Transcript 5. Fields and Electrical Physics

COULOMB’S LAW
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Oppositely charged particles
experience force of attraction;
similarly charged particles
experience force of repulsion F
F – forces
Q1, Q2 – magnitudes of
charges
r – distance between charges F
Each particle experiences
same force as its counterpart,
regardless of which has
greater charge
Charges will move in
directions indicated by arrows
(provided there are no
disturbances)
Magnitude of force F is
directly proportional to the
product of charges Q1Q2, and
inversely proportional to r2
Example A 40µC positive
charge and a 100µC negative
are separated by 50mm. Find
the force of attraction
between the charges.
Q1
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Q2
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F
Q2
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F
r
Q1
r
Q1
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F
F
Q2
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r
Coulomb’s Law
kQ1Q2
F
r2
F = force, Newtons
k = constant, 9109Nm2C-2
Qn = charge, Coulombs
r = distance, meters
Fields and Electrical
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ELECTRIC FIELDS (1)
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Force experienced
by positive charge
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Isolated
negative
charge
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An electric field is represented by the lines that show the
direction of the force on a positive charge placed
anywhere in the field
Positive charges are drawn toward the isolated negative
charge
The lower figures show the electric field lines in the vicinity
of an isolated negative charge and an isolated positive
charge respectively
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ELECTRICAL FIELDS (2)
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The lines shown on the previous slide are a pictorial
way of visualising the reaction of any positive
charge to the presence of a fixed negative charge
There are infinite number of lines, but only a few
can be shown
The negative charge is responsible for an electric
field that are represented by the lines
Electric field lines are used to predict the behaviour
of a positive charge placed anywhere in the field
The charge will experience a force in the direction
shown by a line at the point where the charge is
placed
For a single isolated positive charge, the electric
field lines radiate outward, as any positive charge
placed in the vicinity of it would be repelled away
from it
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ELECTRIC FIELDS (3)
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When two or more charges occupy a fixed locations
in space, electric field pattern depends on the
magnitudes of the charges and their locations with
respect to each other
Left figure shows electric field established by fixed
positive charge in vicinity of fixed negative charge
Lines always originate at a positive charge and
terminate at a negative charge: from +ve to –ve
The lines show the direction a positive charge would
move if placed in the field
Point charges are fixed points of charges
Right figure shows electric field that results when
two sheet charges (charges distributed over
surfaces) are placed in the vicinity of each other
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ELECTRICAL FLUX DENSITY
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An electric field pattern shows the direction of force
on a positive charge placed in field
It can also be used to show the relative magnitude
of the force experienced by a +ve charge placed at
any point in the field
Where the lines are close together (dense), force on
+ve charge is greater than it is in a region where
lines are less dense (see previous slide)
Lines are more closely spaced nearer the charge
Confirmed by Coulomb’s Law: The closer we move
a +ve charge to fixed –ve charge, the greater the
force on it
Further away from fixed negative charge, field lines
are less dense, thus smaller force
Number of lines on field diagram is arbitrary – there
are an infinite number of lines for +ve charges to
move along
For computational analysis, it is convenient to
assume that the number of lines produced by a
charge is the same as the charge in Coulombs
Instead of lines, the term electric flux is used, ψ,
and has Coulombs as units
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FLUX DENSITY
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The notion of flux, and its units, serves only as a
convenient basis for mathematical computations
We could not draw a field diagram showing one
one-millionth of a line corresponding to 1µC of flux!
Electrical flux gives us a basis for defining a
numerical quantity that reflects how closely spaced
the lines are in an electric field
We must visualise the lines in 3D space around a
charge
Flux density D is defined to be flux per unit area
D = ψ/A Coulombs/meters2 (Cm-2)
For the above formula to be valid, the surface area
used in the computation has to be perpendicular to
the flux lines at every point where the flux
penetrates the area
Depending on the field pattern, the area A may be a
curved surface
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FLUX DENSITY EXAMPLE
Small value of D = ψ/A
Flux ψ
Large value of
D = ψ/A
+
Same size area A
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Example One of the areas between two charged
surfaces measures 6mm by 8mm, and the flux
penetrating it is 96µC. Each of the charged surfaces
measures 2.5cm by 4cm. What is the flux density in
the region between the charged surfaces? What is
the total flux in the region between the charged
surfaces?
Suppose that each dimension of the two charged
surfaces is doubled, but the total charge on each
surface remains the same. What, then, is the flux
density in the region between the surfaces?
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ELECTRIC FIELD INTENSITY
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Electric field intensity, also called electric field
strength, is the ratio of the force experienced by the
charge placed in the field to the magnitude of the
charge itself: E = F/Q Newtons/Coulomb (NC-1)
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The stronger the field, the greater the force a given
charge will experience when placed in the field
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Values for E will be different at different points of the
field
The field intensity in a region close to a positive
charge is greater than it is at a long distance from
the charge (see previous slide)
Where the flux is denser in the region near to a fixed
charge, the force experienced is greater
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PERMITTIVITY
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Points on previous slide suggest there is a
relationship between D and E
Each is related to the magnitude of the force
experienced by a charge placed in an electric field
The denser the flux (greater D), the greater the
force on such a charge
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The greater the field intensity E, the greater the
force on the charge
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D and E are proportional to each other: D = E
 is a constant whose value depends on the
material in which the field is established (air, glass,
water, etc.). It is called the permittivity of the
material
Typical values are: 8.8410-12 for a vacuum,
6.610-8 for certain ceramics
Example When a 1000µC charge is placed in a
certain electric field, it experiences a force of 28.2N.
If the field exists in a vacuum, find the field intensity
at the point where the charge is placed. Also find the
flux density at the point where the charge is placed
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FIELD INTENSITY AND VOLTAGE
+Vvolts
E
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d
A voltage always exists between two oppositely
charged regions
A field is said to be uniform when the flux density of
the electric field in a region between two charged
surfaces is the same everywhere (assuming two
perfectly parallel surfaces as above)
The electric field intensity between two charged,
parallel surfaces can be computed from the voltage
difference v across the surfaces
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The units of E (N/C) are equivalent to V/m
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For two charged parallel surfaces: E = V/d Vm-1
Example Two parallel surfaces are separated by
12mm. A 600µC charge placed between them
experiences a force of 7.2N. What is the voltage
difference between the surfaces?
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