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The electric field between the two plates shown below has a
magnitude of 500N/C. The plates are separated by a distance
d of 2 cm. What is the potential difference VBA between the
two plates? What is the speed of a proton released from plate
B just before it hits plate A if the system is in a vacuum?
(assume that the surface area of the plates is much larger
than the distance between the plates.)
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Question 1
Question 2
Question 3
Question 4
Question 5
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Question 6
Question 7
Question 8
Question 9
Question 10
Paired Question
1. Which of the following physics principles
is the best to use to solve this problem?
A: Charge conservation principle
B: The principle of conservation of
mechanical energy
C: Gauss’ Law
Choice: A
Incorrect
There is no exchange of charge
in this problem.
Choice: B
Correct
Mechanical energy will be
conserved in this system.
Choice: C
Incorrect
This law is used to find the
electric field, which is of no
interest to us here.
2. Which plate is at the higher
potential?
A: A
B: B
C: They are both at the same potential.
Choice: A
Incorrect
Potential decreases in the
direction of the electric field E,
which is directed from the
positively charged plate to the
negatively charged plate.
Choice: B
Correct
The positive plate has a higher
potential.
Choice: C
Incorrect
The charge separation will
cause a potential difference
between the two plates.
3. What is meant by an electric potential
difference VBA between two points?
A: It is the electric potential energy with respect to
infinity of a unit positive charge.
B: It is the work that must be done to move a unit
charge slowly from b to a against an electric
force.
C: It is the negative of the work done by the
electric forces in taking an object from a given
reference point to a point p.
Choice: A
Incorrect
Electric potential energy is not the same as
electric potential.
Electric potential is electric potential energy per
unit charge associated with a test charge q at
that point.
Choice A is the electric potential energy where
one has placed the reference point at infinity.
Choice: B
Correct
This is expressed
mathematically as:
W UB  UA

 (VB  VA )  VBA
q
q

Choice: C
Incorrect
This is the electric potential
energy at point p.
4. When released, what type of path will
the proton follow?(assume that the electric
field between the plates is uniform)
A: Parabolic
B: Linear
C: Oscillatory
Choice: A
Incorrect
The proton is not shot perpendicular
to the electric field so the path will not
be parabolic.
Choice: B
Correct
This is similar to an object in a free
fall in a gravitational field.
Choice: C
Incorrect
The charge will move in the direction of
the electric field, which has only one
component of direction. The path will be
linear.
5. As the proton moves through the
potential difference, what kind of
energy does it lose?
A: Gravitational energy (GPE)
B: Kinetic energy (K)
C: Electric potential energy (EPE)
Choice: A
Incorrect
The gravitational potential energy
of this system is much less than
the electric potential energy and
can be ignored.
Choice: B
Incorrect
Initially, there is no kinetic
energy. The proton gains kinetic
energy as it speeds up from rest.
Choice: C
Correct
It loses electrical potential
energy, which is the analog of
gravitational potential energy
when a ball is dropped from the
roof of a building.
6. What is the change in electric
potential energy UBA equal to, as the
proton moves between the plates?
A:
Ed
B:
qE
C:
qVBA


UB UA  UBA

Choice: A
Incorrect
This is the magnitude of the
potential difference.
Choice: B
Incorrect
This is the electrical force.
Choice: C
Correct
If the same reference point is used, potential
energy U and potential difference V are
related by the following equation:
U  qV
…and since potential energy can be defined
as the capacity for doing work, We see that:
W  UA  UB   UAB  UBA
UBA  q(VB  VA )  qVBA
7. As the proton moves between the
charged plates, it gains which type of
energy?
A: gravitational potential energy
B: kinetic energy
C: both kinetic and gravitational potential
energy
Choice: A
Incorrect
Changes in gravitational potential
energy can be ignored, because the
gravitational interactions involved are
much weaker than the electrical
interaction.
Choice: B
Correct
This proton will hit plate A with
some speed v. Since it starts at
rest and gains speed it also gains
kinetic energy.
Choice: C
Incorrect
Changes in gravitational potential
energy can be ignored. Notice
that the proton will gain kinetic
energy as it speeds up from rest.
8. Which of the following equations describes the
conservation of mechanical energy for this
system?
1
A: qE  mv2
2
1 2
B: VBA  mv
2
1 2
C: qVBA  mv
2
Choice: A
Incorrect
qE has the units of force, not energy.
Using the law of conservation of energy in this
situation, we should equate the sum of the
kinetic energy and electric potential energy of
the proton before it is released, to the sum of
the kinetic energy and electric potential energy
just before it hits the negatively charged plate.
Choice: B
Incorrect
VBA is potential difference not
potential energy. We must have
units of energy on each side of
the equality.
Choice: C
Correct
There is no
initial velocity.
0
1
1
2
2
mvi  qVB  mv f  qVA
2
2
1
2
qVB  qVA  q(VB  VA )  qVBA  mv f
2
1
1 2
2
qVBA  mv f  mv
2
2
9. From the information given in the original
problem statement, what is magnitude of the
potential difference VBA between the plates?
A: 1000 V
B: 10V
C: 250V
Choice: A
Incorrect
Make sure to convert the distance between
the plates to meters.
VBA  Ed  (500N/C)(0.02m) 10J/C 10V
Choice: B
Correct
VBA  Ed  (500N/C)(0.02m) 10J/C 10V
Choice: C
Incorrect
Try Again using the
following equation:
VBA  Ed

10. Solve the equation found in
question 8 for the speed of the
proton v.
answer
Final
Answer:
1 2
qVBA  mv
2
2qVBA
 v2
m
2qVBA
v
m
2(6.02x1019 C)(10.0V)
v
(1.67x1027 kg)
v  8.5x104 m/s