#FocusOnExams : Engineering Science GCE Tecnical – 16th May 2020
ENGINEERING SCIENCE
Continuation of electrostatic
Exercise
In the diagram above, A and B form an ironic dipole with charges of +1.6 x 1019c and 1.6 x1019c respectively and at a separation of 2 x1010m. calculate; 
 What is the electric field strength due to the Find the electric field of the dipoles at Y
 Find the ratio of these two field strength given that 1/4= 9×10^{9} Nm^{2} c^{2}
dipole at X midway between A and B
B X A Y
. .
2 x1010m 50 x 1010m
In the diagram above, A and B form an ironic dipole with charges of +1.6 x 1019c and 1.6 x1019c respectively and at a separation of 2 x1010m. calculate;
What is the electric field strength due to the dipole at X midway between A and B
Find the electric field of the dipoles at Y
Find the ratio of these two field strength given that 1/4πεo = 9×109 Nm2 c2
Correction of assignment
Solution
data
rAx= 1 x 10^{10}m, QA = +1.6 x 10^{19}c
rBx = 1×10^{10}m, QB= 1.6x 10^{19}c
 Calculation of electric field strength (E)
E_{Ax}=
 =
E_{Ax} = +1.44 x 10^{11}vm^{1}
E_{Bx}^{ =}
 =
E_{Bx} = 1.44 x 10^{11}vm^{1}
Therefore;
EX = E_{Ax} – E_{Bx}
= +1.44 x 10^{11}vm^{1} – 1.44 x 10^{11}vm^{1}
EX= 2.88 x 10^{11}vm^{1}
 E_{AY} =
 =
E_{AY} = 5.76 x 10^{7}vm^{1}
E_{BY }_{= }
 =
E_{BY =}5.33 x10^{7}vm^{1}
Then Ey= E_{AY }+ E_{BY}
= 4.3 x 10^{6}vm^{1}
Since they are in opposite directions
 Ratio = =
= 0.67 x 10^{5}
Chapter 8: Capacitors and Inductors
Objectives
At the end of this lesson students should be able to:
 Explain how a capacitor stores energy
 Appreciate Q It, E , D , Q CV and D E 0r and perform calculations using these formulae
 State the unit of capacitance
 Define a dielectric
 state the value of 0 and appreciate typical values of r
 Perform calculations involving the parallel plate capacitor
 Define dielectric strength
 Determine the energy stored in a capacitor and state its unit
 Describes practical types of capacitor
 Define inductance and state its unit
 State the factors which affect the inductance of an inductor
 Describe examples of practical inductors and their circuit diagrams
 Calculate the energy stored in the magnetic field of an inductor and state its units
8.1 Capacitors and capacitance
A capacitor is a device capable of storing electrical energy.
Figure 8.1 shows a capacitor consisting of a pair of parallel
Metal plates X and Y are separated by an insulator, which could be air. Since the plates are electrical conductors each will contain a large number of mobile electrons. Because the plates are connected to a d.c. supply the electrons on plate X, which have a small negative charge, will be attracted to the positive pole of the supply and will be repelled from the negative pole of the supply on to plate Y. X will become positively charged due to its shortage of electrons whereas Y will have a negative charge due to its surplus of electrons.
The difference in charge between the plates results in a p.d. existing between them, the flow of electrons dying away and ceasing when the p.d. between the plates equals the supply voltage. The plates are then said to be charged and there exists an electric field between them. Figure 8.2 shows a side view of the plates with the field represented by ‘lines of electrical flux’.
If the plates are disconnected from the supply and connected together through a resistor the surplus of electrons on the negative plate will flow through the resistor to the positive plate. This is called discharging. The current flow decreases to zero as the charges on the plates reduce. The current flowing in the resistor causes it to liberate heat showing that energy is stored in the electric field
8.1.1 Summary of important formulae and definitions
From the above lesson charge Q is given by:
Q = I x t coulombs

where I is the current in amperes, and t the time in seconds.
A dielectric is an insulating medium separating charged
surfaces.
Electric field strength, electric force, or voltage gradient,
E =
E = volts/m

Electric flux density,
D =C/m^{2}

Charge Q on a capacitor is proportional the applied voltage,
V, i.e. Q V
Q = CV

Where the constant of proportionality, C, is the capacitance
Capacitance,
C =

The unit of capacitance is the farad, F (or more usually F =10^{6}F or pF = 10^{12} F), which is defined as the capacitance of a capacitor when a p.d. of one volt appears across the plates when charged with one coulomb. Every system of electrical conductors possesses capacitance. For example, there is capacitance between the conductors of overhead transmission lines and also between the wires of a telephone cable. In these examples the capacitance is undesirable but has to be accepted, minimised or compensated for. There are other situations, such as in capacitors, where capacitance is a desirable property. The ratio of electric flux density, D, to electric field strength, E, is called absolute permittivity, , of a dielectric.
Thus =
Permittivity of free space is a constant, given by
0 = 8.85 x 10^{12} F/m.
Relative permittivity,
(r has no units). Examples of the values of r include: air =1, polythene = 2.3, mica =3–7, glass =5–10, ceramics = 6–1000.
Absolute permittivity, =0r, thus = 0r
8.1.2 The parallel plate capacitor
For a parallelplate capacitor, experiments show that capacitance C is proportional to the area A of a plate, inversely proportional to the plate spacing d (i.e. the dielectric thickness) and depends on the nature of the dielectric:
Capacitance,
C = farads

Where 0 = 8.85 x 10^{_}^{12} F/m (constant),
r = relative permittivity,
A = area of one of the plates, in m^{2},
d = thickness of dielectric in m
and n = number of plates
Example 1. (a) A ceramic capacitor has an effective
plate area of 4 cm^{2} separated by 0.1mm of ceramic of relative permittivity 100. Calculate the capacitance of the capacitor in picofarads. (b) If the capacitor in part
(a) is given a charge of 1.2C what will be the p.d. between the plates?
Solution
(a) Area A = 4 cm^{2} =4 x 10^{_}^{4} m^{2},
d = 0.1mm = 0.1 x 10^{_}^{3} m, 0 =8.85 x 10^{_}^{12} F/m,
r = 100 and n = 2
 Capacitance, C = farads
 = F
C= 3540pF
 Q = CV thus V =
NA . V= v
V = 339v
Exercise1. A waxed paper capacitor has two parallel plates, each of effective area 800 cm^{2}. If the capacitance of the capacitor is 4425 pF determine the effective thickness of the paper if its relative permittivity is 2.5
Exercise2
 Define
 Capacitance
 Electrical field strength
(2marks)
 State two practical uses of capacitors (2marks)
 A 300v battery is connected across capacitors of 3µf and 6 µf
 In parallel
 In series
Calculate the charge and energy stored in each capacitor in (i) and (ii) (4marks)
 A capacitor of capacitance 4µf is charge to a potential of 100v and another of capacitance 6µf Is charge to a potential of 200v. these capacitors are now join with plates of like charges connected together. Calculate
 The potential across each after joining (3marks)
 The total energy stored before joining (3marks)
 The total energy stored after joining (3marks)