Download II. Electric Force III. Electric Field IV. Electric Potential

Physics A
Unit 11
Electrostatics
From unit 11 through unit 13 you will study another of the major branches of physics,
electromagnetism. In this unit we consider electric forces for particles at rest, known as
electrostatics. Be familiar with the material outlined below before test 301.
I. Charge
A. Defined
B. Methods of Charging
1. By Friction
2. By Contact
3. By induction
II.
Electric Force
A. Between two particles
B. Among multiple, collinear particles
C. Among multiple, coplanar particles
III. Electric
Field
A. Defined
B. Particles interacting with the field
C. Particles as sources of field
IV. Electric
Potential
A. Definition of the Volt
B. Field and Volt relation
C. Particle Accelerator
D. Electric Potential for Point Charges
This unit should take approximately eight days to complete as follows. Day 1 is I. Day 2
is II. Day 3 is III. Day 4 is IV.A-C. Day 5 is IV.D. Day 7 is practice test day and day 8
will be the test. A lab mapping E fields is also necessary.
Lesson 3-01
Charge Defined (pp. 557-563)
Particles that react to electromagnetic forces have the property of charge associated with
them. One could construct the analogy that mass is to gravity as charge is to electricity.
At the end of this unit you should be able to write a wonderful compare and contrast
paper between electricity and gravity.
If you have already completed a course in chemistry then the concept of charge
should be familiar to you. Unlike in chemistry where +1, 0 and -1 were good enough for
conserving charge we need to be a little more exact. The symbol for charge is q. Charge
is measured in Coulombs or C.
You have been taught that all
Name
Symbol Mass (Kg) Charge (C)
matter is composed of protons,
Neutron
n
1.67E-27
0
neutrons and electrons. The
symbol, mass and charge for
Proton
p+
1.67E-27
+1.6E-19C
these basic particles are listed
Electron
e9.11E-31
-1.6E-19C
to the right. You should
refer to this table as you
complete problems in this unit.
Most electrical charges are very small in value. Because the values are small one
needs to become reacquainted with the Latin prefixes for E-3 (milli or m), E-6 (micro or
µ), E-9 (nano or n) and E-12 (pico or p). You should know these for the next test.
The most common way to charge an object is by friction. Let two, neutral
surfaces rub together; electrons can be transferred from one surface to another. Since you
cannot transfer a fraction of an electron one should recognize that charge is always a
whole number multiple of the electron charge.
q=Ne
where e = 1.6E-19C
The object gaining electrons will become
N = 0, ±1, ±2, ±3, …
negatively charged. The object losing
electrons becomes positively charged.
The second most common way to charge is by contact. Suppose that you drag
your foot across a new carpet to build up a charge. That would be charging by friction.
Then you sneak behind somebody and touch them gently on the ear lobe until there is a
loud pop followed by a mild scream. Since you were charged and they were not some of
the charge that was on the surface of your body transferred to them. When a charged
object touches a neutral object and charge is transferred we call that charging by contact.
We finish the second method of charging with an example and an after thought.
Example #1
Shown below are two identical, conducting spheres that are mounted on glass stands.
Each sphere is charged as shown below. If the two spheres are made to touch by a person
holding the glass stands or made to connect with an insulated wire what will be the final
charge on each stand?
+24 µC
-10 µC
Since both spheres are conducting then the charges are free to move between the surfaces
of the two spheres. The negative charge on the right side is totally neutralized by the
positive charge on the left. The net charge remaining is +14µC. Since the spheres are
identical the net charge will evenly distribute between the two spheres. The final charge
distribution is +7µC on either sphere.
The final thought about charging by contact has to do with where the electrical
discharge will occur. Excess charge has a tendency to transfer at the tip of a pointed
surface like your fingernail or even a car key. If you are always getting shocked during
the winter when the air is dry and electrically insulating try the following. Before
touching a surface that would accept excess charge from you, take out your car keys and
grasp them firmly in the palm of your hand. Touch the tip of one key gently to the
surface of the neutral object. The charge transfer should be concentrated at the tip of the
key while being evenly distributed over the palm of your hand. No more pain! Also note
that if you are caught in an electrical storm you are encouraged to keep your hands inside
of your armpits. Do you see why? Umbrellas have recently been modified to have blunt
ends instead of sharp, metallic points. Thank you Benjamin Franklin!
The third way to charge is by induction. It is called “induction” because you use
a charged object to polarize and then separate charge from neutral objects. Reconsider
the conducting spheres from the previous page. In this thought experiment both start out
electrically neutral, figure A below. A positively, charged rod is brought near one of
spheres, figure B below. Since both spheres are conductors, electrons on either sphere
will move closer to the rod and leave the side of the sphere farther from the rod more
negative. This is known as polarization. While polarized the two spheres are made to
touch allowing the negative charge from one sphere to neutralize the positive charge of
the other sphere, figure C. Finally, the spheres are once again separated and then the
charged rod is removed, figure D. Now two neutral objects have charged each other in
the presence of a charged object. This is known as induction because the charged object
induced the two neutral objects to become charged.
q=0
q=0
Fig A
++++
−
−
Fig C
++++
−
−
+
+
−
−
+
+
Fig B
+
+
−
−
Fig D
+
+
Homework problems for lesson 3-01
1. A deuteron is a subatomic particle composed of a proton and neutron stuck together.
What are the approximate mass and charge of this particle? An alpha particle is two
deuterons stuck together. What are the approximate mass and charge of an alpha?
2. Which of the following charges is not possible?
A) +3.2E-19C B) -4.0E-19C C) -4.8E-19C D) +5.6E-19C E) -4.0E-18C
3. What is the mass and charge of 2.8E+3 electrons?
4. If the charged, conducting spheres of example 1 were +24µC and +10µC what would
be the final result for the example? Repeat with -24µC and + 10µC.
Lesson 3-02
Coulomb’s Law (pp. 564 – 571)
The force between two point charges can be determined with a single equation. The
direction of the force is expressed by recall of the statement that opposite charges attract
and like charges repel. The equation can also
FE = k q q′/ r2
be used to determine the direction if you plug
in the polarity of each charge. A positive force
1) q is the charge pushing
is repulsive and a negative force is attractive.
2) q′ is the charge getting pushed
The magnitude of the force is found using the
3)
k = 9.0E+9 Nm2/C2
equation in the text box to the right. You can
store the value of k = 9 E+9 in your calculator
using the sequence of key strokes: [9], [2nd], [,], [9], [STO], [ALPHA], [(], [ENTER].
The above equation is known as Coulomb’s Law after the discoverer, Charles Coulomb.
Use the above equation with absolute value of charges to find the magnitude of
the force vector. Use the italicized statement to find the direction of the force vector.
Example #2
A +8µC charge is placed at the origin. A −5µC charge is placed on the x-axis at
x=+20cm. Find the force that each charge exerts on the other.
+8µC
−5µC
|
|
|
|
|
|
|
x-axis
0
4
8
12 16
20
24
cm
|F| = (9E9 Nm2/C2)(8E-6C)(5E-6C)/ (0.2m)2 = 9.0 N Notice that the polarity of each
charge is ignored in finding the magnitude. Since opposites attract, the 8µC will
experience a 9 N force towards the other charge. The 8µC charge experiences a force of
9 N to the right. The −5µC charge experiences a force of 9 N to the left.
Collinear Charges
If multiple charges are on the same line then you find the force that each charge (q) exerts
on to the charge in question (q′). Consider the figure below. What is the net force on the
charge located at the origin? In this case, q′ = +8µC. You know that the force from the
−5µC charge exerts a force of 9 N to the right. The +7µC charge exerts a force of 4.92N
to the left. Subtracting gives the net force of 4.08 N to the right.
+8µC
−5µC
+7µC
|
|
|
|
|
|
|
|
|
x-axis
0
4
8
12 16
20
24
28
32
cm
You should recognize that these electricity problems are nothing more than using
Coulomb’s Law to get the magnitude and direction of force vectors from charge pairs
then using the steps of vector addition to arrive at an answer.
Coplanar Charges
With coplanar charges you take the same steps as in the previous example except that the
force vectors will have to be broken into x parts and y parts then added to form a triangle.
You may want to make use of symmetry considerations to save steps.
A Charge and a Neutral Object
A neutral object can experience an attractive
force towards a charged object. This is the actual
cause of all of the neutral dust particles
−
+
around your TV and other electrical appliances.
++++
−
+
The mechanism for the attractive force arises
−
+
due to polarization of the neutral object.
Consider bringing a charged rod next to a
neutral object as shown in the figure to the right.
The charged rod will induce a polarization effect on the neutral object. The side nearer to
the rod will have the opposite charge and an attraction to the rod. The side away from the
rod will have like charges and experience a repulsive force. Since the attractive side is
closer then the attractive force is stronger; the repulsive side being farther away creates a
weaker repulsive force. The result is a net, attractive force. The polarization will occur
regardless of whether or not the neutral object is a conductor or a non-conductor.
Homework Problems
In each problem you see a charge (q′) and its coordinates with two or more charges (q)
and coordinates. For each set determine the net force on q′ with magnitude and direction
Problem 3
Problem 1
charge (x, y) in cm
charge (x, y) in cm
(0,0)
q′ 12 µC
(0,0)
q′ 12 µC
q
(-20,0)
8µC
q
(-20,0)
8µC
q −6 µC
(0, 15)
q −6 µC
(15, 0)
Problem 2
charge
q′ 9 µC
q 10µC
q −6 µC
(x, y) in cm
(0,0)
(0,-12)
(0, 12)
Problem 4
charge
q′ 15 µC
q
8µC
q −8 µC
(x, y) in cm
(0,0)
(-15,20)
(15, 20)
Problem 5
charge (x, y) in cm
(0,0)
q′ -10 µC
q
(15,8)
8µC
q −6 µC
(-12, 5)
Problem 6
charge
q′ 18 µC
q 9 µC
q −6 µC
q 9 µC
(x, y) in cm
(0,0)
(-20,0)
(15, 0)
(20,0)
Problem 7
charge
q′ 18 µC
q 9 µC
q −6 µC
q −6 µC
q
Answers to homework:
1) 50.4 N to the right
4) 20.7 N to the right
Lesson 3-03
9 µC
(x, y) in cm
(0,0)
(-20,0)
(15, 0)
(-15,0)
(20,0)
2) 90.0 N up
3) 36.0 N @ 53.1° above +x-axis
5) 51.5N to the right 6) 43.2N up
7) 0N
Electric Field (pp. 572 – 579)
Consider taking a second look at the problem in example #2.
A +8µC charge is placed at the origin. A −5µC charge is placed on the x-axis at
x=+20cm. Find the force that each charge exerts on the other.
+8µC
−5µC
|
|
|
|
|
|
|
x-axis
0
4
8
12 16
20
24
cm
The conclusion according to Coulomb’s Law is that the force each charge exerts on the
other is 9.0 N. How can something at one point in space exert a force at some other point
in space without being in direct contact with the second particle? In other words, how
does the 8 µC charge at x=0 push on the -5 µC charge without touching it? This is our
second encounter with action at a distance. Gravity was the first. In this case we would
say that the 8 µC charge is radiating an electric field through all points in space. The
electric field passing through (20, 0) is what exerts the force on the negative charge.
Likewise, the -5 µC charge is also radiating an electric field through all points in space.
The electric field passing through (0, 0) is what exerts a 9.0 N force on the charge at the
origin. So the electric field at a point exerts a force on any charge placed at that point. In
this lesson we have two objectives:
1) What is the relationship between the electric field at a point and the charge placed at
that same point?
2) What is the relationship of a charge at one point in space and the electric field that it
creates at some other point in space?
Electric Field Defined
The symbol for the electric field is E. The electric field
FE = q′E
is a vector quantity that has units of Newtons/Coulomb.
If a charge, q′ is placed in an electric field it will experience
a force, F that is described in the above box. If charge q′ is positive then F and E are in
the same direction; if q′ is negative then F and E are antiparallel.
The +8 µC charge at the origin at the beginning of our lesson experienced a force
of 9.0 N to the right. Using our last equation we can find the electric field at the origin
from the other charge- 9.0 N to the right = +8E-6 C* E. Solving for E gives a solution of
E = 1.125 E +6 N/C to the right. We conclude that the +8 µC charge at the origin
experiences a force of 9.0 N to the right because it is in the presence of a 1.125E+6N/C
electric field at that point. We can also conclude that the -5 µC charge at x = 20 cm is
radiating an electric field that has a value of 1.125E+6N/C at the origin.
The -5 µC charge at x=20cm experiences a force of 9.0 N to the left. We can
calculate the electric field present at x = 20 cm that is necessary to provide this much
force. E = F / q′ = (9.0 N to the left) / (−5.0 E -6 C) = 1.8E+6 N/C to the right. We
conclude that the -5 µC charge experiences a force of 9.0 N to the left because it is placed
in an electric field of 1.8 E+6 N/C to the right. Notice that negative charges get pulled
against the field. Also notice that we get the direction of “to the right” in the above
calculation because we had a “to the left” divided by a negative sign. And of course we
all know that “negative to the left” is merely a complicated way of saying “to the right”.
Homework (Part 1 of 2)
1) An electron and a proton are both placed at rest in an electric field of E = 2000 N/C to
the right. Find the force acting on either particle. Find the acceleration of either
particle.
2) An alpha particle is placed in an electric field of E = 30,000 N/C to the right. What is
the electric force and resulting acceleration of the alpha particle? See the first
problem of homework for lesson 3-01 for mass and charge of the alpha particle.
3) Return to the homework problems of lesson 3-02. For the first four problems
determine the electric field at the origin that is acting on q′ at the origin.
Electric Field of a Point Charge
All point charges will radiate an electric field. You are responsible for getting the electric
field at a specific point due to point charges located elsewhere. For positive charges the
electric field will radiate outward as shown in figure A below. For negative charges the
electric field points inward towards the charge as shown in figure B below.
Fig A - E field for +q
Fig B – E field for −q
E fields point away from positive charges and towards negative charges.
Check to see if this is consistent with the calculations in the first two paragraphs of this
page. You should see that the answers from the paragraphs do indeed agree with the
diagrams above. The concept of the electric field was proposed by Michael Faraday. But
we still have a problem. How do you calculate the strength of an electric field near a
point charge?
The value of the electric field can be determined by
E = kq/ r2
using the equation in the box to the right. This equation is
E points outward if q
constructed by combining Coulomb’s Law with the first
is positive and inward
equation of this lesson. I highly recommend that you use
if q is negative
the boxed equation to find the magnitude of the E and
the checked statement from the previous page to get the
direction of the E.
Again, let us return to the example problem that started this lesson. We choose to
find the electric field strength 20 cm to the right of the 8 µC charge. Using the above
equation you get |E| = 9E9 (Nm2/C2)* 8E-6 (C)/ (0.20m)2 = 1.8E+6 N/C away from the
+8µC charge. So the electric field at x = 20cm is to the right because the charge that is
the source of this E field is positive. This equation agrees with our previous calculation.
Problems of this type are cleverly disguised vector addition problems. You use
the above boxed equation to find the magnitude of each vector. You use the checked
statement at the bottom of the last page to get the direction of each vector. Then you add
the vectors.
Example Problem
charge
(x, y) in cm
Consider the charges and their locations in
q -10 µC
(16,0)
the table to the right. Determine the electric
field at the origin due to the presence of these
q
(15,8)
8µC
charges only.
q −6 µC
(-12, 5)
Step 1
Find the E at the origin from each
charge.
E1 = 3.52 E+6 N/C to the right.
E2 = 2.49 E+6 N/C @ 28.1°
below –x axis
E3 = 3.20 E+6 N/C @ 22.6°
above –x axis.
Step 2
Draw and label the electric field vectors
as rays pointing from the origin.
3.20E6
3.52E6
22.6°
28.1°
2.49E6
Step 3
Add the vectors just as you did in unit 1.
The space provided below is left for
your calculations.
Homework (part 2 of 2)
4) Return to the first four problems of lesson 3-02. Assume that q′ is no longer present
in the table. Use the method outlined at the bottom of the previous page to calculate
the electric field vector at the origin. Check to see if you have the same answers as
found in problem (3) of this same homework assignment.
Lesson 3-04
The Volt (pp. 593 – 601)
The consideration of the electric force is nice but it leads to variable accelerations in most
instances. These accelerations cannot be analyzed without the use of the calculus. As we
have seen in the previous semester, energy considerations can be used in order to avoid
the calculus. In today’s lesson we define the concept, volt that will enable us to analyze
situations in terms of energy rather than force.
A Volt is defined to be Joule of energy per coulomb of charge; 1 Volt = 1 J/C. It
is very critical that you know how the units of Volt break down into more fundamental
values. A 9 Volt battery connected to a circuit will deliver 9 Joules of energy to a circuit
for every Coulomb of charge moved through the circuit.
As the electric field moves a charge from one point in space to another, the field
will do work on the charge. The work that the field does on the charge depends on how
much the voltage changes from the beginning point
to the end point. The equation used to find the work
W = -q(Vf – Vo) = -q∆V
done by the field on the charge is shown in the boxed
equation to the right. The negative sign in front of the
charge is placed there when finding work done by the field on the charge. If a person
moves a charge against the field then work done by the hand to move the charge is +q∆V.
Example Problem #1
A 9 mC charge is moved from rest a point in space where the voltage is 14V to
rest at a place where the voltage is 84V. How much work was done by the hand while
moving the charge? What was the net work done on the charge? How much work was
done by on the charge by the field?
Using W = +q∆V = 9mC(84V-14V) = 9mC(70 J/C) = 630 mJoules.
The net work done is zero since the kinetic energy does not change between
beginning and end points.
This also means that the work done on the charge by the hand is equal and opposite to
the work done on the charge by the field. In other words, the charge is being pushed
against the electric field so that the field is turning kinetic energy into potential
energy as fast as the hand is attempting to give the charge kinetic energy. The result
is that the work done by the field on the charge is -630 mJ. You can confirm this by
using the above boxed equation.
Notice in the above example that the path of the charge is not specified; only the end
points are mentioned. This is the power and beauty
WNET = ½ m (vf2 – vo2)
of working with voltage rather than field concepts.
Do not forget that net work can also be defined in
terms of changes in kinetic energy. A review of that definition is placed in the above box
as a reminder. Combining the two boxed equations will take us from electricity problems
to mechanics problems. The following practice problems will demonstrate this point.
Homework
1. A proton is moved at constant speed from a point where the voltage is 100 Volts to a
point where the voltage is 400 volts. How much work was done by the mover? How
much work was done by the field on the charge?
2. A proton is released from rest at a point where the voltage is 900 Volts. The proton
moves freely until it is at a point where the voltage is zero. How much work did the
field do on the proton? What is the final speed of the proton?
3. An electron is released from rest at a point where the voltage is zero. The electron
moves freely to a place where the potential is +1200 volts. How much work is done
on the electron by the field? What is the final speed of the electron?
4. Show that if a particle of charge q and mass m is released from rest and moves freely
through a voltage difference of ∆V then it will have a final speed that can be
calculated by using v = √ (-2q∆V/m) . Also apply the equation to the previous two
problems in order to verify your answers.
5. An alpha particle is moved from rest at a place where V= -300 Volts to resting at a
point where V = +900 Volts. How much work is done on the alpha particle by the
mover? If the alpha is released from rest and moves freely from +900 Volts to a point
where V=0 Volts what will be its speed?
6. Circle the correct choices in the following statement- Positive charges move freely in
the direction of increasing/decreasing voltage while negative charges move freely in
the direction of increasing/decreasing voltage.
1) 4.8E-17J; -4.8E-17J 2) +1.44E-16J; 4.1E5 m/s 3) +1.92E-16J; 2.05E+7 m/s = 0.068c
5)3.84E-16J; 2.94E5 m/s 6) decreasing; increasing
Lesson 3-05
Particle Accelerators
A particle accelerator is a machine that can take very small (sub-atomic) particles from
rest up to near light speeds. In general, the machine uses electric fields to give the
particles their near-light speeds. In a later unit (303) we will see how combining
magnetic and electric fields enables us to find things like mass and charge for these same
particles. This lesson is a first approximation of the particle accelerator.
We start with the rough draft of a particle accelerator, charged parallel plates.
Consider the figure below showing a high voltage source and two parallel plates. The
parallel plates must be within an almost pure vacuum environment.
Near Vacuum
+
High Voltage Source
Protons would accelerate from the left-hand, positive plate to the right-hand, negative
plate. Electrons would accelerate in the opposite direction. There are three different
reasons why the protons and electrons move in the direction that they do. 1) The positive
plate on the left repels positive charges and attracts negative charges. 2) The negative
plate on the right repels negative charges and attracts positive charges. Both of these
explanations are according to our rule- opposite charge attract and like charges repel. A
third explanation has to do with the existing
electric field between the plates. The electric
field will always point from high voltage
(positive plate) to the low voltage (negative plate).
So the third explanation is that the electric field
between the plates moves the charges. From a
previous lesson we see that positive charges are pushed in the direction of the field while
negative charges tend to move freely against the field. The direction of the electric field
lines are shown in red in the above figure. So now we have a new rule that relates the
direction of the electric field to the potential difference.
Electric field lines point in the direction of decreasing voltage.
Ok, you know about the direction of the electric field but what about the strength of
the electric field? The answer really depends on the shape of the plates as demonstrated
in lab. For parallel plates the electric field between
the plates is uniform in value and has parallel field
lines. Over short distances where the lines have little
E = - ∆V / ∆x
chance to diverge or converge the equation may also
be applied. The negative sign in the boxed equation is
there to remind the reader that the direction of the electric field is in the direction of
decreasing voltage. The equation also indicates a second set of dimensions or units for
the electric field vector. Electric field has units of Newtons/Coulomb or Volts/meter.
Example #1
Two parallel plates are separated by a distance of 25 cm. The left plate is at +1000 Volts
while the right plate is grounded (connected to Earth) and is at zero volts. Find the
electric field strength between the plates, the acceleration of an electron between the
plates and maximum speed of the electron between the plates assuming it starts from rest.
E = 1000 V/ 0.25m = 4000 V/m or 4000 N/C
From F = q′E = ma you get a = 1.6E-19C(4000N/C)/ 9.11E-31Kg = 7.025E+14m/s2
Maximum speed occurs when the electron accelerates over the total distance of 25cm;
using the kinematics vf2 = vo2 + 2ad or vf = √ (2*7.025E14*0.25) = 1.87E& m/s.
Alternately, the proof from the previous lesson can be used, v = √ (-2q∆V/m).
Example #2
While getting out of a car to pump gasoline a learned physics student grounds herself to
the car before grabbing the handle of the gasoline hose. She notices that her finger was
approximately 2.0 mm from the car when a spark occurred. Realizing that on a cold, dry
day dielectric breakdown takes place in our atmosphere at about 3 million volts per meter
she approximates her potential difference (voltage) before discharge. What is that value?
|∆V| = E∆x = 3E6 V/m * 2.0 E-3 m = 6000 Volts
The trouble with our parallel plate accelerator is that the particle beam can only go as far
as the opposite plate. In many accelerators rings rather than solid plates are used with the
beam line moving through the holes in the center of the rings. Another problem is that
not all particles move through the total voltage difference between plates. This causes a
spread of particle velocities that would rapidly lead to a divergent beam. The solution to
the latter problem is a velocity selector that will clean up the particle speeds in the beam.
We shall come back to the particle accelerator again in unit 303 where we will be able to
fully explain the velocity selector.
Homework
1. Two parallel plates are raised to a potential difference of 800 Volts. The plates are
separated by a distance of 40 cm. What is the electric field strength between the
plates? What is the maximum speed that a proton can attain from rest between the
plates? How long would it take a proton to go from rest on the positive plate to
maximum speed at the negative plate?
a) 2000 V/m
b) 391.5 km/s
c) 2.04 µsec
2. Two parallel plates are separated by a distance of 33.3 cm. The electric field strength
between the plates is 4800 V/m. What is the potential difference between the plates?
What acceleration would an electron experience between the plates? What maximum
speed could an electron attain starting at rest between the plates?
c) 2.37E7 m/s or 0.079 c
a) 1600 V
b) 8.4E14 m/s2
3. A deuteron (charge of proton but double the mass) is accelerated from rest to a speed
of 1.8E6 m/s in between two parallel plates. The deuteron starts from rest on the
positive plate. How much kinetic energy does the particle have at the negative plate?
Through what potential difference (voltage difference) was the deuteron moved?
What is the plate separation distance if field strength between plates is 50,000 V/m?
a) 5.41E-15J
b) 33.8 kilovolts
c) 67.6 cm
4. A dental x-ray machine typically accelerates electrons from rest through a potential
difference (voltage difference) of 110,000 volts or 110 kV. The fast moving electrons
then slam into a heavy lead or gold target where x-rays are generated. Ignoring
relativity effects what is the upper limit of speed for the electrons? If the electrons
are accelerated over a distance of 50 cm what is the electric field strength inside the
x-ray machine? What acceleration do electrons experience?
a) 2.0E8 m/s or 0.66c b) 220,000 N/C
c) 3.86E+16 m/s2
5. A nuclear physicist wishes to move an alpha particle (twice proton charge but four
times the mass) from rest to 0.2c or 20% the speed of light, 6 E7m/s. He needs to do
this over a distance of 120 centimeters between parallel plates. What is the potential
difference between the plates? What electric field strength will exist between the
plates? What force will the field exert on the alpha particle?
a) 37.58 Megavolts
b) 31,310,000 N/C
c) 10 picoNewtons
Lesson 3-06
Point Charge Potentials
In the two previous lessons the voltage at a point in space was given without telling the
reader how that value was determined or measured. The voltage at a point is space is also
known as the electric potential. Voltage difference is also known as the electric potential
difference. In this lesson you will learn how to calculate the electric potential or voltage
at a point in space due to the presence of one or more nearby point charges. You will
also end this lesson with an example problem that is a nice segue into the next unit.
The electric potential due to a point charge is determined using the formula in the
box to the right. The equation is scalar in nature
but can lead to both positive negative voltages due
V = kq/r
to the sign on the source charge. Do not ignore or
Where k = 9.0 E +9 Nm2/C2
omit the sign on the source charge.
Suppose that you wanted to know the electric potential at a point 50 cm from a
+40 nC charge? V = 9E9* 40E-9 / 0.5 Nm/C = 720 volts. Note that a direction is not
indicated. In fact, for an isolated charge you could draw a circle or a sphere around the
charge with a radius of 50 cm and the electric potential at every point on the sphere
would be 720 Volts. For multiple charges one merely adds potentials due to each charge.
Example #1
Find the electric potential on the x-axis at x = 10.0 cm.
+8µC
−5µC
+7µC
|
|
|
|
|
|
|
|
|
x-axis
0
4
8
12 16
20
24
28
32
cm
V8 = 9E9*8E-6/0.10m = +720 kVolts
V5 = 9E9*-5E-6/0.10m = −450 kVolts
V7 = 9E9*7E-6/0.22m = +286 kVolts
VTOT = 720 kV – 450 kV + 286 kV = 556 kV
Problems are worked similarly for coplanar charges except that you have to use
Pythagoras’s theorem to find the values of r. Look to use symmetry in order to reduce
the number of terms in a problem. In the previous example steps 1 and 2 could have been
combined by recognizing that the -5µC charge would cancel 5/8 of the 8 µC charge. You
could replace the 8E-6 in step 1 with 3E-6 and skip step two completely because both
charges are equidistant from the point in question.
You should also be warned about reading a problem. When you read phrases
such as “very far away” you should be aware that the write is describing a place where r
has a large value and V is approaching 0 volts. Consider the following problem.
Example #2
A +6 nC charge is placed at (20cm, 0) on the x-y plane. A +8 nC charge is placed at (33cm, 0) on the x-y plane. A) What is the electric potential at the origin due to the
presence of these two charges? B) What will be the final speed of a proton that is
released from rest at the origin after it has moved very far away from the two initial
charges?
V = 488 volts
∆V = 0V – 488 V = -488V. Using v = √ (-2q∆V/m) = 305 km/s
Homework (part 1 of 2)
For each of the following problems calculate the electric potential (voltage) at the origin.
All charges will be in nanocoulombs and coordinates in centimeters.
1. q1 = +9 nC at (+20cm, 0), q2 = +12 nC @ (0, 25cm) and q3 = -9nC @ (-20cm, 0)
2. q1 = +8 nC at (+20cm, 0), q2 = -12 nC @ (0, 20cm) and q3 = +8nC @ (-20cm, 0)
3. q1 = +10 nC at (+20cm, 0), q2 = +12 nC @ (0, 25cm), q3 = -10nC @ (-20cm, 0) and
q4 = -9 nC @ (0, -25cm)
4. q1 = -9 nC at (+20cm, 0), q2 = +12 nC @ (0, 25cm) and q3 = -9nC @ (-20cm, 0)
5. q1 = +9 nC at (+20cm, +15cm), q2 = +12 nC @ (-8, -15cm)
and q3 = -9nC @ (-20cm, +15cm)
Answers
1. 432 V
2. 180 V
3. 108 V
4. -378 V
5. 635 V
Charged Conducting Spheres
It happens that any charge conducting sphere can be treated as a point charge as long as
you are outside of the sphere or at the surface of the sphere for both electric fields and for
electric potential. Suppose for example, that an aluminum sphere has a radius of 20 cm
and has a charge of 60 nC deposited on its surface. Under static conditions the charge
will be uniformly distributed over the surface. In order to develop this much charge on
the sphere you would have to raise the sphere to a potential of +2700 Volts according to
V = kq/r = 9E+9 Nm2/C2*60E-9C/0.20m = 2700 Nm/C or 2700 J/C.
This leads us to a very interesting and also very important final concept in static
electricity. Recall from the first lesson the idea of charge distribution for different
charges on identical conducting sphere. In that case the conservation of charge was all
that was needed to solve the problem. See below.
Example #3
Shown below are two identical, conducting spheres that are mounted on glass stands.
Each sphere is charged as shown below. If the two spheres are made to touch by a person
holding the glass stands or made to connect with an insulated wire what will be the final
charge on each stand?
+24 nC
+10 nC
How would the answer change if the spheres were not identical? Suppose that the lefthand sphere has a radius of 30cm and the right-hand sphere has a radius of 50 cm. If the
spheres are connected by a wire what will happen to the charge distributions? Why? The
final outcome is discussed on the next page. Be prepared to calculate the final charge
distribution and explain why in paragraph form on test day.
Charges will redistribute until each sphere is at the same potential or voltage.
As long as the spheres are at different voltages there will be an electric field
established inside of the wire. That electric field will move electrons from one sphere
to another causing the difference in potential to reduce.
At the instant the charges have moved to create equal potentials on the surface of the
sphere there is no longer an electric field inside of the wire so that charges no longer
have the force needed to move them.
Solution to example #3
Blue conducting wire is shown connecting spheres.
+24 nC
+10 nC
The left-hand sphere is raised to a potential of 720 Volts. The right-hand sphere has a
potential at the surface of only 180 volts. According to the equation E = - ∆V / ∆x there
will be an electric field established inside of the wire that has a strength of about
540Volts divided by the length of the wire. Although the real value of the E field is more
complicated than that due to other factors you should still see that the potential difference
between the wires creates an E field. If we could peer inside of the wire with a magic
magnifying glass that shows E fields and electrons we would see the following:
E field lines would point from left to right inside of the wire since E fields point from
high voltage to low voltage.
Electrons would be moving “upstream” all along the wire (right to left).
As electrons leave the right-hand sphere its positive charge would increase and so
would its voltage thus reducing the potential difference on the wire and weakening
the E field inside of the wire.
At the same time, electrons entering the left-hand sphere would be reducing the net
charge on that sphere and the electric potential. This would also help to reduce the
electric field strength inside of the wire.
Eventually charges are redistributed so that both spheres are at the same potential
causing the E field to instantly disappear. Without the E field the charges can no
longer move within the wire.
In order to find the final charge distributions we recognize two facts. First, charge must
be conserved- qL + qR = 34 nC. Second, we recognize that the voltages on the spheres
must be the same so that the following is also true- kqL/30cm = kqR/ 50cm. You now
have two equations and two unknowns. From the second equation you can cancel the k’s
and the cm to get qL = (3/5)qR. Subbing for qL into first equation leads to (8/5)qR = 34nC
or qR = 21.25 nC. This leaves 12.75nC on the left hand sphere. Both spheres are at a
final potential of 382.5 Volts.
As long as opposite ends of a wire are at different potentials then charges will move
through the wire. We call this electric current and that is what unit 302 is about!
Expect a practice test tomorrow and the real test to soon follow. Listed below are the
objectives of each lesson.
Lesson 3-01 objectives: The learner will …
1. Know Latin prefixes, symbols and values for mega, kilo, milli, micro, nano, pico.
2. Identify possible charges based on whole number values of e=1.6E-19C
3. Be able to describe and identify charging by friction, contact and induction.
4. Explain electrical polarization
5. Use conservation of charge to find the final charge distribution for two, identical
conducting spheres that are brought into contact with initially different charges.
Lesson 3-02 objectives: The learner will …
6. Write Coulomb’s Law for the electric force between two point charges
7. Describe how the magnitude of the force between two point charges will change
when the distance between charges or value of either charge is (are) altered.
8. Use Coulomb’s Law to find the net force on a single charge due to the presence of
one or more other collinear or coplanar charges.
9. compare and contrast the electric and the gravitational forces
Lesson 3-03 objectives: The learner will…
10. Write the equation that describes the relation among force, charge and electric field.
11. Combine the equation in #10 along with Newton’s Second law to find the
acceleration of a charge at a point in and electric field of known strength.
12. Sketch the electric field lines around several charges using the idea that positive
charges are sources of E field lines while negative charges are sinks.
13. Find the electric field strength at a specific point in space when given one or more
static charges at other fixed points in space for both collinear and coplanar conditions.
Lesson 3-04 objectives: TLW…
14. Define the volt in terms of work and charge and list the possible different units
15. Use the definition of potential difference to calculate work done on a charge by an
external agent or by electric field when moving a charge through potential difference.
16. Combine equation from #15 along with the work-energy theorem in order to find the
final speed of a charge that has moved freely through a known potential difference.
Lesson 3-05 objectives: TLW
17. Describe the relationship between the electric field and the potential difference for the
special case of charged, conducting parallel plates.
18. Analyze the motion of a particle in a uniform E field between parallel plates using the
kinematics equation along with acceleration from #11 and/or final speed from #16.
19. Describe the unique relationship between the direction of the E field and equipotential
lines. This was also demonstrated in lab #1.
Lesson 3-06 objectives: TLW
20. Write the equation for electric potential at a point in space due to a single charge
21. Determine the electric potential at a point in space due to the presence of other
collinear or coplanar, stationary charges.
22. Use the equation from #20 to find the potential on the surface of a conducting sphere
of known radius and excess charge.
23. Describe the conditions that occur when two, different spheres from #22 are brought
into contact by a conducting wire.
24. Calculate the final charge and voltage of the spheres mentioned in #23.