1. Electrical Capacitance
Lecture 6
Electrical Capacitance

2. Intended Learning Outcomes
After completing this course, students are expected to be able to derive expression for the capacitance of:
A parallel-plate structure
A two-wire overhead transmission line
A coaxial cable
An isolated sphere

3. Concept of Capacitance
Capacitance is the ability of an object to hold an electrical charge.
The amount of charge an object can hold for a given voltage is a measure of its capacitance.
Quantitatively, the capacitance (C) of an object is defined as ratio of the electric charge of the object (Q, measured in coulombs) to the voltage across the object (V, measured in volts). That is,
The unit of measure for capacitance is a farad. One farad (F) is equal to one coulomb per volt.
A capacitor is a device specifically designed to hold an electrical charge.

4. Basic Configuration Of A Capacitor
Capacitors vary in shape and size, but the basic configuration is two conductors carrying equal but opposite charges (see figure below).
In the uncharged state, the charge on either one of the conductors in the capacitor is zero. During the charging process, a charge Q is moved from one conductor to the other one, giving one conductor a charge Q+, and the other
one a charge Q-. A potential difference (V) is created, with the positively charged conductor at a higher potential than the negatively charged conductor. From the definition of capacitance, we have the capacitance (C) of the structure

5. PARALLEL-PLATE CAPACITOR
The simplest example of a capacitor consists of two conducting plates of area, which are parallel to each other, and separated by a distance d, as shown in the figure below.
Figure showing a parallel-plate capacitor
The top plate carries a charge +Q while the bottom plate carries a charge –Q. The charging of the plates can be accomplished by means of a battery which produces a potential difference.

6. To find the capacitance C, we first need to know the electric field between the plates. A real capacitor is finite in size. Thus, the electric field lines at the edge of the plates are not straight lines, and the field is not contained entirely between the plates. This is known as edge effects, and the non-uniform fields near the edge are called the fringing fields. In the figure shown below the field lines are drawn by taking into consideration edge
effects. However, in what follows, we shall ignore such effects and assume an idealized situation, where field lines between the plates are straight lines.

7. In the limit where the plates are infinitely large, the system has planar symmetry and we can calculate the electric field everywhere using Gauss’s law:
By choosing a Gaussian “pillbox” with cap area A′ to enclose the charge on the positive plate (see figure), the electric field in the region between the plates is
Figure showing Gaussian surface for calculating the electric field between the plates.
Figure Gaussian surface for calculating the electric field between the plates.

8. Rearranging terms, gives us the expression
The potential difference between the plates is
Since E is constant, we get

9. Hence, from the definition of capacitance, we have
The capacitance is always a positive number, for if the charge on an
isolated conductor is positive, its potential is positive; if the charge on an isolated conductor is negative, its potential is negative.
Note also that C depends only on the geometric factors A and d. The capacitance C increases linearly with the area A since for a given potential difference VAB, a bigger plate can hold more charge. On the other hand, C is inversely proportional to d, the distance of separation because the smaller the value of d, the smaller the potential difference VAB for a fixed Q.

10. Worked Example
A p.d. of 10 kV is applied to the terminals of a capacitor of two circular plates each having an area of 100 sq. cm. separated by a dielectric 1 mm thick. If the capacitance is 3 x 10-4 microfarad, calculate the electric flux density and the relative permittivity of the dielectric.
Solution
A = 100 cm2
10 kV
d = 1 mm r D

11. Solution (continued)
Magnitude of electric field established between the plates by the battery,
Total charge accumulating on the positive plate,
Surface charge density,
Electric flux density,

12. Solution (continued)
Relative permittivity,

13. TRANSMISSION LINES
Two Types:
Overhead Lines
Underground cables

14. Overhead/Underground Transmission Lines Comparison
Overhead Lines
Lower construction cost and cable cost.
Advantage of air for cooling and insulation of the line.
Vulnerable to strong wind and severe weather.
Negative visual impact.
Easier maintenance/repair work
Underground Cables
Expensive pipe work and cable cost (because of the special insulations)
Less vulnerable to the severe weather because it is buried underground.
Environment and aesthetic advantage.
Tedious and costly maintenance/repair work.

15. Concept of Capacitance of Overhead Transmission Line
Electric charge is a source of electric fields. This gives capacitance property.
Electric current is a source of magnetic fields. This gives inductance property. I
Two-wire Overhead Transmission Line

16. SINGLE-PHASE SINGLE-LINE MODEL
Basic parameters of transmission line model
Resistance R to represent resistive loss along the line.
Conductance G to represent leaking currents to the insulator and corona effect.
Inductance L to represent magnetic field.
Capacitance C to represent electric field.
Types, sizes of conductor determine resistance value.
Types, number of insulators determine conductance value.
Conductor spacing, bundling, determines magnetic and electric field strength.

17. When there is more than one conductor, there will be coupling effect from magnetic field and electric field.
This effect is represented by ‘mutual’ inductance and ‘mutual’ capacitance in the conductor model.
The magnitude of mutual inductance and capacitance depends on the distance between conductors.

18. Deriving an expression for the (mutual) line capacitance
Consider a two-wire line shown in the figure excited from a single-phase source. The line develops equal and opposite sinusoidal charges on the two conductors, QA and QB such that QA = -QB. d P x
d - x A B +Q -Q
Figure showing cross-sectional view of a two-wire transmission line
With reference to the figure above, let
d = distance between centres of the wires A and B
r = radius of each wire (r  d)
Q = charge in coulomb/metre of each wire

19. Consider electric field intensity at any point P between conductors A and B. Electric field at P due to +Q coulomb/metre on A is
V/m (towards B)
Electric field intensity at P due to charge –Q coulomb/metre is
V/m (towards B)
Total electric field intensity at P,
V/m (towards B)

20. Hence, potential difference between the two wires is
So,
If the separation distance d is very much greater than the radii of the conductors, i.e. if d >> r, then we can write

21. The line capacitance is then
which is the capacitance per unit length of a single-phase two-wire transmission line.

22. Worked Example
Find the capacitance in terms of F/km of a single-phase line whose conductors are cylindrical with a diameter of 15 mm and spacing in between them is 6 metres.
6 metres
 = 15 mm

23. Solution
In this case r = 7.5 mm = metres
d = 6 metres
Hence

24. Worked Example
A single-phase overhead transmission line has two conductors of diameter 1.5 cm each with a spacing of 1.2 m between their centres. If the dielectric strength of air is 21 kV/cm, determine the voltage for which corona will commence on the line.
Solution
Diameter of the conductor, d = 1.5 cm
Radius of conductor, r = 0.75 cm
Spacing between conductors, d = 1.2 m = 120 cm
Dielectric strength, Emax = 21 kV/cm
The electric field at the surface of the conductor wire is given by

25. Therefore, the amount of surface charge per metre length of the conductor at the onset of breakdown is
Hence, potential difference between the two wires is

26. Therefore, voltage at which corona will commence is

27. LINE TO EARTH CAPACITANCE
The potential difference between each conductor and the ground (or neutral) is one half of the potential difference between the two conductors. Therefore, the capacitance to ground of this single-phase transmission line will be A
CAN N
CBN B

28. MUTUAL CAPACITANCE OF COAXIAL CABLE
Consider a solid cylindrical conductor of radius a surrounded by a coaxial cylindrical shell of inner radius b, as shown in the figure below. The length of both cylinders is L and we take this length to be much larger than b− a, the separation of the cylinders, so that edge effects can be neglected. The capacitor is charged so that the inner cylinder has charge +Q while the outer shell has a charge –Q.

29. To calculate the capacitance, we first compute the electric field everywhere. Due to the cylindrical symmetry of the system, we choose our Gaussian surface to be a coaxial cylinder with length L and radius r where < a < r< b. Using Gauss’s law, we have
Therefore,
where  = Q/L is the charge per unit length.
Notice that the electric field is non-vanishing only in the region a < r < b. For r < a, the enclosed charge is Qenclosed = 0 since any net charge in a conductor must reside on its surface. Similarly, for r > b, the enclosed charge is Qenclosed = +Q - Q =0 since the Gaussian surface encloses equal but oppose charges from both conductors.

30. The potential difference is given by
where we have chosen the integration path to be along the direction of the electric field lines.

31. The total charge:
Mutual capacitance between inner and outer conductors of length L is

32. Area of cross-section of conductor, a = 0.645 cm2
Worked Example
A single-core cable for use on 11 kV, 50 Hz system has conductor area cm2 and internal diameter of sheath is 2.18 cm. The permittivity of the dielectric used in the cable is 3.5. Find (i) the capacitance of the cable per km length and (ii) charging current.
Solution
Area of cross-section of conductor, a = cm2
Diameter of conductor,
Dielectric
Sheath
Internal diameter of sheath,

33. Solution (continued) Capacitance of cable, Charging current, I V C
Dielectric
Sheath

34. ISOLATED SPHERE CAPACITANCE
An isolated sphere can be thought of as concentric spheres with the outer sphere at an infinite distance and zero potential.
The potential V is for obtained by solving the equation
Earlier, we found for r > R,
Thus,

35. Solving the previous equation for V, we obtain
Therefore,
for r > R.
Therefore, the capacitance of a spherical shell is
for r > R.

36. ENERGY STORED BY CAPACITORS
Physically, a capacitor is used to store electrical energy in a circuit for use at a later time. How much energy is stored in a capacitor? We can calculate this by using the definition of capacitance and the relation between potential energy and electrical potential.
Let us consider charging an initially uncharged parallel plate capacitor by transferring a charge Q from one plate to the other, leaving the former plate with charge –Q and the later with charge +Q. Of course, once we have transferred some charge, an electric field is set up between the plates which opposes any further charge transfer. In order to fully charge the capacitor, we must do work against this field, and this work becomes energy stored in the capacitor. Let us calculate this energy.

37. Suppose that the capacitor plates carry a charge q and that the potential difference between the plates is dV. The work we do in transferring an infinitesimal amount of charge dq from the negative to the positive plate is
dW = qdV
In order to evaluate the total work W done in transferring the total charge from one plate to the other, we can divide this charge into many small increments dq, find the incremental work dW done in transferring this incremental charge, using the above formula, and then sum all of these works.
The only complication is that the potential difference V between the plates is a function of the total transferred charge. In fact, V = q/C, so

38. Integration yields
Note, again, that the work W done in charging the capacitor is the same as the energy stored in the capacitor. Since C = Q/V, we can write this stored energy in one of three equivalent forms:
These formulae are valid for any type of capacitor, since the arguments that we used to derive them do not depend on any special property of parallel plate capacitors.

39. END