the capacitance between concentric cylinders,C= 2πε0εr ln(b/a)
and calculateC given values of radii
a and b
• calculate dielectric stress E= V r ln(b/a)
• appreciate cable
• show that the C= πε0εr ln(D/a)
and calculate C given values of a and D
• calculate energy
• show that the L= μ0μr 2π
( 1 4 + ln b
a
) and calculate L given values of
a and b
• show that the L= μ0μr π
( 1 4 + ln D
a
) and calculate L given values of
a and D
• calculate energy
Electric fields, magnetic fields and conduction fields (i.e. a region in which an electric current flows) are analogous, i.e. they all exhibit similar characteristics. Thus they may all be analysed by similar processes. In the following the electric field is analysed. Figure 43.1 shows two parallel plates, A and B. Let
the potential on plate A be +V volts and that on plate B be −V volts. The force acting on a point charge of 1 coulomb placed between the plates is the electric field strengthE. It ismeasured in the direction of the field and itsmagnitude depends on the p.d. between the plates and the distance between the plates. In Figure 43.1, moving along a line of force from plate B to plate A means moving from −V to +V volts. The p.d. between the plates is therefore 2V volts and this potential changes linearlywhenmoving fromone plate to the other. Hence a potential gradient is followedwhich changes by equal amounts for each unit of distance moved. Lines may be drawn connecting together all points
within the field having equal potentials. These lines are called equipotential lines and these have been drawn in Figure 43.1 for potentials of 23 V,
The zero equipotential line represents earth potential and the potentials on plates A and B are respectively above and below earth potential. Equipotential lines form part of an equipotential surface. Such surfaces are parallel to the plates shown in Figure 43.1 and the plates themselves are equipotential surfaces. There can be no current flow between any given points on such a surface since all points on an equipotential surface have the same potential. Thus a line of force (or flux) must intersect an equipotential surface at right-angles. A line of force in an electrostatic field is often termed a streamline. An electric field
der capacitor is shown in is set up in the insulating conductors. Any volt drop usually be neglected
the insulation since the conductors have a high conductivity. All points on the conductors are thus at the same potential so that the conductors form the boundary equipotentials for the electrostatic field. Streamlines (or lines of force) which must cut all equipotentials at right-angles leave one boundary at right-angles, pass across the field and enter the other boundary at right-angles. In amagnetic field, a streamline is a line so drawn that
its direction is everywhereparallel to the direction of the magnetic flux. An equipotential surface in a magnetic field is the surface over which a magnetic pole may be moved without the expenditure of work or energy. In a conductionfield, a streamline is a line drawnwith
a direction which is everywhere parallel to the direction of the current flow. A method of solving certain field problems by a
form of graphical estimation is available which may only be applied, however, to plane linear fields; examples include the field existing between parallel plates or between two long parallel conductors. In general, the plane of a field may be divided into a number of squares formed between the line of force (i.e. streamline) and the equipotential. Figure 43.3 shows a typical pattern. In most cases true squares will not exist, since the streamlines and equipotentials are curved.However, since the streamlines and the equipotentials intersect at right-angles, square-like figures are formed, and these are usually called ‘curvilinear squares’. The squarelike figure shown in Figure 43.3 is a curvilinear square since, on successive sub-division by equal numbers of
figures are seen to approach a true square form. When sub-dividing to give a field in detail, and in
some cases for the initial equipotentials, ‘Moore’s circle’ technique can be useful in that it tends to eliminate the trial and error process. If, say, two flux lines and an equipotential are given and it is required to draw a neighbouring equipotential, a circle tangential to the three given lines is constructed. The new equipotential is then approximately tangential to the circle, as shown in Figure 43.3. Consider the electric field established between two
parallel metal plates, as shown in Figure 43.4. The streamlines and the equipotential lines are shown sketched and are seen to form curvilinear squares. Consider a true square abcd lying between equipotentials AB and CD. Let this square be the end of x metres depth of the field forming a flux tube between adjacent equipotential surfaces abfe and cdhg as shown in Figure 43.5. Let l be the length of side of the squares. Then the capacitance C1 of the flux tube is given by
C1 = ε0εr (area of plate)plate separation
i.e. C1 = ε0εr(lx) l
= ε0εrx (1)
Thus the capacitance of the flux tube whose end is a true square is independent of the size of the square. Let the distance between the plates of a capacitor be divided into an exact number of parts, say n (in Figure 43.4, n=4). Using the same scale, the
number of parts (which is not always an integer value), say m (in Figure 43.4, m=10, neglecting fringing). Thus between equipotentials AB and CD in Figure 43.4 there are m squares in parallel and so there are m capacitors in parallel. For m capacitors connected in parallel, the equivalent capacitance CT is given by CT =C1+C2+C3+ ·· · +Cm. If the capacitors have the same value, i.e.C1=C2=C3= ·· · =Cm=Ct , then
CT = mCt (2) Similarly, there are n squares in series in Figure 43.4 and thus n capacitors in series. For n capacitors connected in series, the equivalent capacitance CT is given by
1 CT
= 1 C1
+ 1 C2
+ ·· ·+ 1 Cn
If C1=C2= ·· · =Cn=Ct then 1/CT =n/Ct , from which
CT = Ct n
(3)
Thus if m is the number of parallel squares measured along each equipotential and n is the number of series squares measured along each streamline (or line of force), then the total capacitanceC of the field is given, from equations (1)–(3), by
C = ε0εr xmn farads (4)
For example, let a parallel-plate capacitor have plates 8mm×5mm and spaced 4mm apart (see Figure 43.6). Let the dielectric have a relative permittivity 3.5. If the distance between the plates is divided into, say, four equipotential lines, then each is 1mm apart. Hence n=4. Using the same scale, the number of lines of force
from plate P to plate Q must be 8, i.e. m=8. This is, of course, neglecting any fringing. From equation (4), capacitance C=ε0εrx(m/n), where x=5mm or 0.005m in this case. Hence
C = (8.85× 10−12)(3.5)(0.005)(84)= 0.31pF
plate capacitor,
C = ε0εrA d
= (885× 10 −12)(3.5)(0.008× 0.005)
0.004 = 0.31pF
The capacitance found by each method gives the same value; this is expected since the field is uniformbetween the plates, giving a field plot of true squares.) The effect of fringing may be considered by estimating the capacitance by field plotting. This is described below. In the side view of the plates shown in Figure 43.7,RS is themedial line of force ormedial streamline, by symmetry. Also XY is the medial equipotential. The field may thus be divided into four separate symmetrical parts.
