44 CHAPTER
Key Ideas10
Stage 1 Physics
10.3 The Electric Field
10.3.1 Concept
The electrostatic force between two charged bodies is an example of action at a distance. That is,
two bodies separated by some distance exert forces on each other, without ever coming into
contact. There are other examples of “action at a distance” forces – namely the gravitational force
between two masses, and the magnetic force between the north and south poles of bar magnets.
Before the nineteenth century the electrostatic forces acting between two charges (as envisaged by
Coulomb), and gravitational forces acting between two masses (as envisaged by Newton), were
considered to be direct and instantaneous interactions between the two bodies.
However, this did not stand up to close scrutiny because, if some distance separates two bodies,
how does one body “know” about the presence of the other in order to exert a force on it? And if
the separation of the bodies is increased, how is this information communicated from one body to
the other so that each body “knows” to exert a smaller force?
These and other such questions bedevilled scientists for over
150 years. In fact, both Isaac Newton and Charles Coulomb
recognised these difficulties, but neither could explain how the
two bodies managed to exert a force on each other.
In the middle of the 19th century, the great English experimental
physicist Michael Faraday developed the concept of an electric
field to explain the interaction between electric charges.
Faraday suggested that each charged body somehow modifies
the properties of the space around itself, creating an electric field
in that space. Each body comes into contact with the field of the
other, and the information is thus transmitted from one to the
other via their fields.
Michael Faraday (1791 – 1861)
Thus, the electric field plays an intermediary role in the forces between charges.
A charge is thus thought to interact with a field – it does not interact directly with the other charge
to exert a force on it.
Therefore, when two charges q1 and q2 exert a force on each other, we visualise this interaction as
the charge q1 setting up an electric field around itself, and this field (which is in contact with
charge q2) exerting a force F on q2. Similarly q2 sets up a field and this field acts on q1, producing
a force F on it. The situation is completely symmetrical, each charge being immersed in the field
associated with the other charge.
Presence of a Field
A charged body modifies the space surrounding itself, i.e. has an electric field in this space. When
another charge is placed in this space, the electric field exerts a force on it.
Therefore, we use the effect of a field on a charged body to define the presence of the field in a
region of space. Specifically, we define the presence of an electric field as follows.
An electric field exists in a region of space if a charged body, placed in that region, experiences a
force because of its charge.
Electric charges establish an electric field E in the surrounding space. The electric field at any
point causes a force on an electric charge placed at that point.
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Electric Fields
Electric Fields 45
Testing for a Field
We test for the presence of an electric field in a region of space by using a test charge, which is
actually a small charged object. The charge on this object is usually quite small, as a large charge
would affect the field we wish to test.
We place the test charge at some point in that region of space. If it experiences a force, due to its
charge, we conclude that there is an electric field in that region. If it does not experience a force,
then there is no electric field in that region of space. This is because the electric force on a
charged body is due to the interaction between that charged body and an electric field.
We can also use the force on the test charge to get an idea of the strength of the electric field and
to define its direction.
Strength
We use the magnitude of the electrostatic force on the test charge placed at some point in space to
determine the strength of the electric field at that point.
The stronger the electrostatic force on the test charge, the stronger the electric field is at that
point.
Direction
We use the direction of the force on the test charge to define the direction of the electric field at a
point in space. If a test charge is placed at some point in an electric field, the direction of the force
on the test charge depends on the sign of the charge. The direction of the force on a negative
charge will be opposite to that on a positive test charge.
We define the direction of the electric field at some point as the direction of the force on a positive
test charge placed at that point.
Note that an electric field is an example of a vector field, for it applies forces of calculable
magnitude and specific directions to charges placed within it.
10.3.2 Pictorial Representation
A convenient way of visualising electric field patterns is by drawing lines of force – these lines
represent the direction and the strength of a field.
Lines of Force – Direction
With electric fields, the lines of force at any point in the field are in the direction of the force on a
small positive (test) charge placed at that point in the field. The direction of the field at any point
is at a tangent to the field line at that point.
Therefore, if we see a pictorial representation of an electric field with a charge placed in it, we
immediately know that
•
if the charge is positive, the direction of the force on it is
in the same direction as the field line at that point;
F2
•
if the charge is negative, the direction of the force on the
charge is opposite to the direction of the field line at that
point.
F1
Fig 10.2
This is depicted in Figure 10.2. This diagram shows the direction
of the force F1 on a positive charge placed in an electric field and the direction of a force F2 on a
negative charge placed in the field.
Essentials Text Book
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46 CHAPTER
Key Ideas10
Stage 1 Physics
Lines of Force – Density
In all electric fields, the lines of force are drawn so that the number of lines of force per unit crosssectional area (perpendicular to the field near the point) is directly proportional to the field
strength of the given field. Where the lines are close together, the field is strong; where they are
far apart, the field is weak or weaker.
Thus, considering Figure 10.2 again, in the region of space where the negative charge lies, the
electric field is stronger than in the region of space where the positive charge lies. Consequently,
if the two charges are of equal magnitude, the force F2 on the negative charge is stronger than the
force F1 on the positive charge.
Drawing Force Lines
1.
Lines of force are always drawn away from positive charges and towards negative charges.
The direction of the line of force at any point in the field is the direction of the force on a
positive charge placed at that point – and a positive charge placed at any point will be
repelled by other positive charges and attracted towards negative charges.
Sketching electric fields is just a matter of common sense. To determine the direction of a
field line at any point, just imagine placing a positive charge at that point and try to visualise
the direction of the net force on it using the fact that like charges repel and unlike charges
attract.
2.
Lines of force always leave, or arrive at, conducting surfaces at right angles to the surface.
If the electric field was not perpendicular to the surface,
there would be a force, at some angle θ to the surface, on
charges on the surface of the conductor. This force would
have a component Fx parallel to the conducting surface.
Thus, charges would move around the conducting surface,
until their distribution was such that there was no longer
any force parallel to the surface. The forces on the charges
at this time, and hence the lines of force, would then all be
perpendicular to the surface.
Fy
F
Fx
Fig 10.3
This is depicted in Figure 10.3.
3.
Lines of force never touch or cross each other.
If they did, it would mean that a charge placed at that point in the field would experience two
different forces – one in the direction of each of the field lines.
But the net force on a body is the vector sum of the
forces on the body. Thus, if we were trying to detect the
field at a point, we would only be able to observe this
one net force on a charge placed at that point. By
definition, the direction of this one force is the direction
of the field at that point.
F1
F
F2
Fig 10.4
We would never be able to detect two different forces on
a charge, and so we could never map two different field lines at any point in space.
This is depicted in Figure 10.4, which shows the single force F that we would actually
observe if we placed a positive test charge at a point in space where two field lines cross each
other.
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Electric Fields 47
Electric Fields
10.3.3 Examples
Isolated Point Charge
+.
−.
Fig 10.5a
Fig 10.5b
Note that the lines of force at each point are in the direction of the force on a positive charge
placed at that point. The lines of force come closer together as they approach the charge that
creates the field, where we expect the electric field to be stronger.
These are examples of radial fields, and they are depicted in Figures 10.5a and 10.5b.
Electric Dipole – Two Point Charges
To determine the appearance of this field, imagine placing a small positive test charge in the space
near the two charges and visualise the net force acting on this test charge at any point.
•
Figure 10.6a shows the field due to two bodies carrying equal charges of opposite sign.
•
Figure 10.6b shows the field due to two bodies carrying equal positive charges.
The field due to negatively charged bodies would be the same as in Figure 10.6b, except that the
directions marked on the lines of force would be reversed.
Fig 10.6a
Fig 10.6b
Charged Hollow Sphere
++++
+
+
+
+
+
+
+
+
++++
.
−−−−
−
−
−
−
−
−
−
−
−− − −
Fig 10.7a
Fig 10.7b
.
On an isolated, hollow conducting sphere, all the charge resides on the outside. There is no charge
on the inside, and hence no electric field inside the sphere. Outside the sphere, the field is radial,
as if all the charge was concentrated at the centre.
Essentials Text Book
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48 CHAPTER
Key Ideas10
Stage 1 Physics
Parallel Plates
Consider two large, flat parallel conducting
plates carrying charges of equal magnitude,
but opposite in sign, as depicted in Figure
10.8.
Electrostatic attraction between the opposite
charges causes them to move to the inside of
the plates, and there is no charge on the
Fig 10.8
outside. Consequently, the field is as shown
in Figure 10.8. There is a parallel field between the plates – and no field outside them.
This is an example of a uniform field. The lines of force are evenly spaced throughout the region
between the parallel plates, indicating that the electric field strength is constant throughout.
End Effects
In, fact the field above is not uniform throughout
the total area of the plates. It is a good
approximation to a uniform field within the two
plates, but not at their ends.
The field lines must leave conducting surfaces at
right angles to the surface, and at the ends the
Fig 10.9
surfaces are not parallel – in reality there is
always some curvature. Thus, the net field between these two plates is as depicted in Figure 10.9.
Non-Uniform Conductor
In all the above examples, and in the previous chapter, the conductor has had a uniform shape. For
instance:
•
a sphere has a constant radius of curvature – hence its smooth and even, round shape.
•
a flat plate, as shown in Figures 10.8 and 10.9, has a uniformly flat surface.
When these objects are charged, the excess charge spreads itself over the whole outside surface of
the object, until all repulsive forces cancel each other out and the charge stops moving. The object
now carries an electrostatic charge. If the surface of the object has a
consistently uniform shape, the charges are spread evenly over this surface
when the object is electrostatically charged.
What about conductors that do not have a consistent shape?
Figure 10.10 depicts a typical such object used in school physics – a pearshaped conductor. It is a conical object with a hemisphere at the base of the
cone. The whole object is covered with a conducting material and is
supported on an insulating stand.
Fig 10.10
When a charge is placed on this conductor, it spreads itself over the surface. However, when we
bring this charged conductor near the plate of an uncharged electroscope, we find that the degree
of divergence of the leaves is greatest when the “pointy” end is brought near the plate, and is least
when the hemispherical section is placed near the plate.
From this we can conclude that the charge is not uniformly spread over the surface as in previous
examples. Further experimentation with this phenomenon leads us to conclude that charges tend
to accumulate at points.
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Electric Fields 49
A point is just a region where the radius of curvature of the surface is very small – when a point is
viewed under a microscope, it is curved.
Thus, the smaller the radius of curvature of a surface, the greater is the concentration of
charge in that region.
Why do charges tend to concentrate at points?
Figure 10.11 shows three charges on a flat, electrostatically
charged conducting surface. The charges are evenly distributed
over the surface because, by Coulomb’s law, in this
configuration the repulsive forces on each charge cancel out.
F
F
Fig 10.11
Now consider Figure 10.12. This shows the “pointy” section of the
pear-shaped conductor, where a relatively flat surface merges with
one that has a small radius of curvature. The charge q is equidistant
from the other two positive charges in the diagram.
Therefore, the charge q experiences two repulsive forces (F1 and F2) –
one from each of the other two charges. These two forces are of equal
magnitude F. However, the force F1 acts along the surface of the
conductor, while force F2 acts at an angle to the surface, as depicted.
Fs
F2
q
F1
Fig 10.12
But the charge q can only move along the surface of the conductor, and so its motion is governed
by forces in the direction of the conductor’s surface. It experiences a force F1 to the right and a
force of Fs to the left, where Fs is the component of F2 along the surface. But, the magnitude of Fs
is less than the magnitude of F2, and so there is an unbalanced net force on charge q, and it will
move to the right until the forces in the direction of the surface cancel each other out.
Thus, the charges are forced towards the point where they will accumulate.
The field around such a pear-shaped conductor will have a configuration as depicted in Figure
10.13.
Fig 10.13
Essentials Text Book
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50 CHAPTER
Key Ideas10
Stage 1 Physics
10.3.4 Corona Discharge
The intensity of an electric field near a sharp projection of a conductor can be large enough to
cause the charge to leak away from the conductor at that point.
This is called a corona discharge.
The intensity (strength) of the field near a point may be enough to polarise molecules in the air.
These polarised air molecules are attracted to the point, and then ionised by contact and charge
transfer when they touch it. The ions are then repelled and move away from the conductor. Thus,
charge effectively “sprays away” from the point. This charge transfer will discharge the
conductor.
As the ions move away, they may collide with other molecules in the air – in turn ionising them.
If this process occurs rapidly, a spark will be produced. Lightning is an extreme example of this
effect.
However, a slow leakage results in a corona discharge. A faint violet glow is sometimes produced
as the ionised gases in the air emit photons of light. Thus, a corona (a glow) appears around the
conductor.
A glow discharge can be observed at night at the tips of the masts of sailing ships – a phenomenon
called Saint Elmo’s fire. The same effect can be seen at the wing tips of planes, especially at the
ends of the conducting rods that are used to leak away the substantial static charges that build up
on the outside of planes – not on the inside of the hollow conductor – as they move through the air.
A similar effect is visible on the wing and tail tips of the Space Shuttle on re-entry.
10.4 Electric Field Strength Vector E
10.4.1 Definition
We test for the presence of an electric field in a region of space by placing a charge there. If the
charged body experiences a force due to its charge, then there is an electric field in that region of
space. The direction of the field is defined as the direction of the force on a positively charged
body.
The stronger the field, the stronger the force on the charge. Thus, we can use the strength of this
force as a measure of the strength of the electric field.
The electric field strength vector E at any point in an electric field is defined as the force per unit
positive (test) charge placed at that point in the field.
i.e.
E=
F
q
The unit of electric field strength is newtons per coulomb (NC−1).
Rearranging this formula gives us F = Eq. Thus, if we know the electric field strength at any point
in an electric field, then we know the force on any charge placed at that point. The direction of the
electric force on a charge is parallel to the electric field if the charge is positive, and antiparallel
(opposite to the direction of the field) if the charge is negative.
We use this relationship to calculate the magnitude of the electric field strength at any point in an
electric field. To find the direction of this vector, consider the direction of the force on a positive
charge placed at that point in the field.
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Electric Fields 51
10.4.2 Isolated Point Charge
To find the electric field strength at a point A in the
field of an isolated point charge (+Q), and at a
distance r from it, place a small positive (test)
charge,+q, at the point A. See Figure 10.14.
Coulomb’s law gives us the magnitude of the force
on this charge, +q, given by
F=
1 Qq
4πε 0 r 2
r
.
+Q
.A
Therefore, the magnitude of the electric field
strength is given by
E=
i.e. E =
F
q
Q
4πε 0 r 2
Fig 10.14
Note the following.
•
This formula gives us the magnitude, E, of the electric field strength vector at any point.
•
The direction of the electric field strength, in the depicted example, is directly away from the
charge +Q. If the charge were negative, the electric field strength would be directed towards
the charge.
•
The magnitude of E is the same for all points equidistant from the charge.
•
The magnitude of E is directly proportional to the magnitude of the charge.
•
The magnitude of E at any point is inversely proportional to the square of the distance from the
charge.
Field Lines
An electric field is a vector field. We can calculate the magnitude and direction of the electric
field strength vector E at any point in the field. We depict fields by drawing field lines (lines of
force).
The relationship between the electric field strength vector and the (imaginary) lines of force, or
field lines, is shown in Figure 10.15.
•
The tangent to a line of force, at
any point, gives the direction of E
at that point.
•
The number of lines of force per Field vector at
unit cross-sectional area is
point A
proportional to the magnitude of
E. Where the lines of force are
close together, E is large. Where
they are far apart, E is small.
Field vector
at point B
B
A
Line of force
Fig 10.15
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52 CHAPTER
Key Ideas10
Stage 1 Physics
Example Calculation – Electric Field Strength
Example 5
(1)
Calculate the electric field strength E at a point P, 3cm from a charge of +6µC in a vacuum.
(2)
(3)
Now, if a charge of +10−12 C is placed at P, what force would it experience?
What other information would you need if you were asked to calculate the acceleration of the masses
due to the Coulombic force?
(4)
(a)
What is the electric field strength 6cm from the 6µC charge?
(b)
What is the electric field strength 1cm from the 6µC charge?
E=
Q
4πε 0 r 2
(1)
=
9 × 10 9 × 6 × 10 − 6
(3 ×10 )
−2
= 6 × 10 7 NC
(2)
2
−1
in a direction away from the charge
F = Eq
= 6 × 10 7 × 10 −12
= 6 × 10 − 5 N away from the 6µC charge
(3)
We would need to know the masses of the bodies on which the charges reside, as a = F/m.
(4)
(a)
As E ∝
1
, if d is doubled to 6cm, then E is reduced to
d2
1
4
of its original value.
∴ E = 1·5×107NC−1 away from the 6µC charge.
(b)
If d is reduced to 1cm ( 13 of its original value), then E is multiplied by a factor of 9.
∴ E = 5⋅4×108NC–1 away from the 6µC charge.
10.4.3 Superposition of Electric Fields
We can calculate the magnitude of the electric field strength at any point in the field of an isolated
charge, using
q
E=
4πε 0 d 2
If we wish to calculate the electric field strength due to two or more charges, we must calculate the
individual field strengths and then add them vectorially.
If the field at a point is the result of a number of fields due to charges, the total field ET is found by
vectorially adding all the individual field vectors.
ET = E1 + E2
Because the electric field is defined as a force per unit charge, the law of superposition also applies
here, as in the determination of electric forces between charged objects, using Coulomb’s law.
When calculating the electric field strength at a point due to two charges in one dimension, we
need to:
52
•
calculate the magnitude of the electric field strength due to each of the charges;
•
determine the direction of the field strength due to each of the charges; and
•
add or subtract the two magnitudes, depending on whether the fields are in the same or
opposite directions.
Essentials Text Book
Electric Fields 53
Electric Fields
Example Calculations – Superposition of Electric Field Strengths
Example 6
One-dimensional superposition
Two charges of +2µC and −3µC are separated by a distance of
20cm in a vacuum.
(1) Calculate the resultant electric field strength at X, the midpoint between the two charges as shown.
(2)
What would be the electric field strength at X, if the −3µC
charge were changed to +3µC?
(1)
q1 = +2µC = 2×10−6C, q2 = −3µC = −3×10−6C
q
E=
4πε 0 r 2
(2)
Therefore the electric field strength E1, due to charge
q1, is given by
9 × 10 9 × 2 × 10 −6
E1 =
(10 −1 )2
X
q1
q2
G
If q2 is made positive, then E2 is directed to
the left. Therefore:
G
G
G
E T = E 2 − E1
G
E T = 9 ⋅ 0 × 10 5 NC −1 , to the lef t.
= 1 ⋅ 8 × 10 6 NC −1 , to the right.
The electric field strength E2, due to charge q2, is 1⋅5
times E1 (by proportionality).
G
∴ E2 = 2⋅7×106NC−1, to the right.
G
G
G
G
∴ ET = E1 + E2
i.e. E T = 4 ⋅ 5 × 10 6 NC −1 , to the right.
Example 7:
Two-dimensional Superposition (Extension)
Two charges q1 = +2µC and q2 = −3µC are separated by a distance
of 50cm in a vacuum.
Calculate the electric field strength at the point Y, which is 40cm
from q1 and 30cm from q2.
q2
q1 = +2µC = 2×10−6C, q2 = −3µC = −3×10−6C
q
E=
4πε 0 r 2
ET2 = E12 + E 22
= 1 ⋅125 2 + 2 2
ET = 5 ⋅ 2656
−1
= 1 ⋅125 × 10 5 NC , to the right.
≈ 2 ⋅ 295
2
tan α =
1 ⋅125
≈ 1 ⋅ 7778
The electric field strength E2 due to charge q2 is
9 × 10 9 × 2 × 10 −6
E2 =
2
3 × 10 −1
)
5
40cm
By Pythagoras’ theorem:
)
(
E1
G
G G
ET = E1 + E2
The electric field strength E1 due to charge q1 is
9 × 10 9 × 2 × 10 −6
E1 =
2
4 × 10 −1
(
30cm
Y
q1
(1)
E2
cm
50
α ≈ 60 ⋅ 6
−1
= 2 ⋅ 00 × 10 NC , up
∴
ET
E2
α
E1
D
ET ≈ 2⋅295×105N, at = 60⋅6° to the line joining
q1 and the point Y.
Essentials Text Book
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