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Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 1
Worksheet 1: Electrostatics
1) Explain why it is electrons and not protons which are
thought to be exchanged in electrostatic interactions.
Electrons are located on the exterior of the atom, while the positive
protons are located in the nucleus. In frictional charging
only the exterior parts of the atoms will be in contact. The
protons are also protected by layers of electrons.
2) How many electrons make up a charge of 100 µC?
(6.25 x 1014 e-)
4 14
19
1 e1.00 x 10 C x = 6.25 x 10 e
1.60 x 10 C
−− −
−
3) The number of electrons (elementary charges) in a coulomb
is 6.25 x 1018. If you could count 4.0 electrons per second,
and if you worked 8.0 h each of the 365 days of a year, how
many years would it take to count this number of electrons ?
(1.5 x 1011 a)
18
3
11
1 s 1 h6.25 x 10 e x x
4 e 3.6 x 10 s
1 d 1 ax x = 1.5 x 10 a
8 h 365 d
−
−
4) What is the charge on a metal sphere if it has:
a) an excess of 1.0 x 1015 electrons?
b) a deficiency of 1.0 x 1015 electrons?
( -1.6 x 10-4 C)
1915 41.60 x 10 C
1.0 x 10 e x = -1.6 x 10 C1 e
−− −
−
1915 41.60 x 10 C
1.0 x 10 e x = +1.6 x 10 C1 e
−− −
−
5) A strip of acetate and a strip of silk are rubbed together.
What can be said about the charges before and after the |
interaction?
The same number of charges are present after the interaction
as before. Charge is conserved. The negative electrons are
merely separated from one atom and attracted to another.
6) What property makes a metal a good conductor and rubber
a good insulator?
Metals have free mobile electrons whereas insulators have
localized electrons.
7) A metal sphere and a plastic sphere of the same size are
supported on insulated stands. Each is now touched to a
charged body. Compare the distribution of charge on each
sphere.
In the metal, the charge will reside evenly on the exterior
surface as electrons repel one another. In the insulator, the
charge will be localized where the object was touched, but
will also reside on the exterior surface.
8) Why do certain types of clothing tend to stick together
when they are removed from a clothes dryer?
In the dryer they are charge by frictional contact. Since they
are insulators, they do not ground out to the metal surface of
the dryer.
9) What is the significance of Benjamin Franklin (1706 - 1790)
in the study of electrostatics? Why was Franklin's designation
of charges as positive and negative better than the terms vitreous
and resinous? What problem did Franklin inadvertently create?
Franklin coined the terms positive and negative and invented a
one fluid theory that more clearly explained what happens in
static charging. He performed numerous electrostatic
experiments including the famous kite-flying experiment that
showed that lightning is an electrostatic phenomena. In designating
excess charge as positive, Franklin started the convention that
positive charges move in an electric field, which gives the
impression that it is protons, not electrons, which are electrically|
transferred.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 2
10) Explain why trucks carrying flammable fluids drag a
chain along the ground?
The chain acts as a ground to dissipate any static electric
charge which might build up on the truck as a result of
frictional charging (metal to air or rubber to ground)
11) Why is there a limit to the quantity of charge which can
be acquired by a body?
As an object becomes more negative (or positive) it will start
to repel like charges and attract opposite charges. This will
change the objects ability to attract further similar charge. A
very large charge will also spontaneously discharge through
the air to the ground.
12) If you charge a pocket comb by rubbing with a silk scarf,
how can you determine if the comb is positively or negatively
charged?
You can test the charge on any object by bringing it close to
an electroscope of known charge. If the leaves start to repel
further, the object has the same charge as the electroscope, and
if the leaves start to drops, it has the opposite charge as the
electroscope.
13) If a strongly negatively charged body is brought near the
terminal of a charged electroscope, the leaves come together.
When the rod is brought still closer, but not touching, the
leaves again diverge. Explain why this might happen.
The induced charge separation will push negative charges
down into the leaves of a positively charged electroscope. If
the object has a sufficiently large charge, it could push so
many electrons down that the leaves actually become
negatively charged and then start to repel one another again.
14) Why does a plastic ruler that has been rubbed with a cloth
have the ability to pick up small pieces of paper? Why is it
difficult to do on a humid day?
The plastic ruler (negatively charged) will cause an induced
charge separation in the molecules of the insulator paper. The
upper surface of the paper will thus be positively charged and
since unlike charges attract, it will be pulled toward the plastic
ruler. (the unlike charges being at a shorter distance will have
a stronger force of attraction than the like negative charges which
repel at a greater distance.) One a humid day, water molecules
in the air will be attracted to charged bodies and cause an
increased gravitational force which works against the electrical
attraction.
15) Can you suggest an experiment that will prove the law
of conservation of charge?
Faraday’s Ice Bucket Experiment shows that charge is
conserved in frictional interactions
16) In charging an electroscope by induction, why is the
finger withdrawn first, then the rod?
If the rod is removed before the finger, charges will again
flow through the ground to neutralize the electroscope and
as a result, there will be no charge on the electroscope.
17) Two identical metal spheres have charges of q1 and q2.
They are brought together so they touch, and then they are
separated.
a) How is the net charge on the two spheres before they
touch related to the net charge after they touch?
The net charge before and after charging will be the same.
Charges are merely rearranged, they are neither created nor
destroyed.
b) After they touch and are separated, is the charge on each
sphere the same? Why?
Since they are conductors, the charge will be equal on the two
spheres and will be located on the exterior of the spheres.
18) Four identical metal spheres have charges of
qA = -8.0 µC, qB = -2.0 µC, qC = +5.0 µC, and
qD = +12.0 µC.
a) Two of the spheres are brought together so they touch
and then they are separated. Which spheres are they, if the
final charge on each one is +5.0 µC?
charge 1 + charge 2Since = +5 C
2
charge 1 + charge 2 = 10 C which
would indicate charges B and D
µ
µ
b) In a similar manner, which three spheres are brought
together and then separated, if the final charge on each
one is +3.0 µC? c) How many electrons would have to be
added to one of the spheres in part b) to make it electrically
neutral?
charge 1 + charge 2 + charge 3Since = +3 C
3
charge 1 + charge 2 + charge 3 = 9 C which
would indicate charges A, C and D
µ
µ
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 3
19) Describe what happens at each of the stages in charging
a neutral electroscope negatively using a positively charged
glass rod.
If an opposite charge is permanently given to the electroscope,
charge by induction is indicated. In the first stage a negative
charge is brought near the electroscope and an induced charge
separation occurs on the electroscope in which the leaves and
far side of the ball are negatively charged and the near side of the
ball is positively charged. Then the ball is momentarily grounded
on the far side and electrons leave to the ground. When the
ground is removed, electrons cannot return. When the charging
body is subsequently removed, there is a net positive charge on
the electroscope. The leaves and all surfaces of the electroscope
are positively charged.
20) Explain why the force of repulsion is the only sure test
of the sign of an electric charge.
Two bodies could be attracted either by an induced charge
separation or by unlike charges, but can only be repelled because
the charges are the same. If two bodies attract, you can’t tell
whether the second body was neutral or of an unlike charge.
21) A charged body when brought near a suspended pith ball
will first attract it. If the charged body and the pith ball then
touch, the ball will be repelled. Explain.
In the first case, there is an induced charge separation on the pith
ball whereby the far side of the ball has a like charge and the near
side has an unlike charge due to the motion of electrons. As it is a
conductor, electrons are free to move throughout the surface of the
pith ball. The near side is at a shorter distance, therefore the
attractive forces of the unlike charges will be stronger than the
repulsive forces of the like charges. Upon touching (conduction),
electrons will be transferred from the more negative body to the
less negative body and the like charges will subsequently repel.
Worksheet 2: Applications of Charging
1) Completely explain the process by which an electrostatic
precipitator removes fly ash from industrial smoke. Include
an appropriate diagram.
In the electrostatic precipitator, the large ash particles are
first removed by passing through a mechanical filter such as
asbestos. The smaller particle then pass through an
electrically charged grid which charges the small ash
particles with a coronal discharge from the closely placed
metallic mesh. The particles are then removed from the air
when they pass through a second grid with opposite charge.
As unlike charges attract, the ash will stick to the second grid,
and can be removed by washing the grid at some later time.
2) State the 5 steps involved in producing a Xerox copy.
Specify the types of charging involved at each step (where
appropriate)
1) The selenium drum is charged one its surface through a
coronal discharge from the corotron. (small sharp points
which pass charge to the drum.
2) A halogen light reflects from the ‘to be copied’ surface to
the drum and discharges the drum wherever the light contacts
(i.e. not where the print is) the drum.
3) Small ‘toner’ beads (insulators) are attracted to the drum
by an induced charge separation and stick to the drum
4) A second corotron, below the paper attracts the toner onto
the paper, where it sticks electrostatically
5) The toner and paper pass through a heat sealer where the
small plastic beads are melted onto the paper and the paper is
discharged.
3) Why is a Van de Graff generator less dangerous than an
electrical outlet box in your home?
The Van de Graff generator (although it develops a large
charge) discharges for only a fraction of a second. The wall
outlet, although at a much smaller voltage (i.e. amount of
charge) discharges continually and causes muscle tissue to
contract and remain contracted
4) Explain the process by which a lightning strike is generated.
Specifically state the type charging involved and electron
motion at each step. Include appropriate diagrams.
In the summer, there is greater heating of air just above the
ground than higher up in the atmosphere. The hotter air rises
in updrafts which tunnel up through the cooler air. This
results in frictional charging between water molecules in
different states (solid, grommel - semisolid, liquid). The
bottom part of the cloud becomes negatively charged as a
result, the middle part positively charged and the upper part
of large clouds negatively charged. In turn, the negative
bottom of the cloud repels electrons in the ground, causing an
induced positive charge on the ground. The higher pointed
objects on the ground then build up step leaders towards the
clouds (areas of positively charged pockets of air) and the
cloud negatively charged step-leaders which move towards the
ground. The lightning becomes luminous when the two
leaders meet somewhere in between and electrons move to the
ground. There is a further ground discharge as the electrons
move out laterally through the positive ground, neutralizing
the ground charge.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 4
5) Jan 2001 - In a physics demonstration, a student inflates a
balloon by blowing into it. The end of the balloon is then tied.
The balloon is rubbed with fur and develops an electrostatic
charge. The balloon is placed against the ceiling and released.
It remains "stuck" to the ceiling.
The teacher then presents the following challenges to the
students:
-explain how the balloon received the electrostatic charge
-explain why the balloon is attracted to the ceiling
-provide a procedure that would determine if the charge on the
balloon is positive or negative. Include a list of any additional
equipment needed.
-provide a procedure that could be used to determine if there is
a relationship between the amount of rubbing and the amount
of charge developed on an inflated balloon. Include a list of
any additional equipment needed.
Using concepts from Physics 30, provide a response to each of
the teacher's challenges.
Marks will be awarded for the physics used to solve this
problem and for the effective communication of your
response.
The balloon becomes negatively charged through frictional
charging when the balloon attracts electrons away from the
fur. The charge on the balloon could be confirmed with an
electroscope of known negative charge. If the balloon is
negatively charged, the leaves of the electroscope will
separate further as the balloon is brought close to, but does
not touch the ball of the electroscope.
The negative balloon then induces a charge separation in the
wall by repelling electrons within the molecules of the wall.
The near surface of the wall gains a positive charge and the
far surface of the wall a negative charge. Since the near
surface is closer to the balloon this electrical attraction is
greater than the repulsion of the two negative surfaces.
To determine if there is a relationship between the amount of
rubbing and the amount of charge developed on the balloon:
Manipulated variable: number of rubs
Responding variable: separation of electroscope leaves
Control variable: distance of the balloon from the
electroscope.
How to change the manipulated variable: Do a series of trials
in which the number of rubs of the balloon is steadily
increased over the trials.
How to measure the responding variable: Measure the angle
of separation between the leaves of the electroscope. A
greater angle indicates a greater electrical force. (diagram)
Worksheet 3: Coulombic (Electrical) Forces
1) Two students are sitting 1.50 m apart. One student has a
mass of 70.0 kg and the other has a mass of 52.0 kg. What
is the gravitational force between them? (1.08 x 10-7 N)
1 2g 2
27
g 2
7
g
Gm mF =
R
(6.67 x 10 )(70)(52)F =
(1.50)
F = 1.08 x 10 N
−
−
2) What gravitational force does the moon produce on the
earth if the centers of the Earth and moon are 3.88 x 108 m
apart and the moon has a mass of 7.34 x 1022 kg?
(1.94 x 1020 N)
1 2g 2
27 22 24
g 8 2
20
g
Gm mF =
R
(6.67 x 10 )(7.34 x 10 )(5.98 x 10 )F =
(3.88 x 10 )
F = 1.94 x 10 N
−
3) Calculate the electric force between two point charges
of 4.00 µC and 3.00 µC when they are 2.00 cm apart.
(2.70 x 102 N)
1 2
2
9 6 6
2
2
kq qF =
R
(8.99 x 10 )(4.0 x 10 )(3.0 x 10 )F =
(0.02)
F = 2.70 x 10 N
e
e
e
− −
4) Two points of equal charge produce an electric force on
each other of 3.40 x 10-2 N when placed 1.00 x 10-1 m apart.
What is the charge on each point? (1.94 x 10-7C)
2
e1
2 1 2
1 9
7
1
F Rq =
k
3.4 x 10 (1 x 10 )q =
8.99 x 10
q = 1.94 x 10 N
− −
−
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 5
5) Two point charged objects produce an electric force on
each other of 4.5 x 10-3 N. What is the electric force if the
charge on both objects triple and the distance between them
doubles? (1.0 x 10-2 N)
1 2
2
new
1 2old2
new old
3
new
2
new
k3q 3q
F (2R) =
kq qF
R
9F = F
4
9F = (4.5 x 10 )
4
F = 1.0 x 10 N
−
−
6) Three point charged objects are placed in a line as shown in
the diagram.
Calculate the magnitude of the net electric force on the
center charge due to the other two charges. (1.1 x 10-1 N)
The net electric force will be the vector sum of the two forces
acting on the charge 3. Since the other two charges are both
positive, these forces will be pushing in opposite directions as
like charges repel
1 2e(13)
2
9 6 2
e(13)2
e(13)
e(23)
e(net) e(13) e(23)
e(net)
1
new
kq qF =
R
8.99 x 10 (2.0 x 10 )F =
(0.40)
F = 0.22475 N East
F = 0.337125 N West
F = F + -F
F = 0.22475 + -0.337125
F = 1.1 x 10 N West
−
−
�
�
�
�
� � �
�
7) Two small spheres have the same mass and volume. One of
the spheres has a charge of 4.00 µC and the other sphere has a
charge of -1.00 µC. If these two spheres are brought into brief
contact with each other and then separated to a distance of
2.00 x 10-1 m, what is the electric force between them at this
distance? (5.0 x 10-1 N)
When the two spheres touch, the total charge is shared
between the spheres
6 6
1 2
6
1 2e
2
9 6 2
e1 2
1e
+ Q 4 x 10 + -1 x 10 =
2 2
= 1.5 x 10 C each
kq qF =
R
8.99 x 10 (1.5 x 10 )F =
(2 x 10 )
F = 5.1 x 10 N repulsive
Q− −
−
−
−
−
�
�
�
8) Three point charged objects are placed at the corner of
a right-angle triangle as shown in the diagram.
Calculate the magnitude of the net electric force on the charge
marked with the x due to the other two charges. (5.0 x 10-1 N)
When charges 1 and 2 are at right angles to charge3, the net
force can be found using a Pythagorean sum of the vector
forces acting on 3. Since all the charges are positive and like
charges repel, F23 will be northward and F13 will be to the
East.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 6
1 2e(13)
2
9 -6 -6
e(13)2
e(13)
e(23)
2 2
e(13) e(23)net
2 2
net
net
e(1
kq qF =
R
8.99 x 10 (3.0 x 10 )(4.0 x 10 )F =
(0.60)
F = 0.29966... N
F = 0.39955... N
F = F + F
F = (0.29966...) + (0.39955...)
F = 0.50 N
F = tanϑ −
�
�
�
�
� �
�
23)
e(13)
1
o
1 oe
F
0.39955... = tan
0.29966...
= 53.1 N of E
F = 5.0 x 10 N @ 53.1 N of E
ϑ
ϑ
−
−
�
�
9) Two small spheres, each with a mass of 2.00 x 10-5 kg,
are placed 3.50 x 10-1 m apart. One sphere has a charge of
-2.00 µC and is fixed in position. The other sphere has a
charge of -3.00 µC but is free to move without friction.
What is the initial acceleration due to the electric force on
the sphere that is free to move? (2.20 x 104 m/s2)
1 2e
2
9 -6 -6
e2
e
kq qF =
R
8.99 x 10 (-2.0 x 10 )(-3.0 x 10 )F =
(0.35)
F = 0.4403... N repulsive
�
�
�
e net
e
5
4
2
Since the electric force is the only force on
the object, it is the net force.
F = F
Fa =
m
0.4403...a =
2.0 x 10
m a = 2.20 x 10 away from the other charge
s
−
� �
�
10) The drawing shows three point charges fixed in place.
The charge at the coordinate origin has a value of q1 = +8.00 µC;
the other two have identical magnitudes. but opposite signs:
q2 = -5.00 µC and q3 = +5.00 µC.
a) Determine the net force (magnitude and direction) exerted
on q1 by the other two charges.
b) If q1 had a mass of 1.50 g and it was free to move, what
would be its acceleration?
e(21) e(31)
The net force will be the vector sum of the two forces
F and F . Since these vectors are at non-right
angles, the x and y components need to be added
separately to determine the net force.
F
� �
�
1 2e(21)
2
9 -6 -6
e(21)2
oe(21)
oe(31)
kq q =
R
8.99 x 10 (8.0 x 10 )(5.0 x 10 )F =
(1.3)
F = 0.21278... N at 23 N of E
direction determined as unlike charges attract
F = 0.21278... N at 23 N of W
direction d
�
�
�
etermined as like charges repel
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 7
e(21)
e(31)
e(net)
e(net)
F ( 0.21278 cos(23), +21278 sin(23))
+ F ( 0.21278 cos(23), +21278 sin(23))
F ( 0 , 0.166280 .... )
F 0.166 N North
+
−
=
�
�
�
�
11) Suppose that we place three small, equally charged metal
spheres A, B and C so that A is 1.2 cm west of B and C is
1.0 cm south of B. It is known that C exerts a force of 4.0 µN
on B.
a) What force does A exert on B? (FAB = 2.8
µN)
b) What is the magnitude and direction of the net force on B?
(4.9 µN at 35o E of N)
In this question, all the forces are repulsive since like charges
repel.
e(AB) e(CB)
1 2
2e(AB)
e(CB) 1
The net force will be the vector sum of the two forces
F and F . Since these vectors are at right
angles, a Pythagorean sum can be used to find the net force.
kq q
F (0.12) =
kqF
� �
�
�
2
22
2
26 6
e(AB) e(CB) e(CB)2
(0.1) =
q (0.12)
(0.1)
(0.1)F = F = 0.69 x 10 N F 2.8 x 10 N
(0.12)
− −=� � �
2 2
e(AB) e(CB)net
6 2 6 2
net
6
net
e(CB)1
e(AB)
61
6
o
6 o
net
F = F + F
F = (2.8 x 10 ) + (4.0 x 10 N)
F = 4.9 x 10 N
F = tan
F
4.0 x 10 = tan
2.8 x 10
= 55 N of E
F = 4.9 x 10 N @ 55 N of E
ϑ
ϑ
ϑ
− −
−
−
−−
−
−
� �
�
�
net
net
2
Since F is the only force, it is also the net force
F = F
Fa =
m
0.166a =
0.0015
ma = 111
s
e
e
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 8
12) Two equally charged identical conducting spheres A and B
repel each other with a force of 0.20 µN. Another identical
uncharged sphere, C, is touched to A and then move next to
but not touching B
a) What is the electric force on A now? (0.15 µN (left or west)
b) What is the net electric force with direction on C if it is now
moved halfway between A and B? (F net = 0.20 µN towards A)
1 2
2new
old 1
When the two conductors are touched, A and C
will share the charge equally between the two
spheres
+ + 0 = =
2 2 2
The charge where B was located will now be
3q + q =
2 2
1 3k q q
2 2F
= kq qF
A CQ Q q q
q
R
�
�
2
2
new old
3 =
4
3 3F = F = (0.20 N) = 0.15 N
4 4
The direction will be West since like charges repel
R
µ µ� �
AC BC
1 2 1 2
2 2
AC BC
old old1 2 1 2
2 2
AC ol
If C is located half way between B and A, the net
force will be the linear vector sum of F and F
1 3 1k q q k q q
2 2 2
1 1
F F2 2 = = 1 and = = 2
kq q kq qF F
F = 1F
R R
R R
� �
� �
� �
� �
d
BC old
net AC BC
net
net
0.20 N East as like charges repel
and F = 2F 0.40 N West as like charges repel
F = F + F
F = +0.20 N + -0.40 N
F = 0.20 N West
µ
µ
µ µ
µ
=
=� �
� � �
�
�
13) Three equally charged metal spheres are located as shown
The electric force exerted by A on B is 2.98 µN
a) What electric force does C exert on B? (11.9 µN)
b) What is the magnitude and direction of the net force on B?
(12.3 µN at 14° E of N)
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 9
2
1
2
CB
2AB 1
2
6CB AB
CB AB
kq
1
F 2 = = 4
kqF
F = 4F 4(2.98 x 10 )= 11.9 N North
as like charges repel
The net force can be found from a Pythagorean
sum of the forces F and F since the charges a
R
R
µ−
=
�
�
� �
� �
2 2
net AB CB
6 2 6 2net
5net
e(CB)1
e(AB)
61
6
o
net
re
at right angles to B
F = F + F
F = (2.98 x 10 ) + (11.9 x 10 )
F = 1.23 x 10 N
F = tan
F
11.9 x 10 = tan
9.98 x 10
= 76 N of E
F = 1.23 x 1
ϑ
ϑ
ϑ
− −
−
−
−−
−
� � �
�
�
�
�
�
5 o0 N @ 76 N of E−
14) A student performed an experiment that verified Coulomb's
Law of Electrostatics by measuring the repulsion between two
charged spheres, A and B, as a function of the separation of the
spheres. The spheres were identical in size and mass. The
measurements are shown in the table of value sand plotted on
the graph below. (Jan 99)
a) Show that the results verify Coulomb's Law by manipulating
the data and providing a new table of values that, when plotted,
will produce a straight-line graph.
b) Plot the new data with the responding variable on the vertical
axis.
c) Calculate the slope of your graph. (7.9 x 10-3 Nm2)
d) Using the slope value, or another suitable averaging techniques
determine the charge on sphere B if the charge on sphere A is
3.08 x 10-7 C. (2.9 x 10-6 C)
e) Determine the magnitude of the force between spheres A and B
when they are at a distance of 2.00 m apart. Use the hypothetical
value of 3.00 x 10-6 C for the charge on sphere B if you were
unable to determine the actual value. (2.0 x 10-3 N)
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 10
L1 = separation (m)
L2 = Force (N)
L3 = 1/separation2 (1/m2)
L3 1/separation2 (1/m2)
1.0 x 102
59
25
6.3
2.8
LinReg(ax+b) L1, L2, Y1
a = 7.9 x 10-3 Nm2
b = 2.0 x 10-3 N
r2 = 0.99974…
min max scl
Window x[ -6.944… 109.72… 20 ]
y[ -0.1085… 0.92056 0.1 ]
By Coulomb’s Law
1 2
e 2
2
e 1 2
2
1
2 9 7
6
2
kq qF = , therefore
R
F R = slope = kq q
slope and hence q =
kq
0.0079299...q =
(8.99 x 10 )(3.08 x 10 )
q = 2.9 x 10 C
−
−
1 2
e 2
9 7 6
e 2
3
e
kq qF =
R
(8.99 x 10 )(3.08 x 10 )(2.9 x 10 )F =
2.00
F = = 2.0 x 10 N
− −
−
Worksheet 4: Electric Fields
1) What is an electric field?
An electric field is the area around a charged particle in
which the charged particle can exert a force on other charged
particles.
2) What are the similarities between an electrical field and a
gravitational field? What are the differences between these
two types of fields?
Similarities: Both are fields that occur at a distance, and both
are inversely related to the square of the distance between the
centers of the objects involved.
Differences: Gravity is directly related to the product of two
masses and electrical is directly related to the product of two
charges. Where gravity can only be attractive, electrical
forces are both attractive (unlike charges) and repulsive (like
charges),
3) Draw the shape of the following fields:
a) around a positive point charge.
b) between a positive and a negative point charge
c) between a vertical positive and a vertical negative
plate.
d) inside a negatively charged, circular conductor.
a) b)
c) d)
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 11
4) What are the two methods to calculate the strength of an
electric field?
1
2
e
2
kqE = using the central charge which
R
creates the field
FE = using the force felt by a charge
q
which comes into the field
��
��
Worksheet 5: Electric Field Intensity
1) What is the electric field strength 7.50 x 10-1 m from a
8.00 µC charged object? (1.28 x 105 N/C)
1
2
9 6
1 2
5
kqE =
R
(8.99 x 10 )(8.0 x 10 )E =
(7.5 x 10 )
NE = 1.28 x 10
C
−
−
��
��
��
2) Calculate the gravitational field strength on the surface of
Mars. Mars has a radius of 3.43 x 106 m and a mass of
6.37 x 1023 kg. (3.61 N/kg)
1
g 2
11 23
g 6 2
g
Gma =
R
(6.67 x 10 )(6.378 x 10 )a =
(3.43 x 10 )
Na = 3.61
kg
−
3) At a point a short distance from a 4.60 x 10-6 C charged
object, there is an electric field strength of 2.75 x 105 N/C.
What is the distance to the charged object producing this
field? (3.88 x 10-1 m)
1
9 6
5
kqR =
E
8.99 x 10 (4.6 x 10 )R =
2.75 x 10
R = 0.388 m
−
��
4) If an alpha particle experiences an electric force of
0.250 N at a point in space, what electric force would a
proton experience at the same point? (0.125 N)
-19
1
2
-19
1
2
p
p
p
kq (1.60 x 10 )F R=
kq (3.20 x 10 )F
R
1F = F
2
1F = (0.250)
2
F = 0.125 N
p
α
α
5) What is the initial acceleration on an alpha particle when it
is placed at a point in space where the electric field strength is
7.60 x 104 N/C? (3.66 x 1012 m/s2)
e 2
4 19
e
14
e
e net
e
14
27
12
2
F = E q
F = (7.60 x 10 )(3.2 x 10 )
F =2.43.. x 10 N
As the electric force is the only force
present, it is the net force F = F
Fa =
m
2.43.. x 10a =
6.367 x 10
ma = 3.65 x 10
s
−
−
−
−
��
6) Calculate the electric field strength mid-way between a
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 12
4.50 µC charged object and a -4.50 µC charged object if the
two charged objects are 5.00 x 10-1 m apart. (1.29 x 106 N/C)
11
2
9 6
1
The net electric field will be the vector sum
of the field from 1 and the field from 2, where
the directions of the fields are defined by a
positive test charge
kqE =
R
8.99 x 10 (4.5 x 10 )E =
(2.5
−
��
��
1 2
51
52
net 1 2
5net
6net
x 10 )
NE = 6.47... x 10 right (see diagram)
C
NE = 6.47... x 10 right (see diagram)
C
E = E + E
E = 2(6.47... x 10 )
NE = 1.29 x 10 right
C
−
��
��
�� �� ��
��
��
7) Calculate the electric field strength mid-way between a
3.0 µC charged object and a 6.0 µC charged object if the
objects are 8.0 x 10-1 m apart. (1.7 x 105 N/C)
The net electric field will be the vector sum
of the field from 1 and the field from 2, where
the directions of the fields are defined by a
positive test charge
11
2
9 6
11 2
51
52
net 1 2
5 5net
net
kqE =
R
8.99 x 10 (3.0 x 10 )E =
(4.0 x 10 )
NE = 1.685 x 10 East (see diagram)
C
NE = 3.371 x 10 West (see diagram)
C
E = E + - E
E = 1.685 x 10 + -3.371 x 10
E = 1.7 x
−
−
��
��
��
��
�� �� ��
��
��
5 N10 West
C
8) Calculate the magnitude and direction of the electric field at
the centre of a square with 10.0 cm sides if the corners taken
in counter-lockwise rotation from top right, have charges of
+1.00 µC, +2.00 µC, +3.00 µC and +4.00 µC. The square is
level with the horizontal. (8.99 x 104 N/C toward the top of the
square)
2 2
2 2
The distance from each of the charges to the
point can be deduced from the Pythagorean
relationship.
R = x + y
R = 0.05 + 0.05
R = 0.7071 m
Each of the charges is at a 45 angle to the point
due to the
o
sides of the triangle in the Pythagorean
relationship being equal.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 13
Since the fields created by the three charges
are not all at right angles to the point, the
net electric field will be the vector sum
of the field from 1, 2, 3 and 4, expressed as
components in the x and
11
2
9 6
12
61
62
3
y dimensions where
the directions of the fields are defined by a
positive test charge.
kqE =
R
8.99 x 10 (1.0 x 10 )E =
(0.07071)
NE = 1.798 x 10 45 S of W
C
NE = 3.596 x 10 45 S of E
C
E =
o
o
−
��
��
��
��
��
6
64
61
N 5.394 x 10 45 N of E
C
NE = 7.192 x 10 45 N of W
C
Using component techniques to resolve the net
electric field.
x y
E = (-1.798 x 10 cos 45, 1.7
o
o��
��
6
6 62
6 63
6 64
net
98 x 10 sin 45)
E = (3.596 x 10 cos 45, 3.596 x 10 sin 45)
E = (5.394 x 10 cos 45, 5.394 x 10 sin 45)
+ E = (-7.192 x 10 cos 45, 7.192 x 10 sin 45)
E = ( 0 , 5.09
−��
��
��
��
6 x 10 )
6net
NE = 5.09 x 10 North
C
��
9) A charged sphere having 2.50 C of excess positive charge
is located 50.0 m due north of a second sphere having 10.0 C
of excess positive charge. What is the net electrical field
intensity at the point(s) 30.0 m from the first and 40.0 m from
the second sphere? (6.15 x 107 N/C 60.8o E of N)
-1
-1
Since this is a 3, 4, 5 triangle, it must have
a right angle as indicated.
o= tan = 90 -
a
30= tan = 90 - 36.869
40
= 36.869 = 53.13o o
ϑ φ ϑ
ϑ φ
ϑ φ
Since the fields created by the two charges
are in a coordinate plane, the
net electric field will be the vector sum
of the field from 1 and 2, expressed as
components in the x and y dimensions where
the directions of the fields are defined by a
positive test charge.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 14
1A
2
9
A2
7A
7B
kqE =
R
8.99 x 10 (2.5)E =
(30)
NE = 2.497 x 10 36.869 S of E
C
NE = 5.618... x 10 53.13 N of E
C
Using component techniques to resolve the net
electric field.
x
o
o
��
��
��
��
7 7A
7 7B
7net
y
E = (2.49... x 10 cos36.869, -2.49... x 10 sin 36.869)
+ E = (5.618... x 10 cos53.13, 5.618... x 10 cos53.13)
E = ( 5.37... x 10 ,
��
��
��
7
net(y)2 21
net net(x) net(y)
net(x)
7
7 2 7 2 1net
7
7net
2.99... x 10 )
EE E + E = tan
E
2.99... x 10E (5.37... x 10 ) + (2.99... x 10 ) = tan
5.37... x 10
NE = 6.15 x 10 = 29.1
C
ϑ
ϑ
ϑ
−
−
=
=
��
�� �� ��
��
��
��
7 0net
NE = 6.15 x 10 at 29.1 N of E
C
o
��
10) June 89. All three objects are positively charged
a) Given that the charges on A and B are each of magnitude
3.00 x 10-6 C, and that the charge on C is of magnitude
4.00 x 10-6 C, determine the magnitude and direction of the
net force acting on object A. Illustrate the answer to part with
an appropriate sketch. (37.5 N at 55.1o S of W)
b) If the mass of A is 3.2 x 10-3 kg, what is the initial
acceleration of A? (1.2 m/s2 at 55.1o S of W)
d) Use a vector sketch to draw in the approximate direction of
the electric field intensity at point P which is half way between
B and C.
BA
CA
1 2BA
2
9 6
BA
The net force will be the vector sum of F
and F where the forces are at right angles
and hence the net force can be determined using
a Pythagorean sum.
kq qF =
R
8.99 x 10 (3.0 x 10 )(3.0 x 1F =
−
�
�
�
�
6
2
BA
CA
0 )
(0.06)
F = 22.475 N West
F = 29.96 N South
Where the directions of the forces are determined
from the law of charges that like charges repel.
−
�
�
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 15
2 2 CA1eNet BA CA
BA
2 2 1eNet
eNet
0eNet
FF F + F = tan
F
29.96F (22.475) + (29.96) = tan
22.475
F = 37.5 N = 53.1
F = 37.5 N at 53.1 S of W
o
ϑ
ϑ
ϑ
−
−
=
=
�
� � �
�
�
�
�
e net
e
3
4
2
As the electric force is the only force operating, it is also
the net force F = F
Fa =
m
37.5a =
3.2 x 10
ma = 1.2 x 10 at 53.1 S of W
s
o
−
Worksheet 6: Electrical Potential and Field Potential:
Point Charges
1) What is the potential at a distance of 6.0 cm from a
2.5 µC charge? (3.7 x 105 V)
1
9 6
5
kqV =
R
8.99 x 10 (2.5 x 10 )V =
0.06
V = 3.7 x 10 V
−
2) Three charges are along a line as shown below:
Find the potential at point P. (-2.0 x 105 V)
1
1
9 6
1
5
1
The voltage at P will be the scalar sum of
the three voltages.
kqV =
R
8.99 x 10 (1.0 x 10 )V =
0.075
V = 1.2 x 10 V
−
5
2
5
3
net 1 2 3
5 5 5
net
5
net
V = -1.5 x 10 V
V = -1.6 x 10 V
V = V + V + V
V = 1.2 x 10 + -1.5 x 10 + -1.6 x 10
V = -2.0 x 10 V
3) Three charges are at the corners of a rectangle as shown
below:
Find the potential at point P. (4.4 x 105 V)
2 2
2 2
The distance from charge 2 to the
point can be deduced from the Pythagorean
relationship.
R = x + y
R = .060 + 0.080
R = 0.10 m
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 16
1
1
9 6
1
5
1
5
2
5
2
net 1 2 3
5
net
The voltage at P will be the scalar sum of
the three voltages.
kqV =
R
8.99 x 10 (1.0 x 10 )V =
0.080
V = 1.12 x 10 V
V = -2.697 x 10 V
V = 5.993 x 10 V
V = V + V + V
V = 1.12 x 10 + -2.697
−
5 5
5
net
x 10 + 5.993 x 10
V = 4.4 x 10 V
4) A -3.5 µC charge is 4.0 cm to the left of a 5.0 µC charge.
What is the field potential at a point 1.0 cm to the right of the
-3.5 µC charge? (-1.6 x 106 V)
1
1
9 6
1
6
1
6
2
net 1 2
6 6
net
net
The voltage at P will be the scalar sum of
the two voltages.
kqV =
R
8.99 x 10 (-3.5 x 10 )V =
0.010
V = -3.1465 x 10 V
V = 1.123 x 10 V
V = V + V
V = -3.1465 x 10 + 1.123 x 10
V = -1.6
−
6x 10 V
5) What is the field potential at a distance of 1.00 x 10-10 m
from a proton? (14.4 V)
1
1
9 19
1 10
1
kqV =
R
8.99 x 10 (1.60 x 10 )V =
1.0 x 10
V = 14.4 V
−
−
6) What is the electrical potential energy of an electron at a
distance of 1.00 x 10-10 m from a proton? (-2.30 x 10-18 J ,
negative indicates attraction)
1 2
el
9 19 19
el 10
18
el
kq qE =
R
8.99 x 10 (1.60 x 10 )(-1.60 x 10 )E =
1.0 x 10
E = 2.30 x 10 J
− −
−
−−
7) What is the potential energy of a proton half way between
two alpha particles separated by 1.00 x 10-14 m?
(1.84 x 10-13 J)
1 2
el1
9 19 19
el1 15
14
el1
14
el2
elnet e
The electrical potential energy at P will be
the scalar sum of the two energies.
kq qE =
R
8.99 x 10 (3.20 x 10 )(1.60 x 10 )E =
5.00 x 10
E = 9.20 x 10 J
E = 9.20 x 10 J
E = E
− −
−
−
−
l1 el2
14
elnet
13
elnet
+ E
E = 2(9.20 x 10 )
E = 1.84 x 10 J
−
−
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 17
8) How much does the potential energy of a 0.030 µC charge
in the field of a 5.0 µC charged object change when it is
moved from R = 45 cm away to R = 15 cm? (6.0 x 10-3 J)
el el(final) el(initial)
1 2 1 2
el
f
el 1 2
f i
9 8 6
el
3
el
∆E = E - E
kq q kq q∆E = -
R R
1 1∆E = kq q -
R R
∆E = 8.99 x 10 (3.0 x 10 )(5.0 x 10 )
1 1 -
0.15 0.45
∆E = 6.0 x 10 J
i
− −
−
9) 4.40 x 10-5 J of energy is used in moving a 3.00 µC charge
at a constant speed from point A to point B. If A and B are
2.4 cm apart, what is the potential difference between A and
B? (14.7 V)
2
5
6
EV =
q
4.40 x 10V =
3.00 x 10
V = 14.7 V
−
−
Λ∆
∆
∆
Worksheet 7: Electrical Potential and Field Potential:
Plates
1) Two parallel plates are connected to a 12.0 V battery. If
the plates are 9.00 x 10-2 m apart, what is the electric field
strength between them? (1.33 x 102 V/m)
2
VE =
d
12E =
0.0900
VE = 1.33 x 10
m
��
��
��
2) The electric field between two parallel plates is
5.0 x 103 V/m. If the potential difference between the
plates is 2.0 x 102 V, how far apart are the plates?
(4.0 x 10-2 m)
3
2
Vd =
E
200d =
5.0 x 10
d = 4.0 x 10 m−
��
3) What is the value of the potential energy of an alpha
particle in the electric field between parallel plates if it is right
next to the positive plate at 500.0 V? (1.60 x 10-16)
el 2
19
el
16
el
E = Vq
E = 500(3.2 x 10 )
E = 1.6 x 10 J
−
−
4) If an electron is 2.00 cm away from the negative plate in a
1.00 x 103 N/C electric field between parallel plates which are
8.00 cm apart, what is the value of the electron’s potential
energy? (9.60 x 10-18 J)
3
V = E d
V = 1.0 x 10 (0.080)
V = 80 V
��
An electron that is 2.00 cm away from the negative plate still
retains ¾ of its potential energy in the 8.00 cm field between
the plates.
el 2
19
el
18
el
E = Vq
E = 60(1.6 x 10 )
E = 9.60 x 10 J
−
−
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 18
5) What is the field intensity between two parallel plates if an
electron is halfway between the plates? The plates have a
potential of 750 V and are 10.0 cm apart. (7.50 x 103 N/C)
3
E is constant between parallel plates of
opposite charge
VE =
d
750E =
0.10
VE = 7.50 x 10
m
��
��
��
��
6) What is the potential difference between two parallel plates
5.00 cm apart that produces an electric force of 3.50 x 10-13 N
on an object containing a charge of 2.00 x 10-16 C when it is
placed between the plates? (8.75 x 101 V)
e
2
13
16
FV = d
q
3.5 x 10V = (0.050)
2.0 x 10
V = 87.5 V
−
−
7) The electric field strength between two parallel plates
is 9.3 x 102 V/m when the plates are 7.0 cm apart. What would
the electric field strength be if the plates were 5.0 cm apart?
(1.8 x 103 V/m)
1 2
el1 k2
k
k
k(new)
new
k(old)old
k(new)
new old
k(old)
17
2
new 17
Energy is conserved as electrical potential
is converted to kinetic energy.
E = E
E = E
Vq = E
E V =
q
E
V q=
EV
q
EV = V
E
9.00 x 10V = (4.2 x 10 )
3.00 x 10
V
−
−
Σ Σ
∴
3
new = 1.26 x 10 V
8) A helium nucleus, with a mass of 6.65 x 10-27 kg, is
inserted between two charged parallel plates. The potential
difference between the plates is 1.5 x 103 V, and the distance
between the plates is 3.5 cm.
What time is required for the helium nucleus to reach the
negative plate?
net
e
2
2
3 19
27
12
2
2
i
F1) a = since F is the only significant force
m
Fa =
m
a =
m
V qa =
d m
1.5 x 10 (3.2 x 10 )a =
(0.035)(6.65 x 10 )
ma = 2.06 x 10
s
12) d = v t + at since v = 0
2
2dt =
a
2(0.035)t =
2.06
e
i
E q
−
−
���
12
7
x 10
t = 1.84 x 10 s−
9) What is the potential difference between two parallel
charged plates that are 7.50 cm apart if a force of
5.30 x10-14 N is needed to move an alpha particle from the
negative plate to the positive plate?(1.24 x 104 V)
e
2
14
16
4
FV = d
q
5.3 x 10V = (0.075)
3.2 x 10
V = 1.24 x 10 V
−
−
10) An alpha particle is placed between two horizontal parallel
charged plates that are 2.00 cm apart. The potential difference
between the plates is 12.0 V.
a) What is the electric force acting on the alpha particle?
(1.92 x 10-16 N)
b) What is the gravitational force acting on the alpha particle?
(6.52 x 10-26 N)
c) Assuming that the electric force and the gravitational force
are acting in opposite directions, what is the net force acting
on the alpha particle? (1.92 x 10-16 N)
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 19
d) What is the acceleration of the alpha particle?
(2.88 x 1010 m/s2)
e) What potential difference would be required between the
plates in order that the alpha particle becomes suspended?
(4.09 x 10-9 V)
a)
e
e 2
e 2
19
e
16
e
Both F and E are constant between parallel
plates of opposite charge
F = E q
VF = q
d
12F = (3.2 x 10 )
0.020
F = 1.92 x 10 N
−
−
��
��
b)
g
27
g
26
g
F = mg
F = 6.67 x 10 (9.81)
F = 6.52 x 10 N
−
−
c)
net el g
16 26net
16net
F = F + -F
F = 1.92 x 10 + -6.52 x 10
F = 1.92 x 10 N
− −
−
� � �
�
�
d)
net
16
27
10
2
Fa =
m
1.92 x 10a =
6.67 x 10
ma = 2.88 x 10
s
−
−
�
�
�
�
e)
el g
2
2
27
19
9
If a particle were suspended between the
plates then forces would be balanced.
F = F
Vq = mg
d
mgdV =
q
6.67 x 10 (9.81)(0.020)V =
3.2 x 10
V = 4.09 x 10 V
−
−
−
� �
Worksheet 8: Work and Energy in Electric Fields
1) A proton is released 2.0 x 10-11 m from the center of a
6.4 x 10-18 C charged sphere. What is the speed of this proton
when it is 0.50 m from this center? (7.4 x 105 m/s)
1 2
el1 k2 el12
21 2 1 2
2
1 2
1 2
2
1 2
9 19
2
Energy is conserved as some of the electrical
potential energy is converted to kinetic energy
E = E
E = E + E
kq q kq q1 = mv +
R 2 R
2kq q 1 1v = -
m R R
2(8.99 x 10 )(1.6 x 10v =
−
Σ Σ
18
27 11
5
2
)(6.4 x 10 ) 1 1 -
1.67 x 10 2 x 10 0.50
mv = 7.4 x 10
s
−
− −
2) The centers of two alpha particles are held 2.5 x 10-12 m
apart when they are released. Calculate the speed of each
alpha particle when they 0.75 m apart. (2.4x 104 m/s)
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 20
1 2
el1 k2 el12
21 2 1 2
2
1 2
1 2
2
1 2
Energy is conserved as some of the electrical
potential energy is converted to kinetic energy
of the two alpha particles
E = E
E = 2E + E
kq q kq q1 = (2) mv +
R 2 R
kq q 1 1v = -
m R R
Σ Σ
9 19 2
2 27 12
5
2
2(8.99 x 10 )(3.2 x 10 ) 1 1v = -
6.67 x 10 2.5 x 10 0.75
mv = 2.3 x 10
s
−
− −
3) An alpha particle gains 1.50 x 10-15 J of kinetic energy.
Through what potential difference was it accelerated?
(4.69 x 103 V)
1 2
el1 k2
k2
k2
15
19
3
Energy is conserved as the electrical
potential energy is converted to kinetic energy
E = E
E = E
Vq = E
EV =
q
1.5 x 10V =
3.2 x 10
V = 4.69 x 10 V
−
−
Σ Σ
4) A proton is accelerated by a potential difference of
7.20 x 102 V. What is the change in kinetic energy of the
proton? (1.15 x 10-16 J)
1 2
el1 k2
k2
2 19
k2
16
k2
Energy is conserved as the electrical
potential energy is converted to kinetic energy
E = E
E = E
Vq = E
E = 7.2 x 10 (1.6 x 10 )
E = 1.15 x 10 J
−
−
Σ Σ
5) What maximum speed will an alpha particle reach if it
moves from rest through a potential difference of
7.50 x 103 V? (8.50 x 105 m/s)
1 2
el1 k2
2
2
3 19
27
5
Energy is conserved as the electrical
potential energy is converted to kinetic energy
E = E
E = E
mvVq =
2
2Vqv =
m
2(7.5 x 10 )(3.2 x 10 )v =
6.67 x 10
mv = 8.48 x 10
s
−
−
Σ Σ
6) A proton is placed in an electric field between two parallel
plates. If the plates are 6.0 cm apart and have a potential
difference between them of 7.5 x 101 V, how much work is
done against the electric field when the proton is moved
3.0 cm parallel to the plates? (0)
If the proton moves horizontally, there is no change in its
potential energy. Hence, there is no work done.
7) In the above question, how much work would be done
against the electric field if the proton was moved 3.0 cm
perpendicular to the plates? (6.0 x 10-18 J)
If the charge moves vertically 3.0 cm out of a total of 6.0 cm,
then it will cross half of the voltage between the plates. From
the work-energy theorem, the work done is equivalent to the
change in potential energy.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 21
el
2
19
18
W = ∆E
W = ∆Vq
W = 37.5(1.6 x 10 )
W = 6.0 x 10 J
−
−
8) A charged particle was accelerated from rest by a potential
difference of 4.20 x 102 V. If this particle increased its kinetic
energy to 3.00 x 10-17 J, what potential difference would be
needed to increase the kinetic energy of the same particle to
9.00 x 10-17 J? (1.26 x 103 V)
1 2
el1 k2
k
k
k(new)
new
k(old)old
k(new)
new old
k(old)
17
2
new 17
Energy is conserved as electrical potential
is converted to kinetic energy.
E = E
E = E
Vq = E
E V =
q
E
V q=
EV
q
EV = V
E
9.00 x 10V = (4.2 x 10 )
3.00 x 10
V
−
−
Σ Σ
∴
3
new = 1.26 x 10 V
9) An alpha particle with an initial speed of 7.15 x 104 m/s
enters through a hole in the positive plate between two parallel
plates that are 9.00 x 10-2 m apart as shown in the diagram.
If the electric field between the plates is 1.70 x 102 V/m, what
is the speed of the alpha particle when it reaches the negative
plate? (8.11 x 104 m/s)
1 2
k1 el1 k2
2 2
1 2
2
1
2
3 -19 -27
Energy is conserved as some of the electrical
potential energy is converted to kinetic energy
E = E
E + E = E
Emv mv + q =
2 d 2
2 E q - mdvv =
md
2(1.7 x 10 )(3.2 x 10 ) - 6.67 x 10v =
Σ Σ
��
��
4 2
27
4
(0.090)(7.15 x x 10 )
6.67 x 10 (0.090)
mv = 8.11 x 10
s
−
10) An electron with a speed of 5.0 x 105 m/s enters through a
hole in the positive plate and collides with the negative plate at
a speed of 1.0 x 105 m/s. What is the potential difference
between the plates? (6.8 x 10-1 V)
1 2
k1 k2 el2
2 2
1 2
2 2
1 2
-31 5 2 5 2
Energy is conserved as some of the kinetic
energy is converted to electrical potential energy
E = E
E = E + E
mv mv = + Vq
2 2
m(v - v )V =
2q
(9.11 x 10 )[(5.0 x10 ) - (1.0 x10 ) ]V =
2(1.
Σ Σ
1960 x 10 )
V = 0.68 V
−
Worksheet 9: Millikan’s Oil Drop Experiment
1) An oil drop weighs 3.84 x 10-15 N. If it is suspended
between two horizontal parallel plates where the electric field
strength is 1.20 x 104 N/C, what is the magnitude of the charge
on the oil drop? (3.20 x 10-19 C)
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 22
e g
2
2
15
2 4
19
2
If the oil droplet is held stationary, then
the upward electric force must balance with
the downward gravitational force.
F = F
E q = mg
mgq =
E
3.84 x 10q =
1.2 x 10
q = 3.20 x 10 C
−
−
��
��
2) An oil drop with a mass of 4.80 x 10-16 kg is suspended
between two horizontal parallel plates that are 6.00 cm apart.
If the potential difference between the plates is 5.90 x 102 V,
how many excess electrons does the oil drop carry? (3)
194.788 x 10 C−
e g
2
2
2
16
2 2
2
If the oil droplet is held stationary, then
the upward electric force must balance with
the downward gravitational force.
F = F
E q = mg
Vq = mg
d
mgdq =
V
4.8 x 10 (9.81)(0.06)q =
5.9 x 10
q = 4.788
−
��
19
19
19
-
2
x 10 C
1 e4.788 x 10 C x
1.6 x 10 C
q = 3e
−
−
−
−
3) An oil drop with a mass of 7.20 x 10-16 kg is moving up at a
constant speed of 2.50 m/s between two horizontal parallel
plates on the inside of an evacuated iron pot like Millikan’s. If
the electric field strength is 2.20 x 104 V/m, what is the
magnitude of the charge on the oil drop? (3.21 x 10-19 C)
If the oil droplet is moving upward at a constant
velocity, then the upward electric force must
balance with the downward gravitational force.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 23
e g
2
2
16
2 4
19
2
F = F
E q = mg
mgq =
E
7.2 x 10 (9.81)q =
2.20 x 10
q = 3.21 x 10 C
−
−
��
��
4) An oil drop whose mass is 3.50 x 10-15 kg accelerates
downward at a rate of 2.50 m/s2 when placed between two
horizontal parallel plates that are 1.00 cm apart. Assuming the
oil drop is negative and the top plate positive, how many
excess electrons does the oil drop carry if the potential
difference between the plates is 5.38 x 102 V? (3)
net e g
2
2
15
2
If the oil droplet is accelerating downward
velocity, then the upward electric force must
be less than the downward gravitational force.
F = -F + F
ma = - E q + mg
md(a - g)q =
-V
3.5 x 10 (0.0q =
−
��
2
19
2
19
19
-
2
1)(2.5 - 9.81)
-5.38 x 10
q = 4.755 x 10 C
1 e4.755 x 10 C x
1.6 x 10 C
q = 3e
−
−
−
−
5) During a Millikan oil drop experiment, a student records the
weight of five different oil drops. A record is also made of the
Potential Difference necessary to hold each drop stationary
between two horizontal parallel plates. (slope = 2/4 x 1014 N)
Mass (10-16 kg) Potential Difference (103V)
8.8 2.03
6.3 1.58
4.5 1.15
3.5 0.81
1.3 0.27
a) Draw a graph showing voltage as a function of weight.
b) What does the slope of the graph represent?
c) If the distance between the plates was a constant 0.40 m,
what is implied about the charge on each of the drops?
(q = 1.6 x 10-18 C)
L1 = Mass (10-16 kg)
L2 = Voltage (V)
L3 = Weight of Fg (N)
L3 Weight (N) Manipulated = Wieght (N)
8.6 x 10-18 Responding = Voltage (V)
6.2 x 10-18 Control = distance between plates
4.4 x 10-18
3.4 x 10-18
1.3 x 10-18
LinReg(ax+b) L3, L2, Y1
a = 2.4 x 1017 V/N
b = 15 V
r2 = 0.9903…
min max scl
Window x[ 5.395… x 10-16 9.3688… x 10-16 5 x 10-15 ]
y[ -29.2… 2329.2 200 ]
In a balanced situation like this, the upward forces will be
equal to the downward forces.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 24
e g
2 g
2 g
g
2
g
2
If the oil droplet is stationary,
then the upward electric force must
equal the downward gravitational force.
F = F
E q = F
Vq = F
d
F dV =
q
Since slope = V/F using the equation above
dslope = and he
q
��
2
2 17
19
2
dnce q =
slope
0.40q =
2.4 x 10
q = 1.6 x 10 C−
6) An oil drop whose mass is 5.70 x 10-16 kg accelerates
upward at a rate of 2.90 m/s2 when placed between two
horizontal parallel plates that are 3.50 cm apart. If the
potential difference between the plates is 7.92 x 102 V, what is
the magnitude of the charge on the oil drop? (3.20 x 10-19 C)
net e g
2
2
2
1
2
If the oil droplet is accelerating upward,
then the upward electric force must be
greater than the downward gravitational force.
F = F + -F
ma = E q + -mg
Vma = q + -mg
d
md(a + g)q =
V
5.7 x 10q =
−
��
6
2
19
2
(0.035)(2.9 - 9.81)
2.92 x 10
q = 3.20 x 10 C−
7) In a Millikan oil drop experiment a student sprayed oil
droplets with a density of 7.8 x 102 kg/m3 between two
horizontal parallel plates that were 4.0 cm apart. The student
adjusted the potential difference between the plates to
4.6 x 103 V so that one of the drops became stationary.
The diameter of this drop was measured to be 2.4 x 10-6 m.
What was the magnitude of the charge on this oil drop?
(Vsphere = 4/3πr3) (4.8 x 10-19 C)
e g
2 g
2
2
3
2
2 6 3
2
If the oil droplet is stationary,
then the upward electric force must
equal than the downward gravitational force.
F = F
E q = F
Vq = mg
d
mgdV =
q
4( r )gd3q =
V
47.8 10 ( [1.2 x 10 ] )9.81(0.040)
3q =
x
ρ π
π −
��
3
19
2
4.6 x 10
q = 4.8 x 10 C−
8) An oil droplet (q = 5e-) is suspended between two parallel
charged plates (V = 175 V). If an oil droplet of the same mass
but a charge of 3e- to be suspended between the same plates,
what potential difference would be necessary (292 V)
If the oil droplet is stationary,
then the upward electric force must
equal than the downward gravitational force.
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 25
e g
2 g
2
2
F = F
E q = F
Vq = mg
d
mgdV =
q
mgdV 53e= =
mgdV 3
5e
5V = V
3
5V = (175)
3
V = 292 V
new
old
new old
new
new
−
−
��
9) In an oil drop experiment similar to Millikan's, an oil
droplet is suspended between two parallel charged plates.
Calculate the magnitude of the charge on the oil droplet
given the following:
• radius of the oil drop 4.2 x 10-6 m
• density of the oil 7.8 x 102 kg/m3
• distance between plates 2.0 cm
• potential difference between plates 99 V
(4.80 x 10-16 C)
e g
2 g
2
If the oil droplet is stationary,
then the upward electric force must
equal than the downward gravitational force.
F = F
E q = F
Vq = mg
d
��
2
3
2
2 6 3
2
19
2
gdq =
V
4r gd
3q =
V
47.8 x 10 ( [4.2 x 10 ] )9.81(0.020)
3q = 79
q = 4.8 x 10 C
Vρ
ρ π
π −
−
10) In Millikan's oil drop experiment an oil drop having a
mass of 6.00 x 10-17 kg is accelerating upward at a rate of
4.60 m/s2 when sprayed between two horizontal parallel
plates 5.00 cm apart. If the potential difference between
the plates is 9.00 x 101 V, what is the charge on the oil
drop? (4.80 x 10-19 C)
net e g
2
2
2
1
2
If the oil droplet is accelerating upward,
then the upward electric force must be
greater than the downward gravitational force.
F = F + -F
ma = E q + -mg
Vma = q + -mg
d
md(a + g)q =
V
6.0 x 10q =
−
��
6
19
2
(0.050)(4.6 + 9.81)
90
q = 4.80 x 10 C−
Worksheet 10: Electric Current
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 26
1) A current of 3.60 A flows for 15.3 s through a conductor.
Calculate the number of electrons that pass through a point in
the conductor in this time. (3.44 x 1020)
19
20
Q = I t
Q = (3.60)(15.3)
1Q = 55.08 C x
1.6 x 10 C
Q = 3.44 x 10 e
e−
−
−
2) How long would it take 2.0 x 1020 electrons to pass through
a point in a conductor if the current was 10.0 A?
(3.2 s)
Qt =
I
32t =
10
t = 3.2 s
3) Calculate the current through a conductor, if a charge of
5.60 C passes through a point in the conductor in 15.4 s.
(3.64 x 10-1 A)
QI =
t
5.6I =
15.4
I = 0.364 A
4) If 5.26 x 1020 alpha particles pass a point in space every
10.0 s, what is the current at this location? (16.8 A)
QI =
t
168.33I =
10
I = 16.8 A
5) If 34.0 x 10-3 A of electron current passes between the
terminals of a 1.50 V battery in 1.20 s, how much energy has
the battery used up in this time? (6.12 x 10-2 J)
2
3
2
E = Vq where Q = It
gives E = VIt
E = 1.5(34 x 10 )(1.2)
E = 6.12 x 10 J
−
−
6) If a current of 3.20 A passes through an electric circuit for
4.00 s, how many electrons have passed any given point in the
circuit? (8.00 x 1019 e)
19
19
Q = I t
Q = (3.22)(4)
1Q = 12.8 C x
1.6 x 10 C
Q = 8.00 x 10 e
e−
−
−
7) There are 7.50 x 1027 protons traveling through a region of
space every hour. What is the electric current at this point?
(3.33 x 105 A)
19
27 9
+
9
3
5
1.60 x 10 C7.5 x 10 p x = 1.20 x 10 C
1 p
QI =
t
1.20 x 10I =
3.6 x 10
I = 3.33 x 10 A
−
+
Physics 30 Electric Forces and Fields
S Molesky @ Notre Dame Page 27
8) When an electric appliance is connected to a
1.20 x 102 V power line, there is a current through the
appliance of 18.3 A. What is the average amount of energy
given to each electron per second by the power line?
(1.92 x 10-17 J/s)
2
3
19
20
3
20
17
-
E = Vq where Q = It
gives E = VIt
E = 120(18.3)(1)
E = 2.196 x 10 J
118.3 C x
1.60 x 10 C
1.14375 x 10
E 2.196 x 10 =
1.14375 x 10
J1.92 x 10
e
e
e
e
−
−
−
−
−
=
=
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