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ELECTRIC FORCES AND ELECTRIC FIELDS
Electric charge is the fundamental quantity that underlies all electrical phenomena. There
are two types of charges, positive and negative, and like charges repel each other, and
unlike charges attract each other. A conductor is a material through which charge can
easily flow due to a large number of free electrons, whereas an insulator does not allow
charge to flow freely through it. The force between charges can be found by applying
Coulomb’s law. The electric field around a charge is the force per unit charge exerted on
another charge in its vicinity.
QUICK REFERENCE
Important Terms
charging by conduction
transfer of charge by actual contact between two objects
charging by induction
transfer of charge by bringing a charged object near a conductor, then grounding
the conductor
conservation of charge
law that states that the total charge in a system must remain
constant during any process
coulomb
the unit for electric charge
Coulomb’s law
the electric force between two charges is proportional to the product of
the charges and inversely proportional to the square of the distance between them
electric charge
the fundamental quantity which underlies all electrical phenomena
electric field
the space around a charge in which another charge will experience a force;
electric field lines always point from positive charge to negative charge
electron
the smallest negatively charged particle
electrostatics
the study of electric charge, field, and potential at rest
elementary charge
the smallest existing charge; the charge on one electron or one
proton (1.6 x 10-19 C)
parallel plate capacitor
capacitor consisting of two oppositely charged parallel plates of equal area, and
storing an electric field between the plates
neutral
having no net charge
test charge
the very small charge used to test the strength of an electric field
1
Equations and Symbols
F
where
kq1 q 2
1 q1 q 2

40 r 2
r2
F = electric force
k = electric constant = 9x109 Nm2 / C2
ε0 = permittivity constant
= 8.85 x 10-12 C2 / Nm2
q (or Q) = charge
r = distance between charges
E = electric field
F kq
1 q
E
 2 
q0 r
40 r 2
Charged Objects and the Electric Force, Conductors and Insulators
Charge is the fundamental quantity that underlies all electrical phenomena. The symbol
for charge is q, and the SI unit for charge is the Coulomb (C). The fundamental carrier of
negative charge is the electron, with a charge of – 1.6 x 10-19 C. The proton, found in the
nucleus of any atom, carries exactly the same charge as the electron, but is positive. The
neutron, also found in the nucleus of the atom, has no charge. When charge is transferred,
only electrons move from one atom to another. Thus, the transfer of charge is really just
the transfer of electrons. We say that an object with a surplus of electrons is negatively
charged, and an object having a deficiency of electrons is positively charged. Charge is
conserved during any process, and so any charge lost by one object must be gained by
another object.
The Law of Charges
The law of charges states that like charges repel each other and unlike charges attract
each other. This law is fundamental to understanding all electrical phenomena.
Example 1
Consider four charges, A, B, C, and D, which exist in a region of space. Charge A attracts
B, but B repels C. Charge C repels D, and D is positively charged. What is the sign of
charge A?
Solution
If D is positive and it repels C, C must also be positive. Since C repels B, B must also be
positive. A attracts B, so A must be negatively charged.
Charge is one of the four quantities in physics that is conserved during any process.
Example 2
Consider two charged spheres of equal size carrying a charge of +6 C and –4 C,
respectively. The spheres are brought in contact with one another for a time sufficient to
allow them to reach an equilibrium charge. They are then separated. What is the final
charge on each sphere?
+6
2
-4
Solution
When the two spheres come in contact with each other, charge will be transferred, but
the total amount of charge is conserved. The total charge on the two spheres is +6 C + -4
C = +2 C, and this is the magnitude of the equilibrium charge. When they are separated,
they divide the charge evenly, each keeping a charge of +1 C.
Conductors, like metals, have electrons which are loosely bound to the outskirts of their
atoms, and can therefore easily move from one atom to another. An insulator, like wood
or glass, does not have many loosely bound electrons, and therefore cannot pass charge
easily.
Charging by Contact and by Induction
We can give an object a net charge two ways: conduction (contact) and induction. In
order to charge an object by conduction, we must touch the object with a charged object.
giving the two objects the same charge sign.
Charging by induction gives us an object charged oppositely to the original charged
object. For example, as shown in your textbook, if we bring a negatively charged rod near
a conducting (metal) sphere, and then ground the metal sphere, negative charges on the
sphere escape to the ground, leaving the sphere with a net positive charge.
Example 3
Show how we can begin with a positively charged rod and charge a metal sphere
negatively. Take a moment to draw the charges on each of the objects in the sequence of
diagrams below.
++++++++
++++++++
I
II
III
Solution
++++++++
++++++++
-
-
+
-
-
+
-
-
-
+
+
ground
I
3
II
III
In figure I a positively charged rod is brought near a neutral metal sphere, separating the
charges in the sphere. When the sphere is grounded, the positive charges escape into the
ground (actually, electrons come up from the ground). When the rod and grounding wire
are removed, the sphere is left with a net negative charge.
Coulomb’s Law
The force between any two charges follows the same basic form as Newton’s law of
universal gravitation, that is, the electric force is proportional to the magnitude of the
charges and inversely proportional to the square of the distance between the charges.
The equation for Coulomb’s law is
Kq1 q 2
FE 
r2
where FE is the electric force, q1 and q2 are the charges, r is the distance between their
centers, and K is a constant which equals 9 x 109 Nm2/C2.
r
F
-q1
+q2
Sometimes the constant K is written as K 
1
4 o
, where o = 8.85 x 10-12 C2 / Nm2.
Example 4
r
+2 μC
-4 μC
Two point charges q1 = +2 μC and q2 = - 4 μC are separated by a distance r, as shown
above.
(a) If the force between the charges is 2 N, what is the value of r?
(b) Where could you place a third charge q3 = +1 μC on the horizontal axis so that there
would be no net force acting on q3? Find an equation which could be solved for x, where
x is the distance from the +2 μC charge to q3. It is not necessary to solve this equation.
4
Solution
(a)
Kq1 q 2
FE 
r2
r

Nm 2
 9 x10 9
C2
Kq1 q 2


FE
2N



 0.19 m
(b) For the force on the third charge to be zero, it would have to be placed to the left of
the +2 μC charge. Let x be the distance from the +2 μC charge to q3. Then the - 4 μC
charge would be (x + r) from q3.
x
q3
r
+2 μC
-4 μC
Kq1q3 Kq2 q3

0
x2
x  r 2
This equation can be solved for x.
F13  F23 
The Electric Field
An electric field is the condition of space around a charge (or distribution of charges) in
which another charge will experience a force. Electric field lines always point in the
direction that a positive charge would experience a force. For example, if we take a
charge Q to be the source of an electric field E, and we bring a very small positive “test”
charge q nearby to test the strength and direction of the electric field, then q will
experience a force which is directed radially away from Q.
Q
q
F
The electric field is given by the equation
F
,
q
where electric field E is measured in Newtons per coulomb, and F is the force acting on
the charge q which is experiencing the force in the electric field. Electric field is a vector
which points in the same direction as the force acting on a positive charge in the electric
field. The test charge q would experience a force radially outward anywhere around the
E
5
source charge Q, so we would draw the electric field lines around the
positive charge Q like this:
E
Electric field lines in a region can also represent the path a positive charge would follow
in that region.
Remember, electrons (negative charges) are moved when charge is transferred, but
electric field lines are drawn in the direction a positive charge would move.
The electric field due to a point charge Q at a distance r away from the center of the
charge can also be written using Coulomb’s law:
 KQq 
 2 
F
r  KQ
E 
 2
q
q
r
where K is the electric constant, Q is the source of the electric field, and q is the small
charge which feels the force in the electric field due to Q.
Electric Field Lines
Drawing the electric field lines around a charge or group of charges helps us to imagine
the behavior of a small charge place in the region of the electric field. The diagrams
below illustrate the electric field lines in the region of a positive charge and a negative
charge. Your textbook has several more diagrams showing the electric field lines around
pairs of opposite charges and pairs of like charges.
E
Positive charge
Negative charge
The above electric fields are not uniform but vary
with the square of the distance from the source
charge. We can produce a uniform electric field by
charging two metal plates oppositely and creating a
capacitor. A capacitor can store charge and electric
field for later use.
+++++++++++++++++++
E
---------------------------
6
The Electric Field Inside a Conductor: Shielding
When charge is placed on a conductor, all of the charge moves to the outside of the
conductor. Consider a metal sphere. If we place positive charges totaling Q on the sphere,
they all go to the outside and distribute themselves in such a way to get as far from each
other as possible.
+
+
Q
+
+
+
R
+
r
+
+
+
+
+
+
Inside the metal sphere (r < R) , the electric field is zero, since all the charge is on the
outside of the sphere. Outside the sphere (r > R), the electric field behaves as if the sphere
KQ
is a point charge centered at the center of the sphere, that is, Eoutside  2 .
r
We can graph electric field E vs. distance from the center r for the charged conducting
sphere:
E
r
0
R
7
REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. When charge is transferred from one
object to another, which of the following
are actually transferred?
(A) electrons
(B) protons
(C) neutrons
(D) quarks
(E) photons
4. Two charges q1 and q2 are separated
by a distance r and apply a force F to
each other. If both charges are doubled,
and the distance between them is halved,
the new force between them is
(A) ¼ F
(B) ½ F
(C) 4F
(D) 8F
(E) 16F
2. Two conducting spheres of equal size
have a charge of – 3 C and +1 C,
respectively. A conducting wire is
connected from the first sphere to the
second. What is the new charge on each
sphere?
(A) – 4 C
(B) + 4 C
(C) – 1 C
(D) + 1 C
(E) zero
5. Two uncharged spheres A and B are
near each other. A negatively charged
rod is brought near one of the spheres as
shown. The far right side of sphere B is
(A) uncharged
(B) neutral
(C) positive
A
B
(D) negative
(E) equally positive and negative.
3. According to Coulomb’s law, if the
electric force between two charges is
positive, which of the following must be
true?
(A) One charge is positive and the other
charge is negative.
(B) The force between the charges is
repulsive.
(C) The force between the charges is
attractive
(D) The two charges must be equal in
magnitude.
(E) The force must be directed toward
the larger charge.
8
A
9. Which of the particles would not
experience a force while between the
plates?
(A) I and II only
(B) II and III only
(C) I only
(D) III only
(E) I, II, and III
B
6. Two charges A and B are near each
other, producing the
electric field lines shown. What are the
two charges A and B, respectively?
(A) positive, positive
(B) negative, negative
(C) positive, negative
(D) negative, positive
(E) neutral, neutral
7. A force of 40 N acts on a charge of
0.25 C in a region of space. The electric
field at the point of the charge is
(A) 10 N/C
(B) 100 N/C
(C) 160 N/C
(D) 40 N/C
(E) 0.00625 N/C
Questions 8 - 9:
Two charged parallel plates are oriented
as shown.
The following particles are placed
between the plates, one at a time:
I.
electron
E
II.
proton
III.
neutron
8. Which of the particles would move to
the right between the plates?
(A) I and II only
(B) I and III only
(C) II and III only
(D) II only
(E) I only
9
+
+
+
Q
+
+
R
+
r
+
Q
+
+
+
+
+
10. An amount of positive charge Q is placed on a conducting sphere. A positive point
charge Q is placed at the exact center of the sphere and remains there. Which of the
following graphs best represents the of electric field E vs distance r from the center?
(A)
(D)
E
E
r
r
R
(B)
R
(E)
E
E
r
r
R
(C)
R
E
r
R
10
Free Response Question
Directions: Show all work in working the following question. The question is worth 15
points, and the suggested time for answering the question is about 15 minutes. The parts
within a question may not have equal weight.
1. (15 points)
y
+Q
a
x
a
2a
P
+Q
Two charges each with charge +Q are located on the y – axis, each a distance a on either
side of the origin. Point P is on the x – axis a distance 2a from the origin.
(a) In terms of the given quantities, determine the magnitude and direction of the electric
field at
i. the origin
ii. point P
iii. a distance x on the x –axis a great distance from the origin (x >> 2a).
(b) On the axes below, sketch a graph of electric field Ex vs. distance x on the +x – axis.
Ex
a
2a
11
A small ball of mass m and charge +q is hung from a thread which is attached to the
ceiling directly above the mark at a distance a from the origin. Charge +q is repelled
away from the origin and comes to rest at a point of equilibrium at a distance 2a from the
origin on the
x – axis.
y
+Q
a
m,
+q
a
x
a
2a
P
+Q
(c) On the diagram below, draw a free-body diagram of the forces acting on the ball when
it is in equilibrium at point P.
(d) Determine an expression for the tension FT in the string in terms of the given
quantities and fundamental constants.
12
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. A
When charge is transferred, electrons move from one object to another.
2. C
Conservation of charge: - 3 + 1 = - 2, which is divided evenly between the two charges,
so each sphere gets – 1 C.
3. B
In the equation for electric force, two positive or two negative charges multiplied by each
other yields a positive force, indicating repulsion.
4. E
F
K (2q1 )( 2q 2 )
 16F
1 2
( r)
2
5. D
The far right side of sphere B is negative, since the negative charges in the sphere are
pushed as far away as possible by the negative charges on the rod.
6. D
Electric field lines begin on positive charges and end on negative charges, thus A is
negative and B is positive.
7. C
E
F
40 N
N

 160
q 0.25C
C
8. D
Only the positively charged proton would move to the right, toward the negatively
charged plate.
9. D
Since the neutron has no charge, it would not experience a force in an electric field.
10. B
KQ
2 KQ
and on the outside is E outside  2 . In
2
r
r
both cases, the electric field follows the inverse square law.
The electric field on the inside is Einside 
13
Free Response Question Solution
(a)
i. 1 point
The electric field at the origin is zero, since a positive test charge placed at the origin
would experience no net force.
ii. 4 points
The net electric field Ex at point P is equal to the sum of the x-components of the electric
field vectors from each of the two charges, since the y-components cancel.
y
r  a2  2a 
+Q
2
a
θ
θ
Ex
a
2a
P
+Q
 KQ  2a 
E x  E1x  E 2 x  2 E cos   2 2  
 r  r 
Substituting for r:

 KQ 
2a
2 KQa

E x  2 2

2 
3
2 
2
 a  2a   a  2a   a 2  2a 2 2
iii. 2 points
If we go out to a point very far away on the x – axis where x >> 2a, the two charges seem
very close together such that they behave as one point charge of magnitude +2Q. Then
the electric field a distance x away is
K 2Q 
E
x2


14
(b) 2 points
Ex
a
2a
(c) 3 points
FT
FTy
φ
FE
FTx
mg
(d) 3 points
Since the system is in equilibrium, ΣF = 0.


2 KQa 

FTx  FE  qE  q
and
3
 2
2 2 
 a  2a  
FTy  mg
Then


FT  FTx  FTy
2

1
2 2



2 KQa
 
 2
 a  2a 2


1


3 
2


2
2
2
 mg  


15
ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC
POTENTIAL
Work must be done to move a charge in an electric field, and the work is related to the
potential difference between two points in an electric field. A surface on which all points
are at the same potential is called an equipotential surface. Several electric charges in the
same vicinity have an electric potential energy due to their mutual attraction or repulsion.
Two equally and oppositely charged conductors, usually metal plates, which are near
each other form a capacitor, in which electrical energy and charge can be stored.
QUICK REFERENCE
Important Terms
capacitor
two oppositely charged conductors used to store charge and energy in an electric
field between them
electric potential
the amount of work per unit charge to move a charge from a very
distant point to another point in an electric field
electric potential difference
the difference in potential between two points in an
electric field; also known as voltage
electric potential energy
the energy stored between two charges as a result of the electric force between
them; also equal to the energy needed to assemble two or more charges to a
separation distance r.
equipotential surface
a surface which everywhere has the same potential
farad
the unit for capacitance equal to one coulomb per volt
volt
the SI unit of potential or potential difference
16
Equations and Symbols
EPE 
V AB
where
kq1 q 2
1 q1 q 2

r
40 r
EPE = electric potential energy (denoted
as UE on the AP Physics exam)
k = electric constant = 9x109 Nm2 / C2
ε0 = permittivity constant =
8.85 x 10-12 C2 / Nm2
q (or Q) = charge
r = distance between charges
ΔVAB = potential difference between
points A and B
WAB = work needed to move a charge
from point A to point B
Vr = potential at a point a distance r away
from a source charge
E = electric field
Δs = displacement along a line between
two points
C = capacitance
A = area of a capacitor plate
d = distance between parallel capacitor
plates
UE = electrical energy stored in a
capacitor
W
EPEB  EPE A
 AB 
q0
q0
Vr  
kq
1

r
40
q
r
V
s
q  CV
 A
C 0
d
1
1
1 Q2
U E  CV 2  QV 
2
2
2 C
E
Potential Energy and The Electric Potential Difference
The electric potential V is defined in terms of the work we would have to do on a charge
to move it against an electric field. For example, if we wanted to move a positive charge
from point A to point B in the electric field shown below, we would have to do work on
the charge, since the electric field would push against us.
+
+
Q
+
+
Work
+
R
+
B
r
+
+
+
+
A
qo
+
+
17
Just as two masses have potential energy due to the gravitational force between them, two
charges Q and qo have electric potential energy EPE due to the electric force between
them. We say that there is a potential difference V between points A and B, and the
equation for potential difference between two points is
EPE  Work

, and is measured in joules/coulomb, or volts.
qo
qo
When we apply potential difference to circuits in a later section, we will often call it
voltage. If we place the charge q at point B and let it go, it would “fall” toward point A.
We say that positive charges naturally want to move from a point of high potential (B) to
low potential (A), and we refer to the movement of the positive charges as current. We
will return to voltage and current in the next chapter.
V 
The Electric Potential Difference Created by Point Charges
As we did for the electric field, we can write the electric potential due to a source charge
Q at a distance r from the source charge:
V 
KQ
r
Unlike electric field, which is a vector quantity, electric potential is a scalar quantity,
that is, there is no direction or angle associated with potential, and potentials may be
added without worrying about components. However, the potential due to a positive
charge is positive, and the potential due to a negative charge is negative.
Example 1
Recall the free-response problem from chapter 18, which includes two positive charges
located on the y – axis. Let Q = 4.0 μC, and a = 0.05 m.
y
+Q
a
x
a
2a
P
+Q
Find the electric potential at
(a) the origin, and
(b) at point P.
18
(c) i. How much work would it take to move a positive charge q = 2 μC and
mass m = 2 x 10-3 kg from a distance very far away to the origin?
ii. If this particle were placed at rest at the origin and then displaced slightly in the x –
direction, find its speed when it is at a very large distance from the origin.
Solution
(a) Unlike the electric field, the electric potential at the origin is not zero.

Nm 2 
4.0 x10  6 C 
2 9 x10 9
2 
C
KQ
KQ


Vo  
2

 1.44 x 10 6 V
r
a
0.05 m
(b) The distanced r is the hypotenuse of the triangle of legs a and 2a:

Nm 2 
2 9 x109 2  4.0 x10 6 C
C 
KQ
KQ
Vo  
2
 
 6.44 x105 V
2
2
2
2
r
0.05 m  40.05 m
a  2a 


(c) i. At a very large distance from the origin, the potential is essentially zero. Thus the
work required to move this charge from a very large distance to the origin is simply the
product of the charge and the potential at the origin:
W  qVo  2 x10 6 C 1.44 x 10 6 V   2.88 J
ii. As the particle accelerates away from the origin, it essentially reaches a maximum
speed when it is a very large distance from the origin. At this distance, the work done on
the particle is equal to its kinetic energy, from which we can find the speed of the
particle.
1
Wo  KE  mv 2 max
2
22.88 J 
v max 
 53.7 m / s
2 x10 3 kg
Equipotential Surfaces and Their Relation to Electric Fields
An equipotential surface is one in which all of the points on the surface are at the same
potential. A charged conducting sphere is an example of an equipotential surface, since it
would require the same amount of work to move a charge to a point anywhere on the
surface.
+
+
Q
+
+
+
R
+
+
+
+
+
+
+
19
In addition, if we placed another charge on the surface of the sphere, it would take no
work to move it around on the sphere. In order for there to be a potential difference
between two points, there must be an electric field to do work against. The electric field
is constant anywhere on the surface of the sphere.
If we draw imaginary concentric spheres around a positive charge, each sphere will be an
equipotential surface.
Equipotential
Surfaces
E
Note that equipotential lines are always perpendicular to the electric field lines in the
region. We would have to do work to move a charge between equipotential lines, but not
along an equipotential line.
Equipotential
Surfaces
-5V
C
B A
0V
E
+5V
Example 2
In the figure above, three equipotential lines are drawn around a charge. Line A is at a
potential of +5 V, line B is at a potential of 0 V, and line C is at – 5 V.
(a) Is the charge positive or negative? Justify your answer.
A charge of 3 μC is placed on line C.
(b) How much work is required to move the charge along line C for one circumference,
returning it to its starting point?
(c) How much work is required to move the charge along line C for half the
20
circumference?
(d) How much work is required to move the charge along the radius of the circles from
line C to line A?
Solution
(a) The charge in the center of the circles is positive, since the electric field lines point
outward away from the charge. We can also see that the charge is positive because the
equipotential lines become more positive as they approach the charge.
(b) It takes no work to move the charge around the circumference of circle C, since the
entire circle is an equipotential line.
(c) It takes no work to move the charge for any distance along the equipotential line C.
(d) W AC  qV A  VC   3x10 6 C  5V   5V   3x10 5 J
If we have two charged metal spheres of unequal size, we can transfer charge between
them by connecting a wire from one of them to another:
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
The charge will redistribute itself so that both spheres will be at the same potential, but
not the same charge. If the two spheres are connected by a wire, they are essentially one
surface which is at one potential, or an equipotential surface.
Capacitors and Dielectrics
We can create a uniform electric field in a region of space by taking two metal plates,
setting them parallel to each other and separating them by a distance d, and placing a
voltage V (like from a battery) across the plates so that one of the plates will be positive
and the other negative.
21
++++++++++++++++++++++
V
E
d
-------------------------------
The positive charges on the top plate will line up uniformly with the negative charges on
the bottom plate so that each positive charge lines up with a negative charge directly
across from it. This arrangement of charges creates electric field lines which are directed
from the positive charges to the negative charges and are uniformly spaced to produce a
uniform (constant) electric field everywhere between the plates. Conducting plates which
are connected this way are called a capacitor. Capacitors are used to store charge and
electric field in a circuit which can be used at a later time.
q
The capacitance of the plates is defined as C  , where q is the charge on one of the
V
plates, and V is the voltage across the plates. The unit for capacitance is the coulomb/volt,
or farad.
One farad is a very large capacitance, so we often use microfarads (F), or 10-6 F. The
capacitance of a capacitor is proportional to the area of each plate and inversely
proportional to the distance between the plates. In symbols,
A
.
d
The constant of proportionality which makes the proportion above into an equation is o,
which is called the permittivity of free space and is equal to 8.85 x 10 – 12 C2 / Nm2. The
permittivity constant gives us an indication of how well space holds an electric field. The
equation for a parallel plate capacitor is
 A
C o
d
Note that the capacitance of a capacitor ultimately only depends on its geometric
dimensions, like area and distance between the plates. This is true for a capacitor of any
shape.
C
We can also change the capacitance of a capacitor by changing the material between the
plates. If we fill the space between the plates with oil or plastic, the capacitance will
22
increase. The oil or plastic is called a dielectric. Only capacitors with air or a vacuum
between the plates are included on the AP Physics B exam.
Example 3
A capacitor has a capacitance C0. What is the effect on the capacitance if the
(a) area of each plate is doubled?
(b) distance between the plates is halved?
Solution
(a) If the area of each plate is doubled (assuming the plates have equal area), we can find
the new capacitance by
(2) A
C
 2C 0 .
d
Twice the area gives twice the capacitance.
(b) If the distance between the plates is halved, that is, the plates are brought closer
together, the new capacitance is
A
C
 2C 0 .
1
d
2
Bringing the plates closer together increases the capacitance, in this case by a factor of
two.
The electrical energy stored in a capacitor can be found by relating its charge q, voltage
V, and capacitance C:
1
q2 1
CV 2 
 qV
2
2C 2
The electric field, voltage, and distance between the plates are related by the equation
V
E
d
UE 
It follows from this equation that the unit for electric field is a volt/meter, which is
equivalent to a newton/coulomb.
These equations are utilized in the review questions that follow.
CHAPTER 19 REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. Electric potential
(A) is a vector quantity.
(B) is proportional to the work done in
an electric field.
(C) is always equal to the electric field.
(D) is zero when a charge is in an
electric field.
(E) is measured in N/C.
23
Questions 2 – 4:
A hollow metal sphere has a radius R
and a charge Q placed on it.
electric field at their surfaces
(D) the spheres will be oppositely
charged
(E) all of the above are true.
2. The electric field inside the sphere is
(A) zero
KQ
(B)
r
KQ
(C) 2
r
KQ
(D) 2
R
KQ
(E)
R
3. The electric potential at the surface of
the sphere is
(A) zero
KQ
(B)
r
KQ
(C) 2
r
KQ
(D) 2
R
KQ
(E)
R
6. Equipotential lines are always
(A) parallel to electric field lines
(B) perpendicular to electric field lines
(C) perpendicular to charged surfaces
(D) circular
(E) positive
Questions 7 - 10
Two parallel conducting plates each of
area 0.004 m2 are separated by a distance
of 0.001 m. A 9 V battery is connected
across the plates.
7. The electric field between the plates is
(A) 9000 V/m
(B) 900 V/m
(C) 9 V/m
(D) 0.009 V/m
(E) 0.00011 V/m
8. The capacitance of the parallel plates
is
(A) 9.0 x 109 F
(B) 4.0 x 10-12 F
(C) 6.3 x 10-6 F
(D) 3.5 x 10-11 F
(E) 7.0 x 1011 F
4. The electric potential at the center of
the sphere is
(A) zero
KQ
(B)
r
KQ
(C) 2
r
KQ
(D) 2
R
KQ
(E)
R
5. Two unequally sized metal spheres
are each charged. A wire is connected
from one sphere to the other. When the
wire is removed,
(A) the spheres will be equally charged
(B) the spheres will have the same
potential
(C) the spheres will have the same
9. The charge on one of the plates is
(A) 3.2 x 10-6 C
(B) 2.1 x 10-9 C
(C) 4.4 x 10-7 C
(D) 5.2 x 10-6 C
(E) 9.0 x 10-10 C
10. If the distance between the plates is
doubled and the area of each plate is
doubled, which of the following is true?
(A) Both the electric field and the
capacitance is doubled
(B) Both the electric field and the
capacitance is quadrupled
24
(C) the electric field is halved and the
capacitance is unchanged
(D) the electric field is halved and the
capacitance is doubled
(E) neither the electric field nor the
capacitance is changed.
Free Response Question
Directions: Show all work in working the following question. The question is worth 15
points, and the suggested time for answering the question is about 15 minutes. The parts
within a question may not have equal weight.
1. (15 points)
+Q
+Q
d
-Q
-Q
Four charges, two positive and two negative, are arranged at the corners of a square of
sides d. Give all answers in terms of given quantities and fundamental constants.
(a) Determine the electric potential at the center of the square.
(b) Determine the magnitude and direction of the electric field at the center of the square.
Eight more positive and negative charges are added to the top and bottom of the square as
shown below. The lines of charge approximate a uniform electric field between them.
+10Q
d
-10Q
(c) On the diagram of the lines of charge above, sketch several electric field lines
between the charges.
(d) The electric field between the charges is E. Write an expression for the capacitance of
the system in terms of the given quantities and fundamental constants.
25
ANSWERS AND EXPLANATIONS TO REVIEW QUESTIONS
Multiple Choice
1. B
The higher the potential difference between two points, the more work is required to
move a charge between the two points.
2. A
All of the charge is on the outside of the conducting sphere, so there’s no charge or
electric field inside the sphere
3. E
V 
KQ KQ

at the surface of the sphere.
r
R
4. E
Since there is no electric field inside the sphere, we don’t have to do anymore work after
moving our charge past the surface of the sphere. In other words, the potential from the
surface of the sphere to the center doesn’t change, because there is no electric field to
work against.
5. B
Since the spheres are not the same size they will not hold the same amount of charge, but
since they were in contact with one another, they will be at the same potential.
6. B
Equipotential lines and electric field lines must be perpendicular to each other so that no
work will be done as a charge is moved along an equipotential line.
7. A
E
V
9V
V

 9000
d 0.001m
m
8. D

C2 
 8.85 x10 12
 0.004 m 2
2 
Nm 
 A
C o 
 3.5 x10 11 F
d
0.001 m
9. A
q  CV  3.5x10 11 F 9V   3.2 x10 10 C


10. C
V
E
2d
and C 


o A
d

 o 2A
2d
26
Free Response Question Solution
(a) 3 points
KQ  KQ  KQ  KQ  KQ
Vo  




0
r
r
r
r
r
(b) 6 points
+Q
+Q
r
d
-Q
-Q
The net electric field is downward by symmetry, and only includes the vertical
components of the electric field vectors from each of the charges. The y-components of
the electric field vectors are 45º from the diagonals of the square.
 KQ 
E  4  2 
 r 


2
KQ
  2 2 
2
2
2 
 d  d 
  2   2 








(c) 3 points
+10Q
E
d
-10Q
(d) 3 points
q 10Q
C 
V
Ed
27