Transcript Chapter 16 Electric Forces and Fields lecture slides

Chapter 16 Review
Coulomb’s Law and Electric Fields
Objectives
•To study quantitatively the nature of forces between electric charges
•To develop the concept of the electric field as a transmitter of electric force
•To learn how work is done and potential energy is stored in an electric field
•To learn how to use electric fields to control and direct the motion of electric charges
Coulomb’s Law
Experiment shows that the
electric force between two
charges is proportional to the
product of the charges and
inversely proportional to the
distance between them.
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Coulomb’s Law
Coulomb’s law:
Where Q1 and Q2 are the amount of charge and k is a
proportionality constant
Charges produced by rubbing ordinary objects
(such as a comb or a plastic ruler) are typically
around a microcoulomb or less:
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Coulomb’s Law
The magnitude of the charge of an electron, on the other hand,
has been determined to be about 1.602 x 10-19 C, and its sign is
negative. This is the smallest known charge, and because of its
fundamental nature, it is given the symbol e and is often referred
to as the elementary charge:
Example:
How many electrons make up a charge of -30.0 micro coulombs
(C)?
N = Q/e = (-30 x 10-6 C)/ (-1.60 x 10-19 C/electrons) = 1.88 x 1014 electrons
What is the mass of 1.88 x 1014 electrons?
Mass = (9.11 x 10-31 kg)(1.88 x 1014 electrons) = 1.71 x 10-16 kg
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Coulomb’s Law
The charges carried by the proton and electron are
equal in size. However, the mass of the proton is
2000 times the mass of the electron.
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Coulomb’s Law

Double one of the charges
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
Change sign of one of the charges
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force stays the same
Double the distance between charges


force changes direction
Change sign of both charges
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
force doubles
force four times weaker
Double both charges

force four times stronger
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Coulomb’s Law vs. Law of
Universal Gravitation

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F = kQ1Q2/r2 vs. F=GM1M2/r2
Both are inverse square laws F1/r2
Both have a proportionality to a product
of each body-mass for gravity, electric
charge for electricity.
A major difference is that gravity is
always
an ofattractive
whereas
the
Comparison
electrical force vs.force,
Gravitational
force #1
electric
force
canforce
bevs. wither
or
Comparison
of electrical
Gravitational attractive
Force #2
repulsive.
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Electrical Force is stronger than
Solving Problems involving Coulomb’s Law
Sample problem
Find the force between two positive 1.0 C charges
when they are 1000m apart?
Solution
q1=q2 = 1.0C
r = 1000m
F = kq1q2/r2 where k = 9.0 x 109 Nm2/C2
After substitution, F = 9.0 x 103 N
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Solving Problems involving Coulomb’s Law
Sample problem
What is the magnitude of the electric force of
attraction between an iron nucleus (q = +26e) and
its innermost electron if the distance between them
is 1.5 x 10-12 m?
Solution
F = kq1q2/r2 where k = 9.0 x 109 Nm2/C2
F = (9.0 x 109 Nm2/C2)(26)(1.6 x 10-19 C)(-1.6 x 10-19
C)/ (1.5 x 10-12 m)2 = -2.7 x 10-3 N
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Solving Problems involving Coulomb’s Law
Sample problem
What is the repulsive electrical force between two
protons in a nucleus that are 5.0 x 10-15 m apart
from each other?
Solution
F = kq1q2/r2 where k = 9.0 x 109 Nm2/C2
F = (9.0 x 109 Nm2/C2)(1.6 x 10-19 C)(1.6 x 10-19 C)/
(5.0 x 10-15 m)2 = 9.2 N
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Solving Problems involving Coulomb’s Law
Sample problem
Two charged balls are 20.0 cm apart. They are
moved, and the force on each of them is found to
have been tripled. How far apart are they now?
Solution
Let F1 = kq1q2/r12 and F2 = kq1q2/r22 where F2 = 3 F1
F2/F1 = r12 /r22
3= [(20.0cm)/r2]2, which gives r2 = 11.5 cm
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The Electric Field
In physics, the space surrounding an electric
charge has a property called an electric field.
This electric field exerts a force on other
electrically charged objects. The concept of
an electric field was introduced by Michael
Faraday.
The electric field is a vector field with SI units
of newtons per coulomb (N C−1) or,
equivalently, volts per meter. The strength of
the field at a given point is defined as the
force that would be exerted on a positive test
charge of +1 coulomb placed at that point; the
direction of the field is given by the direction
of that force.
Michael Faraday (1791-1867)
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The Electric Field

One can think of electric force as establishing a “field” telling
particles which way to move and how fast
Electric “field lines” tell a positive
charge which way to move.
For example, a positive charge itself
has field lines pointing away from it,
because this is how a positively-charged
“test-particle” would respond if placed
in the vicinity (repulsive force).
+
Run Away!
+
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The Electric Field
The direction of the
electric field is
always directed in
the direction that a
positive test charge
would be pushed or
pulled if placed in the
space surrounding
the source charge
+
+
+
+
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The Electric Field
Measuring an electric field is a quite simple process involving a
test charge. To measure the strength of an electric field, first a
test charge must be placed in its vicinity, then calculate the force
the test charge “feels”. The resulting number is the strength of
the electric field. This process is simplified into the following
equation
In this equation,F is the magnitude of the force, as found by using
Coulomb's Law, q is the magnitude of the test charge. The
resulting electric strength is measured in Newton’s per a
Coulomb.
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Electric Field vs. Gravitational Field
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Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8
m/s2 and points straight down. If you threw a mass through the air, you know it would
follow a parabolic path because of gravity. You could determine when and where the
object would land by doing a projectile motion analysis, separating everything into x
and y components. The horizontal acceleration is zero, and the vertical acceleration is
g. We know this because a free-body diagram shows only mg, acting vertically, and
applying Newton's second law tells us that mg = ma, so a = g.
You can do the same thing with charges in a uniform electric field. If you throw a
charge into a uniform electric field (same magnitude and direction everywhere), it
would also follow a parabolic path. We're going to neglect gravity; the parabola comes
from the constant force experienced by the charge in the electric field. Again, you
could determine when and where the charge would land by doing a projectile motion
analysis. The acceleration is again zero in one direction and constant in the other. The
value of the acceleration can be found by drawing a free-body diagram (one force, F =
qE) and applying Newton's second law. This says: qE = ma, so the acceleration is a =
qE / m.
The one big difference between gravity and electricity is that m, the mass, is always
positive, while q, the charge, can be positive, zero, or negative.
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The Electric Field
Sample Problem
A positive charge of 1.0 x 10-5C experiences a force of
0.30N when located at a certain point. What is the electric
field intensity at that point?
Solution
E=F/q = 0.30N / 1.0 x 10-5 C = 3.0 x 104 N/C
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The Electric Field
Sample Problem
A test charge experiences a force of 0.20 N on it when it
is placed in an electric field intensity of 4.5 x 105 N/C.
What is the magnitude of the charge?
Solution
q=F/E = 0.20N / 4.5 x 105 N/C = 4.4 x 10-7 C
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The Electric Field
Sample Problem
A positive charge of 10-5 C experiences a force
of 0.2N when located at a certain point in an
electric field. What is the electric field strength
at that point?
Solution
F= 0.2N
q=10-5C
E= F/q = 0.2N/10-5C = 2 x 104 N/C
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Measuring Potential Difference
(Voltage)
Electric Strength can also be defined in units of work and
energy. This alternative way of measuring the strength of
an electric field entails finding the potential difference,
which exits between any two points. To visualizing this
difference picture an electric field, then trying to force a
test charge in between two points, if the test charge is
repelled, we have to do work to push it into place. This
process is also simplified into the formula
In this formula, W is work, q is the magnitude of charge,
and V is voltage or potential difference.
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Measuring Potential Difference
(Voltage)
Sample Problem
It takes 5.0 x 10-3 J of work to move a positive charge
of 2.5 x 10-4C from point X to point Y on an electric
field. What is the difference of Potential between X
and Y?
Solution
W= 5.0 x 10-3 J
q = +2.5 x 10-4C
V = W/q = 5.0 x 10-3J/2.5 x 10-4C = 20 J/C = 20 volts
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Electronvolt
The joule is too large a unit for measuring the work done
on moving elementary charges, such as electrons,
protons, or the small charges in ions. For this purpose,
the electron volt is a more convenient unit of energy or
work. An electron volt is the work done in moving an
electron or other body having a unit of elementary charge
through a difference of potential of one volt. Thus,
1eV = 1.60 x 10-19 J
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Electronvolt
Sample Problem
The difference of potential between point X and point Y
of an electric field is 100 volts. (a) How much work
done in electronvolts is done by the electric field in
moving a free electron from point X and Y. (b) What
happens to this work? (c) What is this work in Joules
(a) V = 100 V q = +1 electron charge
W= 100volts and 1 electron’s charge = 100 eV
(b) This work is used to accelerate the proton. It is
converted into the KE gained by the proton on
moving from point X to Y.
© 1 ev = 1.60 x 10-19J, so…100eV = 1.6 x 10-17J
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Electrical Potential Energy
In expressing the gravitational potential energy of a body,
we learned that a base level such as the surface of the earth
must be arbitrarily selected as the level corresponding to
zero PE. Similarly, in expressing the PE of a charge in a
an electric field, a position of charge corresponding to the
base or zero potential energy must first be selected.
A charge in the electric field of a point charge is said to
have zero PE when it is at an infinite distance from the
point charge. This is expressed as:
U = K (q1 q2)/r
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Electrical Potential Energy
Sample Problem
Assuming that a hydrogen atom consists of one electron and
one proton separated by a distance of 5.3 x 10-11 m, what is the
PE of the electron in the field of the proton?
Solution
U = K (qprotonqelectron)/r = -4.3 x 10-18 Nm or J
Note that the PE is negative because the charges are opposite
and attract each other.
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Measuring Electric Charge
The quantity of charge (q) on an object is related to the (unbalanced)
number of electrons that have been either gained or lost by the object.
The unit of charge is the coulomb (C).
The charge on one electron is the smallest charge known to exist
independently and has the value of 1.60 x 10-19 coulomb. This value of
the electron charge is known as the fundamental charge, e = 1.60 x 1019 C. Therefore every electron has a charge of -e and every proton (or
positive charge due to the loss of one electron) has a charge of +e.
A rubber balloon becomes negatively charged after you rub the balloon
with a wool cloth. The quantity of charge due to the excess electrons on
the balloon can be found according to the following general relationship:
Quantity of (positive or negative) charge, q = (number of electrons, Ne) x
(fundamental (electron) charge, +e or - e)
So, q = (Ne)(e)
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Measuring Electric Charge
Sample Problem
What is the charge of 1000 electrons?
Q = (Ne)(e) = 1000 (-1.6 x 10-19 C) = - 1.6 x 10-16 C
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Field Lines
The electric field can be represented by field lines.
These lines start on a positive charge and end on a
negative charge.
At locations where electric field lines meet the
surface of an object, the lines are perpendicular to
the surface.
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Field Lines
Electric dipole: two equal charges, opposite in sign:
•Field lines indicate the
direction of the field; the field is
tangent to the line.
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Field Lines
The number of field lines
starting (ending) on a positive
(negative) charge is
proportional to the magnitude
of the charge.
Electric field lines never cross
each other.
The electric field is stronger
where the field lines are
closer together.
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Field Lines
The electric field between two closely spaced,
oppositely charged parallel plates is constant.
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Field Lines
Summary of Field lines Around Charges
•
The magnitude of the field is proportional to the density of the lines.
•
Field lines start on positive charges and end on negative charges
•
Field lines indicate the direction of the field; the field is tangent to the
line.
•
The electric field between two closely spaced, oppositely charged
parallel plates is constant.
•
At locations where electric field lines meet the surface of an object, the
lines are perpendicular to the surface.
•
Electric field lines never cross each other.
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Millikan’s Oil Drop Experiment
One important application of the uniform electric
field between two parallel plates is the measurement
of charge of an electron. This was determined by
American physicist Robert A Millikan in 1909
Millikan’s experiment showed that charge is quantized. This
means that an object can only have a charge with a
magnitude that is some integral multiple of the charge of the
electron (1.6 x 10-19 C).
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Millikan’s Oil Drop Experiment
Sample Problem
In a Millikan oil drop experiment, a drop has been found to
weigh 1.9 x 10-14 N. When the electric field is 4.0 x 104 N/C,
the drop is suspended motionless. (a) what is the charge on
the drop? (b) If the upper plate is positive, how many excess
electrons does the oil drop have?
Solution
(a) When balanced, Felectric = F gravity Thus, qE=mg solving for q, the charge
will be
q=mg/E = 1.9 x 10-14 N/4.0 x 104N/C = 4.8 x 10-19 C
(b) Determine the number of electrons by n=q/e
n=4.4 x 10-19 C/1.6 x 10-19 C/electron = 3 electrons
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Millikan’s Oil Drop Experiment
Sample Problem
A positively charged oil drop weighs 6.4 x 10-13 N. An electric field of
4.0 x 106 N/C suspends the drop. (a) What is the charge on the
drop? (b) How many electrons is the drop missing?
Solution
(a)Q=F/E = 6.4 x 10-13 N/ 4.5 x 106 N/C
= 1.6 x 10-19 C
(b) N = Q/e = 1.6 x 10-19 C/1.6 x 10-19 C/electron =
1 electron
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Summary
•
•
•
•
•
•
•
•
Coulomb’s Law: F = kQ1Q2/r2
Units of Coulomb’s Law is the Newton.
Electric Field: E = F/Q
Units of Electric Field are Newtons per Coulomb
Potential Difference: V = W/Q
Units of Potential difference is Joules per Coulomb
Work (W) = QV
The work done in moving an electric charge can be
expressed in joules or electron volts (eV).
• One electron volt is equal to 1.60 x 10-19 joule
• Potential Energy of a charge: U = kQ1Q2/r
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