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Electric Fields Lecture
Electrostatics
Electrostatics is the study of electricity at rest. Electrostatics involves
electric charges, the forces between them, and their behaviour in materials.
We are familiar with the force of gravity. It attracts us to Earth, and we call
this weight. Now consider an electrical force acting on us that is billions
upon billions of times stronger than gravity. Such a force could compress a
person into an object with the thickness of paper. However thankfully in
addition to this tremendous attractive force there is also an equal repelling
force. So because each of the electrical forces is balanced, we only perceive
the force of gravity.
Electrical forces arise from particles in atoms. In the simple model of
the atom proposed in the early 1900’s by Ernest Rutherford and Niels Bohr,
a positively charged nucleus is surrounded by electrons. The protons in the
nucleus attract the electrons and hold them in ‘orbit’, akin to the Sun holding
the planets in orbit. Electrons are attracted to protons, but electrons repel
other electrons, exhibiting a property called charge. Electrons are negatively
charged and protons are positively charged. Neutrons have no charge, and
are neither attracted nor repelled by charged particles.
Some facts about atoms:
1. Every atom has a positively charged nucleus surrounded by negatively
charged electrons.
2. All electrons are identical; that is, each has the same mass and the same
quantity of negative charge as every other electron.
3. The nucleus is composed of protons and neutrons. (The common form of
hydrogen, which has no neutrons, is the only exception) All protons are
identical, and similarly all neutrons are identical. A proton has nearly 2000
times the mass of an electron, but its positive charge is equal in magnitude to
the negative charge of an electron. A neutron has a slightly greater mass than
a proton and has no charge.
4. Atoms usually have as many electrons as protons, so the atom has zero net
charge.
The fundamental rule when studying electrical phenomena: like
charges repel and opposite charges attract.
Conservation of Charge
In a neutral atom, there are as many electrons as protons, so there is no
net charge. If an electron is removed from an atom, the atom is no longer
neutral. The atom has one more positive charge (proton) than negative
charge (electron) and is said to be positively charged.
A charged particle is called an ion. A positively charged particle is
called a cation, and a negatively charged particle is called an anion. So if an
object has an imbalance in the number of electrons and protons, the object is
electrically charged.
Three basic ways to charge an object are conduction, induction, and
polarization. Conduction is the flow of charge, induction is the charging of
an object due to the presence of a nearby charged object, and polarization is
the lining up of charges in a non-conductor due to the presence of a nearby
charged object.
Forces Between Charges
The amount of charge on a body (labeled Q or q) is determined by the
number of electrons in excess or less than the number of protons. We
measure the quantity of this charge in coulombs (C) which was named after
the French scientist Charles Augustin de Coulomb. Coulomb set the rate as
being 1 coulomb = the charge on 6.24 x 1018 electrons. So for the charge on
one electron:
1 e- = 1/6.24 x 1018 = -1.602 x 10-19 C (neg. for e-, pos. for p)
While a typical lightning bolt can involve One whole coulomb is a
large quantity to work with at the atomic level. The microcoulomb (μC) is
often used.
1 μC = 10-6 C
Coulomb noticed that the forces of electrostatic attraction and
repulsion seemed to act similar to Newton’s law of gravitation. Using
Newton’s formulas, Coulomb constructed his own law and formula for
electrostatics:
The force between two charges is directly proportional to the product
of their magnitudes, and inversely proportional to the square of the distance
between them.
As a formula:
F = kQ1Q2/d2
Where:
F = force of attraction or repulsion (N)
Q = amount/type of charge (C)
d = distance between objects (m)
k = constant
(the formula is sometimes seen with variations in the letters used, where Q =
q and d = D = r = R, just as we say in Newton's Universal Law of
Gravitation formula)
Now k is slightly different for situations in either air or a vacuum, but
for us we will round this value off to be 9.0 x 109 Nm2/C2
Coulomb's Law is sometimes also written as:
F = (1/4πεo)(q1q2/r2)
The quantity εo is called the permittivity of free space. For a deeper
explanation of what this is and how Coulomb's constant is derived, see the
excerpt at the end of this unit booklet.
For problems dealing with objects of opposite charge, if the force is
positive, they repel each other; if it is negative they attract each other.
Example
Suppose two objects are charged to 0.01 C each and are 10 m apart.
Both are positively charged. How much electrostatic force is between them,
and what type of force is it?
Solution
F = kQ1Q2/d2
F = (9.0 x 109 Nm2/C2)(0.01 C)(0.01 C)/(10 m)2
F = 9.0 x 103 N of repulsive force
Example
Two bodies exert a repulsive force on each other, with charges of 60
μC and 50 μC. The repulsive force is 175 N. What is the distance between
these two charges?
Remember: 1 μC = 10-6 C
Solution
F = kQ1Q2/d2
d2 = kQ1Q2/F
d = √(kQ1Q2/F)
d = √[(9.0 x 109 Nm2/C2)(60 x 10-6 C)(50 x 10-6 C)/175 N]
d = 3.92 x 10-1 m
Example
What is the total charge of all the electrons in 1.000 kg of H2O?
Solution
First, some basic chemistry information:
1. Use the periodic table of elements to find the mass of all the
elements found in the molecule in question. This will give you the
molar mass of the element in g/mol.
2. On the periodic table, the number below the element's symbol is
the molar mass; the number above the symbol is the atomic
number (the number of protons in the element's nucleus)
3. 1 mol = 6.02 x 1023 atoms or molecules (Avogadro's Number)
4. # mol of substance = mass of substance/molar mass
Looking at H2O, we see that there are two atoms of H and one of O.
According to the periodic table, we find that H2O has 10 total protons (1 for
each H, and 8 for the lone O). The number of protons equals the number of
electrons in a neutral atom or molecule. Therefore H2O has 10 electrons.
According to the periodic table, each H atom has a molar mass of 1.01
g/mol, and the O atom has a molar mass of 16.00 g/mol. This means H2O
has a total molar mass of 18.02 g/mol.
We can find the number of moles in 1.00 kg of H2O by using the
formula # mol of substance = mass of substance/molar mass:
mol = 1.000 x 103 g/18.02 g/mol = 5.549 x 101
Therefore the number of electrons in 1.000 kg of H2O is:
(5.549 x 101 mol)(6.02 x 1023 molecules/mol)(10 electrons/molecule)
= 3.34 x 1026 electrons.
Each electron has a charge of -1.602 x 10-19. Therefore the total charge
is:
(3.34 x 1026 electrons)(-1.602 x 10-19 C/electron) = -5.35 x 107 C
Fields and Forces
We describe what happens around a positive object by stating that a
positively charged object is surrounded by an electric field. If we bring a
small positive charge close to the source of the electric field, the electric
field will exert an outward force on this positive charge. It would exert an
inward force on a negative charge. The electric field intensity is defined as
the force that the field exerts on a unit of charge. As a formula:
E=F
q
E – electric field intensity (N/C)
F – force (N)
q – charge (C)
Example
What is the electric field intensity at a point where a positive charge of
2.4 x 10-6 C experiences an outward electric force of 3.2 N?
Solution
E=F
q
E = 3.2 N/2.4 x 10-6 C = 1.3 x 106 N/C
There is another way to determine electric field strength. This
involves just the charge on the object creating the electric field on a test
charge and a certain distance from this object in question. This formula is
created by:
F = kQ1Q2/d2
F = kQQtest/d2
F/Qtest = kQ/d2
Since E = F/Q, we can have:
E = kQ
d2
E = electric field intensity
Q = charge (C)
*** Q or q represents charge
d = distance (m)
k = 9.0 x 109 Nm2/C2
Example
What is the electric field strength at a distance of 0.50 m from an
object in air having a charge of 2.5 x 10-5 C?
Solution
E = kQ
d2
E = (9.0 x 109 Nm2/C2)( 2.5 x 10-5 C)/(0.50 m)2
E = 9.0 x 105 N/C
Electric Potential
The force that a given electric field would exert on a charged particle
can be found by multiplying the charge on the particle by the electric field
strength. The equation for this was:
E=F
q
or
F = Eq
This equation is also part of calculating how much force and energy
that would be required to actually move a charged particle against the force
of a given electric field. For example, let’s assume that between the two
plates of a capacitor the electric field strength is 2.0 x 104 N/C and that the
distance between the two plates is 0.01 m.
If we place a positive charge of 1.0 C at the edge of the negative plate,
what force is required to move the positive particle upward to the positive
plate?
First, calculate the force required to even start moving the positive
particle against the force of the electric field by using:
F = Eq
F = (2.0 x 104 N/C)(1.0 x 10-6 C)
F = 2.0 x 10-2 N
This is the force needed to get the particle moving, but how much
work is required to move it the full 0.01 m? We use the following formula
from our study of work and energy:
W = Fd or W = Fd
W – work (J)
F – force (N)
d – distance (m)
So:
W = Fd
W = (2.0 x 10-2 N)(0.01 m)
W = 2.0 x 10-4 J
This answer is the solution to not only the amount of work required to
move the positive particle to the other side of the capacitor; it is also the
solution to how much electric potential energy that the positive particle
gains. The electric potential energy of a certain charged object can be
calculated by comparing the amount of work that a given particle goes
through with the actual charge of the particle. We can write this as a ratio
and give it the name voltage. Voltage is a term meaning how much work is
done per charge. Another term for voltage is electric potential, and this is the
electric potential energy per charge. As a formula:
V=W
q
V – voltage (V, or J/C)
W – work (J)
q – charge (C)
Using the above example again, we could calculate the electric
potential as follows:
V=W
q
V = (2.0 x 10-4 J) /(1.0 x 10-6 C)
V = 2.0 x 102 V
Another useful voltage formula can be derived from:
V=W
q
and
F = Eq
The formula is:
V = Ed
and
W = Fd
Example
The voltage between two points that are 0.10 m apart in an electric
field is 80 V. What is the electric field strength?
Solution
V = Ed
E = V/d
E = (80 V)/(0.10 m)
E = 8.0 x 102 N/C
Coulomb's Law in One Dimension
Example
Charges are lined up as per the diagram below. Calculate the net force
on charge C.
Solution
The force on charge C caused by charge A is:
FAC = kQ1Q2/d2
FAC = (9.0 x 109)(4.0 x 10-6)(6.4 x 10-6)/(0.300 m)2
FAC = 2.56 N right
The force on charge C caused by charge B is:
FBC = kQ1Q2/d2
FBC = (9.0 x 109)(-2.5 x 10-7)(6.4 x 10-6)/(0.100 m)2
FBC = -1.44 N left
FT = FAC + FBC = 2.56 + (-1.44) = 1.2 N with 2 sig. figs.
Coulomb's Law in Two Dimensions: Equilateral
Triangle
Example
There are three negatively charged points, each with a charge of -4.00
-6
x 10 C, fixed at the vertices of an equilateral triangle whose sides are 20.0
cm long. The base of the triangle runs east/west. Find the net force on the
northernmost point.
Solution
Create a triangle as per the description, and label the three vertices
however you prefer. Such a triangle could look like:
Since the charges are all of the same magnitude, the forces are all of
the same magnitude.
F = kQ1Q2/d2
F = (9.0 x 109)(4.00 x 10-6)(4.00 x 10-6)/(0.200 m)2
F = 3.60 N
To determine the direction of the net force, we will need to know the
directions of the two forces acting on the top charge and then add these two
forces by using trigonometry. Since the triangle is an equilateral triangle, the
angles of the forces are at 60.0o.
The force on charge 1 by charge 3 is 3.60 N at 60.0o N of W.
The x component of this force is (-3.60)(cos60.0o) = -1.80 N
The y component of this force is (3.60)(sin60.0o) = 3.11 N
The force on charge 1 by charge 2 is 3.60 N at 60.0o N of E.
The x component of this force is (3.60)(cos60.0o) = 1.80 N
The y component of this force is (3.60)(sin60.0o) = 3.11 N
The sum of the x components:
-1.80 + (1.80) = 0.00 N
The sum of the y components:
3.11 + 3.11 = 6.22 N
Since there is no x component, the total force is 6.22 N north.
Coulomb's Law in Two Dimensions: Square
Example
Four point charges A, B, C, and D, each with a magnitude of 4.00 x
-6
10 C are found at the corners of a square whose sides are 20.0 cm long.
Two of the charges are positive and two are negative as shown in the
diagram below. Determine the net force acting on charge A.
Solution
It can be beneficial to ignore the polarity of each charge and indicate
the influences on charge A via stating the force's direction.
The force on A due to B is:
FAB = kQ1Q2/d2
FAB = (9.0 x 109)(4.00 x 10-6)(4.00 x 10-6)/(0.200 m)2
FAB = 3.60 N east
The force on A due to C is
FAC = kQ1Q2/d2
FAC = (9.0 x 109)(4.00 x 10-6)(4.00 x 10-6)/(0.200 m)2
FAC = 3.60 N south
To determine the force on A due to D, it is first necessary to know the
distance between A and D. This distance is the hypotenuse of the triangle
ABD.
AD2 = AB2 + BD2
AD = √[(AB)2 + (BD)2]
AD = √[(0.200)2 + (0.200)2]
AD = √(8.00 x 10-2) or AD = 2.82 x 10-1 m
The force on A due to D is
FAD = kQ1Q2/d2
FAD = (9.0 x 109)(4.00 x 10-6)(4.00 x 10-6)/(√(8.00 x 10-2))2
FAD = 1.80 N 45.0o N of W (45.0o because this is a square)
To find the total x component of the net force:
Fax = FABx + FACx + FADx = 3.60 + 0.00 + (-1.80)(cos45.0o) = 2.32 N
To find the total y component of the net force:
Fax = FABx + FACx + FADx = 0.00 + (-3.60) + (1.80)(sin45.0o) = -2.32 N
The magnitude of the total force is:
Fr = √[(FAx)2 + (Fay)2]
Fr = √[(2.32)2 + (-2.32)2]
Fr = 3.28 N