Download Physics E1ax Solutions: Assignment for Feb. 3 – Feb. 10

Physics E1bx
Physics E1ax Solutions: Assignment for Feb. 3 – Feb. 10
Homework #1: Electric Potential, Coulomb’s Law, Equipotentials
After completing this homework assignment, you should be able to…
• Understand the difference between electric potential V and electric potential energy Uelec, and
how they are related.
• Understand how objects move when a gradient is present.
• Calculate force from electric potential using the idea of a gradient.
• Apply the superposition principle to find potential and force.
• Use equipotential surfaces to gain information about the potential, potential gradient, and force at
a given location.
• Calculate the electric force between point charges using Coulomb’s law.
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1. Conceptual Questions. (1 pt)
i)
What is the difference between electric potential and electric potential energy?
Electric potential energy Uelec:
The electric force caused by any collection of charges at rest is a conservative force. The work W
done by the electric force on a charged particle moving in a potential gradient can be represented
by the change in the potential energy Uelec.
Electric potential V:
Potential V is potential energy per unit charge and is a scalar field defined at all points in space.
The potential difference between two points equals the amount of work that would be required to
move a unit positive test charge (i.e. a charge of 1 C) between those points. The potential due to
a quantity of charge can be calculated by summing (if the charge is a collection of point charges).
ii)
What is the difference between electric potential and electric potential gradient?
Electric potential V:
See solution 1a) above.
Electric potential gradient ∆V/∆r:
The potential gradient is a vector that indicates the magnitude and direction of the change in
potential. A gradient in potential makes positive charges move from regions of high potential to
regions of low potential.
2. Putting Everything Together. (2 pts)
Consider the following system. The particle on the left has charge +Q and the particle on the right has
charge –Q, and the two charges are separated by a distance 2r.
A
+Q
-Q
2r
a) If location A is halfway between the two particles, what is the potential V at position A?
The electric potential from a point charge is given by the equation:
𝑞
𝑉=𝑘∗
𝑟
where q is the magnitude of the charge and r is the distance away from the charge. At point A,
the potential due to the left charge is:
𝑘𝑄
𝑉!"#$ !"#$ !!!"#$ = 𝑟
At point A, the potential due to the charge on the right is:
𝑘 ∗ −𝑄
𝑉!"#$ !"#!! !!!"#$ = 𝑟
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Thus, using the superposition principle, we can add these two potentials together in order to find
the total potential V at point A:
𝑉! = 𝑉!"#$ !"#$ !!!"#$ + 𝑉!"#$ !"#!! !!!"#$ 𝑘𝑄 −𝑘𝑄
𝑉! =
+ 𝑟
𝑟
𝑽𝑨 = 𝟎 𝑽
b) What would the force F be on a test charge q0 located at position A?
The electric force on a test charge q0 is given by Coulomb’s law:
𝑄𝑞!
|𝑭| = 𝑘 !
𝑟
At point A, the force is purely in the x-direction. The force from the left charge on a test charge
q0 at point A is directed to the right and has magnitude:
𝑄𝑞!
|𝑭𝒇𝒓𝒐𝒎 𝒍𝒆𝒇𝒕 𝒄𝒉𝒂𝒓𝒈𝒆 | = 𝑘 !
𝑟
The force from the right charge on a test charge q0 at point A is directed to the right and has
magnitude:
𝑄𝑞!
|𝑭𝒇𝒓𝒐𝒎 𝒓"𝒈𝒉𝒕 𝒄𝒉𝒂𝒓𝒈𝒆 | = 𝑘 !
𝑟
Thus, using the superposition principle, we can add these two forces together in order to find the
net force F on a test particle q0 located at point A:
|𝑭𝒏𝒆𝒕 | = 𝟐𝒌
𝑸𝒒𝟎
𝒓𝟐
, directed to the right
c) If the electric potential at a given point is zero, does it necessarily follow that the electric force
on a positive charge is zero at that point?
As seen in this problem, if the electric potential at a given point is zero, it does not necessarily
follow that the electric force on a positive charge is also zero at that point. Electric force is
dependent on the gradient of the potential, not on the value of the potential. Hence, it is possible
to have a situation where V is zero but force F is nonzero.
d) If the electric force on a positive charge at a given point is zero, does it necessarily follow that
the electric potential is zero at that point?
If the electric force at a given point is zero, it does not necessarily follow that the electric
potential is also zero at that point. Recall that force is equal to charge times the gradient of
potential:
∆𝑉
𝐹! = −𝑞
∆𝑥
∆𝑉
𝐹! = −𝑞
∆𝑦
Since we are considering a positive charge, we know that q must be nonzero. This means that in
order for F to be equal to zero,
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∆𝑉
=0
∆𝑥
∆𝑉
=0
∆𝑦
Now, if the potential gradient is equal to zero, this means that the potential must be equal to a
constant at that point. However, it does not mean that the value of the constant must be equal
to zero. Without any other information, we cannot determine what the value of the constant is. It
may be equal to zero, or it may not.
3. Putting Everything Together. (2 pts)
Y
A small test charge –q0 travels from point X to point Y along the
circular arc as shown.
a) Is the electric potential difference ΔVXY positive, negative, or
zero? Explain.
To solve this problem, we must determine the potential at
point X as well as the potential at point Y. Let the side length
of each square be r. With this, the potential at point X is equal
to
−𝑄!
−𝑄!
𝑉! = 𝑘
+𝑘
3𝑟
5𝑟
𝑘𝑄! 1
1
𝑉! = −
+
𝑟
5 3
𝑘𝑄!
𝑉! = −0.781
𝑟
Similarly, the potential at point Y is equal to
X
-Q0
−𝑄!
5𝑟
5𝑟
𝑘𝑄! 1
1
𝑉! = −
+
𝑟
5 5
𝑘𝑄!
𝑉! = −0.647
𝑟
To find the change in potential, ΔVXY, we subtract the initial potential (VX) from the final
potential (VY):
∆𝑉!" = 𝑉! − 𝑉!
𝑘𝑄!
𝑘𝑄!
∆𝑉!" = −0.647
− −0.781
𝑟
𝑟
𝒌𝑸𝟎
∆𝑽𝑿𝒀 = 𝟎. 𝟏𝟔𝟔
𝒓
So, the electric potential difference ΔVXY is positive. This means that the potential at point Y
is greater than the potential at point X.
𝑉! = 𝑘
−𝑄!
-Q0
+𝑘
b) The test charge is launched from point X with an initial speed v0 and is observed to pass through
point Y. Is the speed of the test charge at point Y greater than, less than, or equal to v0? Explain
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your reasoning.
To answer this question, we must apply conservation of energy. Recall that electric potential
energy is related to electric potential via
𝑈!"!# = 𝑞𝑉
This means that the potential energy at point X is
𝑈! = −𝑞!
−0.781
𝑈! = 0.781
𝑘𝑄!
𝑟
𝑘𝑄! 𝑞!
𝑟
The potential energy at point Y is similarly
𝑈! = −𝑞!
−0.647
𝑈! = 0.647
𝑘𝑄!
𝑟
𝑘𝑄! 𝑞!
𝑟
Now, we can calculate the total energy of the test charge at point X:
1
𝐸! = 𝑈! + 𝑚𝑣! !
2
𝐸! = 0.781
𝑘𝑄! 𝑞! 1
+ 𝑚𝑣! !
𝑟
2
The total energy of the test charge at point Y is:
1
𝐸! = 𝑈! + 𝑚𝑣 !
2
𝐸! = 0.647
Conservation of energy tells us that
0.781
𝑘𝑄! 𝑞! 1
+ 𝑚𝑣 !
𝑟
2
𝐸! = 𝐸!
𝑘𝑄! 𝑞! 1
𝑘𝑄! 𝑞! 1
+ 𝑚𝑣! ! = 0.647
+ 𝑚𝑣 !
𝑟
2
𝑟
2
1
𝑘𝑄
𝑞
!
!
𝑚 𝑣! ! − 𝑣 ! = −0.166
2
𝑟
1
!
!
𝑚 𝑣! − 𝑣 = −𝑞! ∆𝑉!"
2
In order for this equation to be true, v must be greater than v0. So, since the potential energy of
the test charge at point Y is less than its potential energy at point X, the test charge must be
moving faster at point Y. Hence, the speed of the test charge at point Y is greater than v0.
4. Putting Everything Together. (2 pts)
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Open the “Charges and Fields” web page and click “Run Now”:
http://phet.colorado.edu/en/simulation/charges-and-fields
a) Assemble two positive charges and on negative charge in the configuration shown below and
check the box labeled “Show numbers.”
e–
•
•
•
•
HA
HB
Plot a number of equipotential surfaces at equal intervals.
Print the resulting plot.
Find a point on the plot where the potential gradient is very large and determine the force on
a positive charge at that location.
Is there a point where the force would be zero?
Here is my plot:
When the equipotential lines are very close together, it means that the potential gradient is
large. Here is a spot where the gradient is large:
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A positive charge in a potential gradient will feel a force pushing it towards a lower potential.
In this case, the positive charge will feel a force to the left, towards the negative charge.
The force on a charge is equal to zero when the potential gradient is zero, such as a saddle
point, in this case:
b) On the right-hand side, there are yellow balls called “E-Field Sensors.” Place five or so of these
on various equipotential lines.
•
•
•
Print the resulting plot.
What is the E-Field Sensor is “sensing”? Consider the direction that the arrow is
pointing, and look closely at the SI units reported by the sensor.
How does the spacing of the equipotential lines relate to the value of the E-Field Sensor?
Here is my plot with E-field sensors:
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Some things we notice about this plot:
1) The E-field sensors show an arrow, which means they’re measuring a vector quantity. The
applet also reports the magnitude and direction of the red arrow.
2) The magnitude reported by the sensor is in “volts/meter”, which are the same SI units we use
for potential gradient.
3) The arrow is perpendicular to the equipotential curve the sensor is sitting on.
4) The magnitude reported by the sensor is larger when the equipotential curves are closer
together.
Putting all of this information together, the E-field sensor measures a vector related to the
potential gradient. (The name “E-field sensor” also gives us a clue that this measured quantity is
called the “electric field”. We’ll learn more about electric fields later.)
c) Below you’ll see some equipotential surfaces—“equipotential line art.” The charges that
created these equipotential surfaces have been erased.
• Find an arrangement of charges that creates the same set of equipotential lines.
• Create some other example of “art” using charges and equipotential lines? Print the
results.
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The “eyes” and “mouth” are positive charges, while the “nose” is a negative charge.
(You’ll get the same solution if you use the opposite charge everywhere, i.e., all +
become –, and vice versa.) The eyes have to be the same charge because there is a saddle
point between them. The nose and mouth must be oppositely charged since the
equipotential curves flatten out between them. (Same with the eyes and nose, for that
matter.)
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