Transcript Uniform Electric Fields

Uniform Electric Fields
Electric Fields
Content
• Concept of an electric field
• Force between point charges
• Electric field of a point charge
• Uniform electric fields
Learning Outcomes
Candidates should be able to:
(a) understand an electric field as an example of a field of force and define,
recall and use electric field strength as force per unit positive charge.
(b) use field lines to represent an electric field.
(c) recall and use Coulomb’s law for point charges in a vacuum in the form F =
kQ1Q2 / r2, where k =1 / 4πε0.
(d) recall and use E = kQ / r 2 for the electric field strength of a point charge.
(e) recall and use E = V/d for the magnitude of the uniform electric field strength
between charged parallel plates.
(f) recognise the similarities of, and differences between, electric fields and
gravitational fields.
Electric Field Phenomena
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Rubbed balloon sticking to wall
Lightning
Photocopying
Laser printing
Flue-ash precipitation
Spray Painting
Electric Field Strength
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Gravitational fields are concerned with masses, whereas electric fields are
concerned with charges. An electric field exists in a region of space in which
a stationary charge experiences a force.
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Electric charges create an electric field in the space around them. Unlike
gravitational forces, which are always attractive, electric forces can be
attractive or repulsive.
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The electric field strength E at a point is the force experienced per unit
positive charge on a point charge placed at that point.
E=F/Q
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Where F is the electric force experienced by a particle of charge +Q. Like
force, electric field strength is a vector. Rearranging this equation gives
F = EQ
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Electric field strength has units N C-1.
Problems
Data
k = 1 / 4πεo = 9.0 x 109 N m2 C-2
Charge of electron = 1.6 x 10-19 C
1.
Where the field strength is 1000 NC-1, what is the force on a 1 C charge? On an
electron?
F = EQ = 1000 x 1 = 1000 N
F = EQ = 1000 x 1.6 x 10-19 = 1.6 x 10-16 N
2.
A charged sphere is placed in a field of strength 3 x 104 N C-1. If it experiences a
force of 15 N, what is the charge on the sphere?
Q = F/E = 15 /(3 x 104) = 5 x 10-4 C
3.
What is the field strength if an electron experiences a force of 4.8 x 10-14 N?
E = F/Q = 4.8 x 10-14 / (1.6 x 10-19) = 3 x 105 N C-1
Electric Field Patterns
Electric field patterns are mapped out using electric field lines. They
have the following properties.
– They never start or stop in empty space only from a charge or at
“infinity.”
– They never touch or cross (otherwise a charge placed at that point
would experience forces in different directions).
– They have direction. The tangent to an electric field line gives the
direction in which a positive point charge placed at that point would
move.
– The density of electric field lines indicates the strength of the electric
field. Closely spaced electric field lines indicates greater electric field
strength.
– Parallel and equally spaced electric field line indicate an electric field of
constant field strength (a uniform field).
Demonstration
Semolina grains sprinkled onto the surface
of castor oil align themselves in the
direction of the electric field caused by
differently shaped electrodes/conductors.
• 0.5 cm castor oil + thin layer of semolina
• EHT power supply, 3000-4000 V
Practical
• Plotting equipotentials
Two parallel conductors.
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An isolated metal ring.
Neutral Point
The force exerted on a
charge here would be exactly
equal and opposite
Worksheet
• Field Lines and Equipotentials
Coulomb’s Law
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The magnitude of the force between the charged spheres
is indicated by the deflection of each sphere from the
vertical plane.
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The force can be attractive (unlike charges) or repulsive
(like charges).
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The force increases as the separation decreases.
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The force increases as the charge increases.
F
F
+Q1
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F α Q1Q2 / r2
F = kQ1Q2 / r2
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Where k = 1 / 4πεo
Where εo is the permittivity (how well a material supports
an electric field) of free space = 8.85 x 10-12 F m-1
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Coulomb’s Law of electrostatics states that the magnitude
of the force between two point charges is directly
proportional to the product of the charges and inversely
proportional to the square of their separation.
–ve F = attraction
+ve F = repulsion
+Q2
r
Worked Example
• According to a simple model of the hydrogen atom, the mean
separation between the proton and the electron is 5.0 x 10-11 m.
Calculate the electric force acting on the electron.
• e = -1.6 x 10-19 C
• εo = 8.85 x 10-12 F m-1
• F = - 9.21 x 10-8 N
• The minus sign denotes attraction between the proton and the
electron.
1.
What is the force of repulsion between two electrons held 1 m apart in a vacuum?
By what factor is the electric repulsion greater than the gravitational attraction?
FE / FG = 4 x 1042
2.
By what factor is the electric force between two protons greater than the
gravitational force between them.
FE / FG = 1.2 x 1036
3.
Given the difference in magnitudes of gravitational and electrical forces, why do
you feel gravitational attraction from the earth, but no electrical forces?
Electrically neutral
4.
Human beings are electrically neutral objects to a high degree of accuracy. In this
question you will estimate the force that would exist between 2 students standing
one metre apart if they had just 1% of the electrons in their body removed, leaving
them both positively charged. Take the mass of each student to be 60 kg and as
a rough estimate, assume that humans are 100 % water. The molar mass of H20
(the mass of 6.02 x 1023 molecules) is 18 g.
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How many water molecules do the students contain?
How many electrons are there in a water molecule?
How many electrons are there in total in each student?
Taking 1 % of these away will leave each student with a net positive charge of
1% of their electrons. What is its value?
Now calculate the force between the two students, if they are standing 1 metre
apart and comment.
– How many water molecules do the students contain?
(60/0.018) X 6.02 X 1012 = 2.0 X 1027
– How many electrons are there in a water molecule?
10 (8 from O and 1 from each H)
– How many electrons are there in total in each student?
2.0 x 1028
– Taking 1 % of these away will leave each student with a net positive charge
of 1% of their electrons. What is its value?
1 % = 2.0 x 1026 therefore charge = 2.0 x 1026 x 1.6 x 10-19 = 32 MC
– Now calculate the force between the two students, if they are standing 1
metre apart and comment.
F = 9.3 x 1024 N
This is a huge force on each student, almost the weight of our planet!
Electric Field Strength for a Point Charge
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Consider the force felt by a charge q in the field of another charge Q, where
the charges are separated by a distance r.
F = kQq / r2
+Q
r
+q
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But E = F/q, so
E = k Q / r2
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The electric field strength for a point charge Q is inversely proportional to
the square of the distance from the point charge.
F
Worked Example
Work out the field strengths at the points labelled A and B in the
diagram below. What do you notice about the values, and why is
this? Add arrows at A and B to indicate the electric field strengths
there.
-5C
20 cm
10 cm
B
A
EA = kQ/r2 = -1.124 x 1012 N C-1
EB = kQ/r2 = -4.5 x 1012 N C-1
Values are negative because the charges are negative indicating an
attractive force.
Electric Field Strength Between Charged
Parallel Plates
For a given separation and p.d, the electric field strength
between two parallel plates is constant (uniform field), except at
the edges. A point charge q is moved between the plates and
experiences a constant force F due to the uniform electric field.
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Work done on the charge = energy transformed by the charge
Fd = Vq
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Rearranging this equation, we have
F/q =V/d
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By definition, F/q = E, so
E=V/d
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E is directly proportional to the p.d between the two plates and
inversely proportional to the separation of the plates.
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Units of electric field strength are V m-1 or N C-1
+ + + + + + ++ + + + + + + +
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E = V/d
+q
d
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Worked Example
A charged dust particle is stationary between two horizontal charged metal
plates. The metal plates have a separation of 3.6 cm and the p.d between
the plates is 720 V. The dust particle has a charge of +7e, where e = 1.6 x
10-19 C. Calculate:
a)
The electric field strength between the plates
E = V / d = 720 / 0.036 = 2.0 x 104 V m-1
b)
The weight of the dust particle
The particle is stationary, therefore the net force must be zero.
Weight = Electric force = Eq = 2.0 x 104 x (7 x 1.6 x 10-19) = 2.2 x 10-14 N
To summarise
Gravitational Fields
Electrical Fields
A gravitational field exists in a region of
space in which a stationary mass
experiences a force.
An electric field exists in a region of space
in which a stationary charge experiences a
force.
Gravitational field strength is the force per
unit mass on a point mass placed at that
point.
g=F/m
Electric field strength is the force per unit
positive charge on a point charge placed at
that point.
E=F/Q
F = Gm1m2/r2
F = kQ1Q2/r2
Inverse Square Law
Inverse Square Law
F is attractive only
F is attractive or repulsive
g = Gm/r2
E = kQ/r2
For the field strength due to a point mass
For the field strength due to a point charge