*
AP PHYSICS B
Electrostatics & Capacitance
Teacher Packet
AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not
involved in the production of this material.
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Electrostatics & Capacitance
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions on electrostatics and capacitance.
Standards
Electrostatics is addressed in the topic outline of the College Board AP* Physics Course
Description Guide as described below.
III. Electricity and Magnetism
A. Electrostatics
1. Charge and Coulomb’s Law
2. Electric Field and Electric Potential (including point charges)
B. Conductors, capacitors, dielectrics
1. Electrostatics with Conductors
2. Capacitors
AP Physics Exam Connections
Topics relating to electrostatics are tested every year on the multiple choice and in most years on
the free response portion of the exam. The list below identifies free response questions that have
been previously asked over electrostatics. These questions are available from the College Board
and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com.
2006
2005
2001
2000
1999
Free Response Questions
Question 3
2006 Form B Question 3
Question 5
2005 Form B Question 3
Question 3
2003 Form B Question 4
Question 3
2002 Form B Question 5
Question 2
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Electrostatics and Capacitance
What I Absolutely Have to Know to Survive the AP* Exam
There are two types of charges, positive and negative. Like charges repel each other, and unlike charges
attract each other. A conductor is a material through which charges flow readily due to a large number of
free electrons, whereas an insulator does not allow charges to flow freely through it. Coulomb’s law, an
inverse square law, describes the force between charges. The electric field is an invisible force field that is
produced by charges and can be felt by other charges. It is not a force, but the potential for a force. Work
must be done to move a charge in an electric field. 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. Capacitors have the capacity to store charge and an electric field between
the plates.
Key Formulas and Relationships
Electric force and electric field are vectors (add using vector addition). The direction for the electric
force is determined by whether the force is attractive or repulsive. The electric field is in the direction of the
electric force that would act on a positive test charge.
kq1q2
1
where k =
2
r
4πε 0
F = q0 E
F=
E=
F
(always true)
q0
E=
kq
(specific for single point charge)
r2
F = electric force [N]
k = electric constant = 9x109 Nm2 / C2
ε0 = permittivity constant = 8.85 x 10-12 C2 / Nm2
q (or Q) = charge [C]
r = distance between charges [m]
E= electric field [N/C] or [V/m]
V= electric potential or potential difference [V]
d = distance between parallel capacitor plates [m]
E =
V
(electric field strength between the plates of a capacitor with voltage V and separation d)
d
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Electrostatics and Capacitance
Electric potential energy and electric potential are scalars, they can be + or -. Include + or – signs since
these quantities have no direction, only magnitude.
U E = q0V =
kq1q2
1 q1q2
=
4πε 0 r
r
UE = electric potential energy [J]
kq
(single point charge)
r
1
kq
q
(configuration point charges)
V =∑ =
∑
r 4πε 0
r
V =
V= electric potential or potential difference [V]
Capacitance
Q
C=
V
ε A
C= 0
d
C = capacitance [F] or [C/V]
1
1
1 Q2
U E = CV 2 = QV =
2
2
2 C
d = distance [m]
A = area of a capacitor plate [m2]
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Electrostatics and Capacitance
Important Concepts
•
•
•
•
•
•
•
•
•
The charge on one electron (or one proton) is called the elementary charge (e) and is equal to
1.6×10-19C. Hence, it takes 6.25×1018 electrons to generate one coulomb of charge.
Charge is either positive (deficiency of electrons) or negative (surplus of electrons). The carrier of
negative charge is the electron, with a charge of – 1.6 × 10-19 C. The proton, carries exactly the
same charge as the electron, but is positive. The neutron carries no charge.
Charge is conserved during any process so any charge lost by one object, must be gained by another
object.
The fundamental law of electrostatics states that like charges (+/+) or (–/–) repel each other while
unlike chages (+/-) or (-/+) attract each other.
Conductors, like metals, are materials that allow charges (electrons) to move freely.
Insulators, like wood or glass, are materials that do not allow charges (electrons) to move freely.
A charged object may transfer some or all of its charges to another object through direct contact
with the other object. This is charging by conduction. The second object gets the same type of
charge as the first object originally had.
A charged object brought “close” to a conductor may induce a charge separation in the other object.
This is know as charge polarization. While polarized, a “grounding” connection can be made to
remove some of the separated charge. This is charging by induction. Any charge given to a
conductor resides on its outer surface and the charges tend to concentrate at sharp corners on the
object. The second object gets the opposite type of charge as the first object originally had.
Insulators can be polarized, but it occurs at the atomic level. When a charged object is brought near
a neutral object, polarization will result in opposite charges being closer than like charges, and
therefore, a net attraction between the objects.Any charge given to an insulator stays in place and is
spread throughout its volume, unlike a conductor where the charges placed inside the conductor will
move to its outer surface so that the electric field inside the conductor is zero (shielding effect).
Electric Force (Coulomb’s Law)
• The force between any two charges is proportional to the magnitude of the charges and inversely
proportional to the square of the distance between the charges and follows the same basic form as
kq q
1
Newton’s law of universal gravitation (inverse square law). F = 12 2 where k =
r
4πε 0
• Forces are vectors and have magnitude and direction. The direction of the electric force is
determined by the law of electrostatics, opposite charges attract and like charges repel. If signs are
used for the charges, a negative result indicates attraction and a positive result indicates repulsion.
• For more than 2 charges, Coulomb’s Law must be used multiple times, and then the resulting forces
must be added using vector addition.
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Electrostatics and Capacitance
Electric Field
• Electric fields are measured in N/C or V/m.
• Electric field is a vector which points in the same direction as the force acting on a positive charge
in the electric field. A negative charge would experience a force in the opposite direction.
Field for + charge
•
•
•
•
•
•
•
•
Field for – charge
Field for electric dipole
Field for two like charges
Electric field lines originate on positive charges and terminate on negative charges.
Electric field lines never intersect.
Where electric field lines are closer together, the field (and the force if a charge were at that
location) is stronger.
Objects with more charge will have more lines originating from (or terminating on) them.
Electric field lines are always perpendicular to the surface of a conductor.
For two opposite charges along a line, the field would never be zero between them because the
fields from each charge point in the same direction.
In order for fields from two charges to cancel, they must be equal and opposite. i.e. you must be
closer to the smaller charge.
A parallel plate capacitor develops a uniform electric field between its charged plates and the
direction of the electic field goes from + to –.
++++++++++++++++
E
----------------------
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Electrostatics and Capacitance
•
When charge is placed on a conductor of radius R, the charge quickly spreads to the outside surface,
so that the electric field inside the conductor is zero (this is a shielding effect). Outside the sphere (r
> R), the electric field behaves as if the sphere is a point charge centered at the center of the sphere,
KQ
that is, Eoutside = 2 . Below is the graph of electric field E vs. distance from the center r for a
r
charged conducting sphere of radius R:
E
r
0
R
Electrostatic Potential Energy and Electric Potential
•
•
•
•
•
•
•
The electrostatic potential energy of a system of charges is a scalar, so we find the EPE or UE for
each pair of charges and add them together. Positive and negative signs do not denote direction.
When charged objects are connected by a wire, they will arrive at the same electric potential V so
charge will flow from one object to the other until they are at the same potential, but not the same
charge.
To find the work done in moving a charge in a system from one point to another, it is usually easiest
to find the energy of the system before and after the move. The difference between the two is the
work done.
Electric potential is a relative quantity, so any point can be defined as the reference (0 V). It is the
difference that is important.
The difference in potential (or electric potential difference) is called voltage.
A volt is a Joule per Coulomb, so the work required to move a charge is ΔVq.
When charge is placed on a conductor, it spreads to the surface until the electric field inside the
conductor is zero. This means that no work will be done to move charges inside the conductor, so all
points on the surface and within the conductor will have the same potential V. Below is the graph of
electric potential V vs. distance from the center r for a charged conducting sphere of radius R:
V
r
R
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Electrostatics and Capacitance
•
•
•
Work is done when pushing a positive test charge against the electric field lines, so electric
potentials increase as you move opposite the field lines. The potential or voltage gets larger towards
positive charges and smaller towards negative charges.
A positive charge will accelerate from higher to lower electric potential, or along the electric field
lines. Negative charges do the exact opposite, accelerating from lower to higher electric potential
against the electric field lines.
Equipotential lines (surfaces) are always perpendicular to electric field lines.
Equipotential
Surfaces
E
•
•
We would have to do work to move a charge between equipotential lines, but not along an
equipotential line.
A parallel plate capacitor develops a uniform electric field between its plates from + to -. The
electric potential increases in a linear fashion from one plate to the other, so if a capacitor has 12 V
across it, there is a 6 V equipotential line or surface from one plate to a point halfway between the
two plates.
+++++++++++++++++++
V
E
d
--------------------------
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Electrostatics and Capacitance
•
•
•
•
•
Q
, where Q is the charge on one of the plates, and V
V
is the voltage across the plates. The unit for capacitance is the coulomb/volt, or farad.
The capacitance of a capacitor depends on its geometrical properties and is proportional to the area
ε A
of each plate and inversely proportional to the distance between the plates. C = 0
d
εo, which is called the permittivity of free space (8.85 x 10 – 12 C2 / Nm2 ) provides an indication of
how well space holds an electric field.
The capacitance of a capacitor also depends on the material between the plates or the dielectric
(such as oil, paper, or plastic), but on the exam the dielectric will be air or vacuum.
The electrical energy stored in a capacitor can be calculated by relating its charge Q, voltage V, and
1
Q2 1
2
capacitance C. U E = CV =
= QV
2
2C 2
The capacitance of the plates is defined as C =
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Electrostatics and Capacitance
Free Response
Question 1 (15 pts)
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 at a distance 2a from the origin.
A. In terms of the given quantities and fundamental constants determine the
i. electric field at the origin
ii. electric potential at the origin
(5 points max)
i. Electric field is a vector. The electric field
at the origin is zero, since a positive test
charge placed at the origin would experience
no net force.
ii. The electric potential is a scalar.
kq
V =Σ
r
kQ kQ
V=
+
a
a
Q
2kQ
2Q
V=
=
or V =
a
4πε 0 a 2πε 0 a
®
1 point for an indication that the electric
field is a vector
1 point for an indication that the electric
field is zero at the origin
1 point for an indication that the electric
potential is a scalar
1 point for using V =
kq
r
1 point for the correct answer in terms of
the given quantities and fundamental
units
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Electrostatics and Capacitance
B. A small negative charge q is now placed at point P. On the diagram above, sketch an
arrow indicating the direction of the
i. electric force acting on the negative charge
ii. electric field at the location of the negative charge
(2 points max)
1 point for an indication that the
electric force is directed to the left
1 point for an indication that the
electric field is directed to the right
C. Determine the magnitude of the electric force at point P in terms of the given
quantities and any fundamental constants.
(5 points max)
1 point for recognizing that the
resultant force depends only on the
horizontal components of the forces
between the charges
1 point for using Coulomb’s law
1 point for recognizing the force
depends on two horizontal
components of the electric forces
since the vertical components cancel
1 point for indentifying and
expressing the horizontal component
of the forces in terms of the given
quantities and fundamental constants
1 point for the correct answer in
terms of the given quantities and
fundamental constants
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Electrostatics and Capacitance
2kQ 2
cos θ
r2
r 2 = a 2 + (2a) 2
F = 2 Fx =
r = a 2 + (2a ) 2
2a
cos θ =
F=
F=
(
a + (2a) 2
2
⎛
2a
⎜
2
2
2
⎜
a 2 + (2a) 2 ⎝ a + (2a)
2kQ 2
)
⎞
⎟
⎟
⎠
4kQ 2 a
(a
2
3
+ (2a) 2 ) 2
D. The negative charge is released from point P and allowed to move freely. Describe
the velocity and acceleration of the negative charge for a long time after it is released.
It is not necessary to calculate these quantities.
(3 points max)
The negative charge q is attracted toward the
midpoint of the two large charges and will
accelerate toward the origin. As it reaches
the origin, it attains maximum velocity and
is then pulled and accelerated back toward
the origin and decreases its speed until it
comes to rest at a distance of 2a from the
origin. Then the charge returns to the origin,
starting the cycle over again and oscillates in
simple harmonic motion about the origin.
®
1 point for the correct description of
the velocity of the charge
1 point for the correct description of
the acceleration of the charge
1 point for indicating that the charge
oscillates about the origin
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Electrostatics and Capacitance
Question 2 (15 pts)
A charge is suspended at rest between two charged parallel plates which are connected to
a 12 V battery. The plates are separated by a distance of 0.05 m.
A. Place a check in one of the blanks below, indicating whether the charge q is positive,
negative, or neutral. Justify your answer.
_____ positive
_____ negative
_____neutral
(2 points max)
1 point for indicating that the charge is
The downward gravitational force is balanced negative
by the upward electric force. Thus, the charge
is attracted to the top positive plate, repelled
1 point for a correct justification
by the bottom negative plate, and must be
negative.
B. On the diagram above, draw an arrow to indicate the direction of the electric force on
charge q.
(1 point max)
Electric force vector is directed up toward
the positive plate.
®
1 point for indicating that the electric
force is directed up toward the positive
plate
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Electrostatics and Capacitance
C. The mass of the charge is 5.0×10-4 kg. Calculate
i. the magnitude of the electric force acting on the charge
(3 points max)
ΣF = 0
FE − FW = 0
1 point for any statement that the net
force on the charge is zero
1 point for any indication that the
electric force must equal the weight
force
FE = FW = mg
⎛ m⎞
FE = ( 5.0 × 10−4 kg ) ⎜ 10 2 ⎟
⎝ s ⎠
FE = 5.0 × 10−3 N
1 point for the correct answer
including correct units and reasonable
number of significant digits
ii. the magnitude of the suspended charge q
(4 points max)
FE = qE
1 point for a correct expression
relating charge to electric force and
electric field
FE
E
V
E=
d
q=
1 point for the correct equation for
electric field
−3
FE d ( 5.0 × 10 kg ) ( 0.05m )
q=
=
12V
V
−5
q = 2.1× 10 C
1 point for correct substitution of
distance between plates and voltage
across plates
1 point for the correct answer or
answer consistent with section i
including correct units and reasonable
number of significant digits
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Electrostatics and Capacitance
D. The same charge q is removed from between the parallel plates and replaced with a
proton of mass 1.67×10-27 kg. If the proton is accelerated from rest with the same
accelerating potential, calculate its speed when it encounters one of the plates.
Indicate which plate (top or bottom) the proton will move toward.
(5 points max)
Proton will move downward toward the
1 point for indicating the correct
direction the proton moves
bottom or negative plate
1 point for a statement of
conservation of energy
U E = KE
1 point for relating the electric
potential energy of the proton to its
charge and the accelerating voltage
1 2
mv
2
2qV
v=
m
qV =
v=
2 (1.6 ×10−19 C ) (12V )
1.67 × 10
−27
kg
= 4.8 ×104
or alternate solution
ΣF = ma
ΣF qE qV
a=
=
=
m
m md
v 2 = v02 + 2ax
1 point for relating the kinetic energy
of the proton to the electric potential
energy
1 point for the correct answer
including correct units and
reasonable number of significant
digits
⎛ qV ⎞
v2 = 2 ⎜
⎟d
⎝ md ⎠
v=
m
s
m
2qV
= 4.8 × 104
m
s
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Electrostatics and Capacitance
Multiple Choice
Questions 1-3
1m
+20 μC
+10 μC
1. A 20 µC and a 10 µC charge are placed 1.0 m apart as shown in the diagram above. The
10 μC charge experiences a force of 1.8 N directed to the right. The 20 μC charge will
experience a force of
A)
B)
C)
D)
E)
0.9 N to the right
0.9 N to the left
1.8 N to the right
1.8 N to the left
3.6 N to the right
D
The charges repel each other with equal and opposite forces.
2. What is the direction of the electric field midway between the 20 μC and 10 μC
charges?
A)
B)
C)
D)
E) The electric field is zero
C
The electric field from the 20 μC directed to the right is twice as strong as the
electric field directed to the left from the 10 μC charge.
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Electrostatics and Capacitance
3. If the 20 µC charge is moved to a distance of 2 m from the 10 µC charge, the new
force on the 10 µC charge will be
A)
B)
C)
D)
E)
0.45 N
0.9 N
1.8 N
3.6 N
7.2 N
A
FE = kq1q2/r2
The distance is doubled. Thus the new Electric Force is 1/4 of its original
value.
Questions 4-6
The figure shows three equipotential lines for a particular charge distribution. Point A lies
on a line of potential 0 volts, point B lies on a line of potential +2 volts, and point C lies
on a line of potential +4 volts.
4. At which point is the electric field the greatest?
A)
B)
C)
D)
E)
Point A
Point B
Point C
The electric field is the same at all points
There is no electric field present
A
The electric field is greatest at point A, because the equipotential lines
are closer together near point A, indicating a strong electric field.
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Electrostatics and Capacitance
5. Which of the following arrows best represents the direction of the electric field vector
at point C?
A)
B)
C)
D)
E)
The electric field vector points from higher potential (more positive) to
lower potential (less positive) so the electric field vector would point
away from C and toward A, perpendicular to the equipotential surface.
C
6. Suppose the distance between the points B and C is 0.4 m. Assuming the electric field
between points B and C is relatively constant, the magnitude of the electric field
strength between points B and C is most nearly
A) 0.5
B) 0.8
C) 1.6
D) 2.0
E) 5.0
E
V
m
V
m
V
m
V
m
V
m
E=
ΔV 4V − 2v
V
=
= 5.0
d
0.4m
m
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Electrostatics and Capacitance
Question 7
1m
+4 μC
-2 μC
7. Two charges of +4 µC and -2µC are placed on a line 1 m apart. At which point could
a small positive charge be placed such that the net force exerted on the small positive
charge is zero?
A)
B)
C)
D)
E)
At a point halfway between the two charges
At a point between the two charges but closer to the -2µC charge
At a point to the left of the +4 µC charge
At a point to the right of the -2 µC charge
None of the above
D
Since the electric force between two charges depends on the magnitude of the
charges and the distance between them, a small positive charge would be
repelled strongly by the + 4 µC charge, and attracted weakly by the - 2 µC
charge, both forces could be equal and opposite if the small positive charge
were far from the + 4 µC charge and close to the - 2 µC charge. The only
location where this is possible is to the right of the - 2 µC charge.
Question 8
A force acts on an electric charge in an electric field as shown.
F
E
8. Which of the following statements must be true?
A)
B)
C)
D)
E)
The charge is positive
The charge is negative
The charge is neutral
The electric field is weak
The electric field is strong
B
Since electric field lines are drawn in the direction a positive charge would
experience a force, and the charge experiences a force to the left, the charge
must be negative.
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Electrostatics and Capacitance
9. Consider two charged spheres of equal size carrying a charge of +8 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?
A)
B)
C)
D)
E)
-4 C
-2 C
-1 C
+1 C
+2 C
E
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 +8 C + -4 C = +4 C, and this is the magnitude of the
equilibrium charge. When they are separated, they divide the charge evenly,
each keeping a charge of +2 C.
10. 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)
B)
C)
D)
E)
uncharged
neutral
positive
negative
equally positive and negative
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.
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Electrostatics and Capacitance
11. The electric field lines are sketched around two charges as shown in the diagram
below. What are the signs of charges A and B respectively?
A)
B)
C)
D)
E)
negative, negative
negative, positive
positive, positive
positive, negative
neutral, neutral
C
Electric field lines originate on positive charges and terminate on negative
charges.
12. In an experiment, it is found that a negatively charged balloon will attract the hair of
all 25 students in the classroom. This proves that all 25 students have
A)
B)
C)
D)
E)
positively charged hair
neutrally charged hair
negatively charged hair
positively or neutrally charged hair
positively or negatively charged hair
D
A negative object will attract both positive and neutral objects (due to
polarization).
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Electrostatics and Capacitance
Questions 13-15
A=0.004 m2
+++++++++++++++++++
12 V
E
0.001 m
--------------------------
13. A capacitor is charged by a 12 V battery as shown in the diagram above. Two parallel
conducting plates each of area 0.004 m2 are separated by a distance of 0.001 m. The
capacitance of the parallel plate capacitor is
A)
B)
C)
D)
E)
7.0×10-11 F
3.5×10-11 F
6.0×10-9 F
9.0×10-9 F
4.0×10-12 F
B
C=
ε0 A
d
=
C2
0.004m 2 )
2 (
Nm
= 3.5 × 10−11 F
0.001m
8.85 × 10−12
14. The charge on one of the plates of the parallel plate capacitor is
A)
B)
C)
D)
E)
2.1×10-10 C
3.0×10-10 C
4.2×10-10 C
5.2×10-10 C
9.0×10-10 C
Q
V
Q=CV= ( 3.5 × 10−11 F ) (12V ) = 4.2 × 10−10 C
C=
C
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Electrostatics and Capacitance
15. If the distance between the plates of the parallel plate capacitor is doubled, and the
area of each plate is quadrupled, which of the following is true?
A)
B)
C)
D)
E)
Both the electric field and the capacitance are quadrupled
Both the electric field and the capacitance are doubled
The electric field is halved and the capacitance is quadrupled
The electric field is halved and the capacitance is doubled
Neither the electric field nor the capacitance is changed
D
E=
4ε A
ε A
V
and C = 0 = 2 0
d
2d
2d
®
Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org