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Electric Fields
Two dissimilar charged glass
or ebonite rods repel
Two similar charged glass or
ebonite rods repel
From the above experiments you should be able to see that the force of attraction or
repulsion is evident whenever a second charged body approaches the first one. There is
no need for the two to be in contact. There exists a force of attraction or repulsion, which
causes the rods to move without being touched by each other. The region surrounding the
body is said to be a field. In the case of a charged body it is known as an electric field. An
electric field cannot be seen, felt or smelt. However, its presence can be detected by
means of a charged body. If a small exploratory charged body is brought into a region,
and if the charged body experiences a force then an electric field is present.
It is common to depict such a field with 'lines of force’. These show the direction of force
on the small positively charged body. It is assumed that this body and its charge are so
small that it does not affect or disturb the field it is detecting. We sometimes use the word
polarity when referring to charges. A positive polarity means a positive charge and a
negative polarity a negative charge.
Since like charges repel, then the lines of force due to a positive charge will always be
directed away from the charge. A negatively charged body will have the lines of force
directed towards it. By convention, the stronger the field the greater the number of lines
which are drawn.
Figure 1 shows the electric fields due to charged spheres. They would also be the patterns
given by a cross section through cylindrical charged bodies.
+
Field due to adjacent +ve charges
-
Field due to adjacent -ve charges
1
+
-
Figure 2
Field due to adjacent +ve and -ve charges
Note: Lines of force emanate from a positive charge and finish on a negative charge.
CHARGE
When we looked at the construction of atoms we saw that charge is due to a body having
unequal numbers of electrons and protons. The charge of a single electron is too small to
be detected by normal laboratory instruments. A very large number of electrons would
have to be removed (or added) before we have a measurable charge.
What is required is a practical unit of charge. It could be defined as some given large
number of electrons. However, you will already have met the practical unit of charge
when you first considered electric current. The unit of charge is the coulomb, named after
the French Colonel who gave his name to the law. The coulomb is defined as that charge
which is conveyed by a constant current of one ampere flowing for one second.
This relationship is expressed in relationships you will have already met, namely:
Charge Q = It
Current I = Q
t
The charge, Q, is measured in coulombs provided the current, I, is in amperes and the
time, t, is in seconds. The unit symbol for a coulomb is C.
A negative charge of one coulomb represents 6.28 x 1018 electrons.
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PROGRESS QUESTION
A current of 10 mA flows in a circuit. Calculate how much charge is transferred in a
minute.
Work out, approximately, how many electrons have flowed through the circuit in this
time.
The field between two parallel plates is shown below.
+Q
A
B
-Q
If a small positive charge is placed in the field close to the positive plate, point A, it will
experience a force of repulsion from the positive plate and a force of attraction from the
negative plate. The total force will be the sum of these two forces.
If the same charge is placed nearer to the negative plate, at point B, it will again
experience two forces. The force of repulsion will be reduced due to the increased
distance from the positive plate. The force of attraction, however, will be increased due to
the reduced distance from the negative plate. The total force will be the same as at point
A. This is true for all points between the plates. If the force remains the same at all points,
the strength of the field must be the same at all points. The field is said to be uniform.
This is illustrated by the lines of force being equidistant.
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Colombs Law
A force exists between two charged bodies that are directly proportional to the product of
the two charges and inversely proportional to the square of the distance between them.
Electric Field Strength
An electric field can be detected by using a small charge and seeing if it experiences a
force. Consider an electric field due to an isolated point charge, Q1. If the exploratory
charge is q2 then the force experienced is found using Coulomb's law:
F = k Q1q2
d2
where d is the distance between Q1 and q2.
F is in newtons, providing Q1 and q2 are in coulombs and d in metres.
As you can see this force, F, is proportional to q2, the exploratory charge.
To give a standard measure of field strength, independent of the size of the exploratory
charge, we define the electric field strength as the force per unit charge and denote it by
E.
Thus E = F and the units of E will be q newtons/coulomb or N/C.
q
If a charge, say q ',is introduced into the electric field it will experience a force, F = q' E.
PROGRESS QUESTION B.4
a)
A small body having a charge of 1 mC is placed in an electric field. If it
experiences a force of 4 mN, what is the electric field strength?
b)
If the is body is replaced by one having a charge of 10 C what force would it
experience?
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Electric Field Strength and Potential Gradient
We can use is an electro-motive force from, for example, a battery to cause current, and
hence electrons, to move round a circuit. If we can arrange that the electrons are removed
from one body and deposited on another, we should obtain two equally charged bodies
but of opposite polarity.
Consider a battery connected to two conducting parallel plates as illustrated in Figure
Electric field
Positive charge
+Q
Negative charge
-Q
Figure
When the battery is charged energy is stored in the form of an electric field that is
established by the opposite charges on the two plates. This is represented by the lines of
flux.
Experiments show that if the value of the applied emf is altered then the field between the
plates alters, e.g. if the capacitor is charged to one voltage, and then the supply voltage is
increased, the capacitor voltage will charge to the new supply voltage and the field
between the plates will also increase. Electric field strength is proportional to the voltage.
E V
If the distance between the plates, say d, is varied, again the field strength is altered, (the
further charges are apart the less effect they have on each other). Electric field strength is
inversely proportional to the distance that is:
El
d
Combining these two results we get:
EV
d
E = kV
d
5
The constant of proportionality, k, has a value of 1 if V is measured in volts and d is
measured in metres. So we have E as volts per metre, V/m.
But earlier we said E was newtons per coulomb, N/C.
Which statement is correct?
Actually both are correct, that is: one newton per coulomb is equivalent to one volt per
metre.
The electric field strength, E, can be expressed in either unit. However, in engineering
use the unit volts per metre, V/m, is much more common.
Potential Gradient
The figure below shows the two parallel plates of a capacitor with the distance between
the plates shown as d metres.
Electric field
Positive charge
+Q
Figure
When the charging current has ceased one plate has a positive charge, +Q, and the other
an equal but negative charge, -Q. The potential on the plates will be equal to the battery
emf V. If the distance between the plates is d metres, then V/d is called the potential
gradient between the plates
As we've seen, this is the same as E, the electric field strength. Another name sometimes
used is electric stress
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PROGRESS QUESTION
A potential of 50 V exists between two parallel plates spaced 1 mm apart.
Calculate:
a)
b)
the field strength between the plates
the potential gradient between the plates.
Electric Flux Density
Lines of flux
-Q
+Q
Cross-section area of
flux path - A
One line of flux is said to emanate from a positive charge and finishes on a negative
charge. If there are Q coulombs then there will be Q lines of flux.
The above figure shows the two charged plates. The diagram shows a three-dimensional
representation of the electric flux between the two charged parallel plates.
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If we ignore the fringing, the electric field is restricted to the region between the plates.
Elsewhere the flux is zero.
The cross-sectional area of the flux path is the same as the area of each plate, A m2, and
so the flux density D at any point between the plates is:
D = Q coulombs/square metre (C/m2)
A
where Q is the charge on either plate.
PROGRESS QUESTION B.7
Two parallel square plates have each charges of 0.4 C, one being positively charged
and the other negatively.
If the sides of the plates are 200 mm, calculate the flux density between the plates.
RELATIONSHIP BETWEEN D AND E
If the charge producing an electric field is increased then both the electric field strength
and the electric flux density will increase. A doubling of the charge will result in a
doubling of both the strength and the flux density. We may conclude, therefore, that the
two are proportional to one another, that is:
DE
D = kE
where k is the constant of proportionality.
For a given charge distribution and a given geometry the value of D is independent of the
medium.
However, a word of caution: it depends upon the circumstances whether or not the charge
remains constant. In the case you looked at earlier of the parallel plates connected to a
battery, if the battery remains connected, then E will be constant and the charge Q and
hence D may vary if the medium between the plates is changed. If the battery is
disconnected then the charge Q and hence D cannot change.
The property of the medium which affects the relationship between the electric field
strength and the flux density is called the permittivity of the medium. It is represented by
 (the Greek letter 'epsilon') and it replaces the k we used earlier so we can now write:
D = E
8
We can define the permittivity of any medium by:
= D
E
The units in which  is measured are farads per metre, F/m.
The reason for these units will be explained in a later section when we will define the
farad.
Permittivity of Free Space
When an electric field is produced in a vacuum the ratio D is known as the permittivity
E
of free space, symbol o.
The value of the permittivity of free space is found to be: 0 = 8.854 x 10-12 F/m
PROGRESS QUESTION B.8
The electric field strength received by the antenna of a capsule on the surface of the
Moon is 10 V/m. Calculate the flux density at the capsule's antenna.
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Relative and Absolute Permittivity
Consider two parallel plates, connected to a voltage source, as shown in Figure
Area of plate, A
Area of plate, A
Dielectric of
relative permittivity
d
V
When a voltage, V, is applied to the two plates, the plates become charged and a flux is
produced between them.
The value of the electric field strength, E, is determined by V and the separation of the
plates, say d.
E=V
d
The flux density D = E
The permittivity  relates to the medium between the plates. This medium is an insulator
and known as a dielectric. The greater the value of , the greater the flux density D, since
E is constant in this case.
It is found that any dielectric gives a greater value of D than does a vacuum. The increase
in flux for a given dielectric compared with that when vacuum separates the plates is
called the relative permittivity and given the symbol r.
The absolute permittivity, , of a dielectric can then be expressed as:
 =  r o
For the configuration shown in Figure we have the following equations.
For a dielectric, other than vacuum, then:
D = E = r o E With vacuum between the plates, then:
Do = o E
The ratio
flux with dielectric = D = r o E = r
flux with vacuum
Do o E
10
The relative permittivity is dimensionless, it's just a number. Typical values for
dielectrics are given in Table B.1.
Table B.1: Values of relative permittivities
Medium
Relative Permittivity r
Air
1.0006
Polythene
2.2
Paper
2.5
Rubber
3.0
Mica
5.0
Bakelite
5.0
Glass
8.0
Barium titanate
6000.0
Note that the relative permittivity of air is 1.0006, but for all practical purposes it is assumed to be 1.0 - the
absolute permittivity of air is taken to be the same as that of free space.
PROGRESS QUESTION
Two parallel plates have areas of 0.02 m2 and are separated by 1 mm. The space between
the plates is filled with mica. The potential difference between the plates is 100 V.
Calculate the charge on the plates.
Check on page 37
11
CAPACITANCE
In the previous section, we looked at two parallel plates connected to a voltage source
which were depicted in Figure 12. These parallel plates form a useful charge storing
device. We are going to study their properties. To help, Figure12 is reproduced here as
Figure 13.
Area of plate, A
Area of plate, A
d
V
Figure B.13
You have seen that when a voltage is applied across the plates, the plates become
charged. The charge is proportional to the voltage, that is:
QV
or
Q=CV
where C is a constant
The constant C = Q coulomb
V volt
represents the number of coulombs produced per volt.
C is therefore a measure of the capacity of the device to store a charge. This property of
storing charge is given the name capacitance.
The capacitance, C, has a unit called a farad, with symbol F.
Any device which has this charge storing property is known as a capacitor.
A capacitor is said to have a capacitance of 1farad if 1 volt produces a charge of 1
coulomb.
Capacitance = Q farad
V
The farad turns out to be a very large unit for most practical purposes and capacitance is
usually expressed in sub-multiples, namely:
micofarad
nanofarad
picofarad
(1 F = l-6 F)
(1 nF = l0-9 F)
(1 pF = l0-12F)
12
You must remember to take care with these negative indices in your calculations.
Note: The letter C is used to represent capacitance. It is also used for the unit of charge
or flux (coulomb). Again, you need to take care to avoid confusing these two uses.
The Parallel Plate Capacitor
One of the simplest forms of capacitor is the device you're already familiar with, the
parallel plate capacitor. The space between the plates is filled with a dielectric. Figure 14
shows this representation again.
Area of plate, A
Area of plate, A
Dielectric of
relative permittivity
d
V
Figure 14
You've just seen that the capacitance C is given by:
C = Q farads.
V
We can use equations you've met in earlier sections to find an expression for the
capacitance in terms of its own properties.
We had
D = Q  Q = DA
A
Also
D = r o E
 Q = r oEA
but
E = V and so Q = r o VA
d
d
Substituting for Q in C = Q gives:
V
C = r o V/dA = C = r o A
V
d
C will be in farads if A is in m2 and d in metres.
13
Thus the capacitance of a parallel plate capacitor is:

directly proportional to the area of the plates, that is:
C  A,

inversely proportional to the distance between the
plates, that is:
C1
D

directly proportional to the relative permittivity of the dielectric, that is: C  r
PROGRESS QUESTION B.1O
A parallel plate capacitor consists of two plates, each 20 cm square. They are separated
by 0.1 mm. The space between the plates is filled with bakelite, which has a relative
permittivity of 5.0. The potential difference between the plates is 1000 V.
Calculate the capacitance of the device and the charge on the capacitor.
Check on page 51
14
Multiplate Capacitor
We have seen that the capacitance of a parallel plate capacitor is directly proportional to
the area of the plates. If a larger capacitance is required then larger plates could be used.
There is a practical limit to this as the device becomes unmanageably large. One way of
overcoming this is to use multiple plates.
Consider the three-plate device shown in Figure l5:
d
d
+ V Figure 15
Each plate has an area A and the separation between adjacent plates is d. The outer two
plates are connected to one side of the supply (the +ve) while the inner plate is connected
to the -ve side of the supply.
Each side of the inner plate is in contact with dielectric while only the inner sides of the
outer plates are. Flux passes from the positive plates to the negative, as shown, giving an
effective flux cross-sectional area of 2A.
The capacitance C = r o 2A
d
Thus the capacitance has been doubled compared with a two-plate device.
PROGRESS QUESTION 11
Draw a diagram, similar to Figure 14, for a four-plate capacitor. Suggest a suitable way
of connecting the plates and show that:
the capacitance C = Er E0 3A
d
Repeat this for four-, five-plates and so on until you are sure that for N plates:
the capacitance C = Er E0 (N - 1)A
d
15
Spirally Wound Capacitors
We have already noted that one way of increasing capacitance is to increase the area of
the plates. Another way would be to reduce the gap between the plates as the capacitance
is inversely proportional to the distance the plates are apart. The increase in area was
somewhat limiting when thinking of solid metal plates. However, a practical capacitor is
possible by using strips of metal foil for the plates and paper or plastic film for the
dielectric. A large area can be accommodated in a relatively small volume by winding the
metal strips and dielectric spirally into a cylindrical roll as shown in Figure16.
16 (a)
16 (b)
Figure 16
When the strips and dielectric are wound into a tight roll it is necessary to include a
second dielectric strip to prevent the two conducting strips, X and Y, from touching. This
is shown in Figure16 (a). This necessary addition also has the advantage of doubling the
capacitance since flux exists on both sides of the outer strip X. This is shown in Figure 16
(b) and is a similar effect to that in the three-plate capacitor described earlier.
Hence C = r o 2A
D
16
Energy stored by a capacitor
Consider an uncharged capacitor being charged by a constant current, I, for a time t
seconds.
The total charge passing round the circuit will be:
Q = It
This total charge will then reside on the plates of the capacitor. The charge on the
capacitor is given by:
Q = CV
 VC = It
Or V = It
C
As I is a constant, and so of course is C, then:
V C
During the charging time, t, the pd across the capacitor will increase linearly with time as
shown in the figure below.
Capacitor
Pd
V volts
V/2
0
Average Voltage
Time t seconds
T
After T seconds the voltage across the capacitor rises to a voltage of V volts as shown in
Figure above
The average voltage over this time will be V volts
2
The circuit energy will be:
Average power x time
17
Since power is voltage x current, and this case current is constant, then:
Average power = average voltage x current,
=VxI
2
 circuit energy = V x I x t joules
2
But Q = It , the charge on the capacitor.
 circuit energy W = V x Q joules
2
Substituting Q = CV we get:
circuit energy W = V x CV joules
2
W = ½ CV2 joules
18