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H3 Electric Potential (23.1-5)
Due: 11:59pm on Monday, September 21, 2015
You will receive no credit for items you complete after the assignment is due. Grading Policy
Electric Potential Energy of Three Point Charges
Description: Calculate the electric potential energy of three identical charges at the corners of an equilateral
triangle.
Part A
Three equal point charges, each with charge 1.70 µC , are placed at the vertices of an equilateral triangle whose
sides are of length 0.350 m . What is the electric potential energy U of the system? (Take as zero the potential
energy of the three charges when they are infinitely far apart.)
Use ϵ 0 = 8.85×10−12
2
C
for the permittivity of free space.
N⋅m2
Hint 1. How to approach the problem
Use the equation for the electric potential energy between two point charges to calculate the energy for each
interaction between two of the three point charges. The sum of these energies will be the total electric
potential energy. Be careful to avoid double counting.
Hint 2. Find the electric potential energy of one pair
Assume that one charge is interacting with a second charge, ignoring any effects from the third charge. What
is the electric potential energy U 12 for this single interaction?
Express your answer in joules to three significant figures.
Hint 1. Electric potential energy of a pair of charges
Recall that the electric potential energy U between two charges q 1 and q 2 separated by a distance r
is given by the formula
U=
1 q1 q2
.
4πϵ 0 r
ANSWER:
= 7.42×10−2
U 12 =
J
Since the charges are at the vertices of an equilateral triangle, each pair of charges will be 0.350 m
apart, no matter which two charges are selected. This fact, coupled with the fact that the charges are
identical, means that all pair interactions are identical.
Hint 3. How many interactions are there?
How many pair interactions are there for the three charges?
Hint 1. Double counting
It is important to keep in mind that a pair of charges can interact only once, so if the first charge is
interacting with the second charge for one pair, the interaction of the second charge with the first
charge cannot also be used, since the pair has already been counted.
ANSWER:
3
ANSWER:
U=
= 0.223
J
The potential energy is usually written
U = ∑ i<j
qi qj
.
4πϵ 0 rij
This means that all pairs of charges (1-2, 1-3, and 2-3) will interact, but no charge can interact with itself (
i = j ), nor can any pair be counted twice as a result of the condition i < j for all possible pairs. For example,
i = 1, j = 2 will be counted, while i = 2, j = 1 will not.
Exercise 23.1
Description: A point charge with a charge q_1=## mu C is held stationary at the origin. A second point charge with
a charge q_2=## mu C moves from the point x= ## m, y=0 to the point x= ## m, y= ## m . (a) How much work is
done by the...
A point charge with a charge q 1 = 2.80 µC is held stationary at the origin. A second point charge with a charge q 2 =
-4.10 µC moves from the point x = 0.170 m , y = 0 to the point x = 0.230 m , y = 0.260 m .
Part A
How much work is done by the electric force on q 2 ?
ANSWER:
= -0.310 J
Electric Potential Ranking Task
Description: Short conceptual problem involving electrical potentials of point charges. (ranking task)
In the figurethere are two point charges, +q and −q. There
are also six positions, labeled A through F, at various
distances from the two point charges. You will be asked about
the electric potential at the different points (A through F).
Part A
Rank the locations A to F on the basis of the electric potential at each point. Rank positive electric potentials as
higher than negative electric potentials.
Rank the locations from highest to lowest potential. To rank items as equivalent, overlap them.
Hint 1. Definition of electric potential
The electric potential surrounding a point charge is defined by
V = kqr ,
where q is the source charge creating the electric potential and r is the distance between the source charge
and the point of interest. If more than one source is present, determine the electric potential from each source
and sum the results.
Hint 2. Conceptualizing electric potential
Because positive charges create positive electric potentials in their vicinity and negative charges create
negative potentials in their vicinity, electric potential is sometimes visualized as a sort of "elevation." Positive
charges represent mountain peaks and negative charges deep valleys. In this picture, when you are close to
a positive charge, you are "high up" and have a higher positive potential. Conversely, near a negative charge,
you are deep in a "valley" and have a negative potential. The utility of this picture becomes clearer when we
begin to think of charges moving through a region of space containing an electric potential. Just as particles
naturally roll downhill, converting gravitational potential energy into kinetic energy, positively charged particles
naturally "roll downhill" as well, toward regions of lower electric potential, converting electrical potential
energy into kinetic energy.
ANSWER:
Exercise 23.23
Description: A uniform electric field has magnitude E and is directed in the negative x direction. The potential
difference between point a (at x= x_a) and point b (at x= x_b) is V. (a) Which point, a or b, is at the higher potential?
(b) Calculate the...
A uniform electric field has magnitude E and is directed in the negative x direction. The potential difference between
point a (at x = 0.70 m ) and point b (at x = 0.85 m ) is 340 V .
Part A
Which point, a or b, is at the higher potential?
ANSWER:
a
b
Part B
Calculate the value of E .
Express your answer using two significant figures.
ANSWER:
E=
= 2300
N/C
Part C
A negative point charge q
electric field.
= −0.200 µC is moved from b to a. Calculate the work done on the point charge by the
Express your answer using two significant figures.
ANSWER:
W=
= −6.8×10−5
J
Potential Difference and Potential near a Charged Sheet
Description: Find the potential difference between two points near a charged sheet. Find the potential for a point
near a charged sheet for different choices of reference points (of zero potential).
Let A
= (x 1 , y1 ) and B = (x2 , y2 ) be two points near and on the same side of a charged sheet with surface charge
σ
⃗
density +σ. The electric field E due to such a charged sheet has magnitude E =
everywhere, and the field points
2ϵ
0
away from the sheet, as shown in the diagram.
Part A
= V A − V B between points A and B?
Express your answer in terms of some or all of E , x 1 , y1 , x 2 , and y2 .
What is the potential difference V AB
Hint 1. Formula for potential difference
⃗
The formula for the potential difference V AB
is
= V A − V B between two points A and B in an electric field E ⃗
A
V AB = − ∫ B E⃗ ⋅ dl ⃗,
where the line integral can be taken along any path ℓ from B to A.
Hint 2. Calculating the line integral
The line integral from B = (x 2 , y2 ) to A
independent of x and y), is given by
= (x 1 , y1 ) of C ⃗ = C x ^ı + C y ^ȷ , a constant vector field (i.e.,
∫ B C ⃗ ⋅ dℓ ⃗ = ∫ x21 Cx dx + ∫ y 21 Cy dy
A
x
y
= C x (x 1 − x 2 ) + C y ( y 1 − y 2 ) .
ANSWER:
V AB = V A − V B =
Also accepted:
Note that the expression E(y2 − y1 ) will not yield the correct potential if you apply it to two points on opposite
sides of the sheet. For example, the expression does not indicate that two points on opposite sides of the sheet
and the same distance from it are at the same potential (V AB = V A − V B = 0 ), which is clear from the
symmetry of the situation. If you take care in carrying out the integration to observe the change in the direction
of the electric field as you pass from one side of the sheet to the other, you will find that the potential difference
between A and B is actually given by
V AB = E(|y2 | − |y1 |) .
Part B
If the potential at y = ±∞ is taken to be zero, what is the value of the potential at a point V A at some positive
distance y1 from the surface of the sheet?
Hint 1. How to approach the problem
Substitute appropriate values for V B and y2 in the equation V A
ANSWER:
∞
−∞
0
−E ⋅ y1
− V B = E(|y2 | − |y1 |).
Part C
Now take the potential to be zero at y
distance y1 from the sheet?
= 0 instead of at infinity. What is the value of V A at point A some positive
Hint 1. How to approach the problem
Substitute appropriate values for V B and y2 in the equation V A
− V B = E(|y2 | − |y1 |).
ANSWER:
∞
−∞
0
−E ⋅ y1
Note that the potential is zero everywhere on the sheet, that is, at every point whose y coordinate is zero. You
always have the freedom to choose a convenient location and reference potential with respect to which other
potentials are measured, since it is potential differences and not absolute potentials that actually matter when
one is doing something with charges in the real world. Potentials, however, are a useful calculational and
bookkeeping tool. For example, if there were four points of interest in an electrical unit, there would be six
possible potential differences, so it would be easier to keep track of the four potentials corresponding to the four
points instead of working with potential differences.
For the case of a charged sheet, it is clear that choosing the potential at the sheet to be zero is a more
convenient choice than choosing the potential to be zero far away from the sheet. In this way, the potentials of
points near the sheet remain finite. The opposite is true for a point charge.
Exercise 23.26
Description: A total electric charge of q is distributed uniformly over the surface of a metal sphere with a radius of r.
The potential is zero at a point at infinity. (a) Find the value of the potential at r_1 from the center of the sphere. (b)
Find the value...
A total electric charge of 4.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 24.0 cm .
The potential is zero at a point at infinity.
Part A
Find the value of the potential at 60.0 cm from the center of the sphere.
ANSWER:
V =
Part B
= 59.9
V
Find the value of the potential at 24.0 cm from the center of the sphere.
ANSWER:
V =
= 150
V
Part C
Find the value of the potential at 11.0 cm from the center of the sphere.
ANSWER:
V =
= 150
V
Electric Fields and Equipotential Surfaces
Description: Find the work done to move a unit charge from and to given points on a diagram showing equipotential
surfaces, and compare the magnitude of the electric field at these points.
The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1 V increments.
Part A
What is the work done by the electric force to move a 1 C charge from A to B?
Express your answer in joules.
Hint 1. Find the potential difference between A and B
What is the potential difference V A
− V B between point A and point B?
Express your answer in volts.
Hint 1. Equipotential surfaces
Recall that an equipotential surface is a surface on which the electric potential is the same at every
point.
ANSWER:
VA − VB = 0 V
Hint 2. Potential difference and work
Recall that the electric potential energy difference between any two points is equal to the negative of the work
done by the electric force as a charged object moves between those two points. If we combine this with the
relationship between electric potential energy and electric potential we have:
WA→B, by electric force = −∆V A→B Q ​.
ANSWER:
0
J
Part B
What is the work done by the electric force to move a 1 C charge from A to D?
Express your answer in joules.
Hint 1. Find the potential difference between A and D
What is the potential difference V D
− V A between point A and point D?
Express your answer in volts.
ANSWER:
V D − V A = -1 V
Hint 2.
Potential difference and work
Recall that the electric potential energy difference between any two points is equal to the negative of the work
done by the electric force as a charged object moves between those two points. If we combine this with the
relationship between electric potential energy and electric potential we have:
WAD, by electric force = −∆V AD Q
ANSWER:
1
J
Part C
The magnitude of the electric field at point C is
Hint 1. Electric field and equipotential surfaces
Since the diagram shows equal potential differences between adjacent surfaces, equal amounts of work are
done to move a particular charge from one surface to the next adjacent one. It follows then that if the
equipotentials are closer together, the electric force does the same amount of work in a smaller displacement
than if the equipotentials were farther apart. Therefore, the electric force, as well as the corresponding
electric field, has a larger magnitude.
ANSWER:
greater than the magnitude of the electric field at point B.
less than the magnitude of the electric field at point B.
equal to the magnitude of the electric field at point B.
unknown because the value of the electric potential at point C is unknown.
Exercise 23.30
Description: An infinitely long line of charge has linear charge density lambda. A proton (mass 1.67*10^-27 kg,
charge +1.60*10^-19 C) is r from the line and moving directly toward the line at v. (a) Calculate the proton's initial
kinetic energy. (b) How...
An infinitely long line of charge has linear charge density 4.00×10−12 C/m . A proton (mass 1.67× 10−27
+1.60× 10−19 C ) is 18.0 cm from the line and moving directly toward the line at 3.20×103 m/s .
Part A
Calculate the proton's initial kinetic energy.
Express your answer with the appropriate units.
ANSWER:
Ki =
= 8.55×10−21
Part B
How close does the proton get to the line of charge?
kg , charge
Express your answer with the appropriate units.
ANSWER:
= 8.56
d=
Also accepted:
= 8.56
,
= 8.56
,
= 8.56
Exercise 23.43
Description: In a certain region of space, the electric potential is V( (x, y, z) ) = Axy - Bx^2 + Cy, where A, B, and C
are positive constants. (a) Calculate the x-component of the electric field. (b) Calculate the y-component of the
electric field. (c)...
In a certain region of space, the electric potential is V (x, y, z)
constants.
Part A
Calculate the x -component of the electric field.
Express your answer in terms of the given quantities.
ANSWER:
Ex =
Part B
Calculate the y-component of the electric field.
Express your answer in terms of the given quantities.
ANSWER:
Ey =
Part C
Calculate the z -component of the electric field.
Express your answer in terms of the given quantities.
ANSWER:
= Axy − Bx 2 + Cy, where A, B , and C are positive
Ez = 0
Part D
At which point is the electric field equal to zero?
ANSWER:
x = 0, y = 0, z = 0
x = −C/A , y = 0 , z = −2BC/A2
x = −C/A , y = −2BC/A2 , z = C/A
Problem 23.68
Description: A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is
distributed uniformly along the rod. (a) Calculate the potential at the center of curvature of the arc if the potential is
assumed to be zero at...
A thin insulating rod is bent into a semicircular arc of radius a , and a total electric charge Q is distributed uniformly along
the rod.
Part A
Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
V =
Also accepted:
,
± Equipotential Surfaces in a Capacitor
Description: ± Includes Math Remediation. Find the work done by a charge moving along an equipotential surface,
and use that idea to find the distance between the charged plates in a parallel-plate capacitor such that there is a
given potential with a given electric field between the plates.
Part A
Is the electric potential energy of a particle with charge q the same at all points on an equipotential surface?
Hint 1. Formula for electric potential energy
For a particle with charge q at potential V , the electric potential energy is qV .
ANSWER:
Yes
No
For a particle with charge q on an equipotential surface at potential V , the electric potential energy has a
constant value qV .
Part B
What is the work required to move a charge around on an equipotential surface at potential V with constant speed?
Hint 1. A formula for work
The total work done on an object is equal to the change in its energy (potential + kinetic).
ANSWER:
Work = 0
Since the speed of the charge is constant as it moves along the equipotential surface, and the electric potential
energy is constant on that surface, there is no change in the total energy of the charge. This also means that no
work is done by the charge as it moves along the equipotential surface.
Part C
What is the work done by the electric field on a charge as it moves along an equipotential surface at potential V ?
Hint 1. Work done by an electric field on a charged particle
The force due to an electric field is a conservative force. As such, the work done by such a force is equal to
the change in the potential energy of the particle it is acting on.
ANSWER:
Work done by the electric field = 0
Just as in Part B, since there is no change in the electric potential energy, no work is done by the electric field
as the charge moves along the equipotential surface.
Part D
⃗
⃗
The work W E done by the uniform electric field E in displacing a particle with charge q along the path d is given by
WE = qE ⃗ ⋅ d ⃗ = q|E ⃗||d |⃗ cos θ ,
⃗
⃗
where θ is the angle between E and d . Since in general, E ⃗ is not equal to zero, for points on an equipotential
surface, what must θ be for W E to equal 0?
Express your answer in radians.
ANSWER:
θ = 1.57 rad
You have shown that equipotential surfaces are always perpendicular to the electric field at their surface.
Now assume that a parallel-plate capacitor is attached across the terminals of a battery as shown in . The electric field E
in the region between the plates points in the negative z
direction, from higher to lower voltage.
Part E
Find the electric potential V (x, y, z) at a point X
system O = (0, 0, 0) is at potential 0.
= (x, y, z) inside the capacitor if the origin of the coordinate
Express your answer in terms of some or all of the variables E , x , y, and z .
Hint 1. The relation between electric potential and the electric field
⃗
ield E in a region of space is
V = − ∫ E⃗ ⋅ dℓ ,⃗
ℓ
⃗
where the line integral may be taken along any path ℓ.
Hint 2. Expressing an infinitesimal length element
In general, a small length vector along the path of choice can be written as
dℓ ⃗ = dx ^ı + dy ^ȷ + dz k^ .
Substitute this expression into the integral for V .
Hint 3. Analysis of the equation
^ are perpendicular (where w
^ is one of the Cartesian coordinate axes), then
Recall that if E ⃗ and w
E⃗ ⋅ w
^ = 0 , since θ = π/2. According to the setup of this part, only one of the directions (^ı , ^ȷ , k^ ) will not be
perpendicular to the electric field as defined.
ANSWER:
V (x, y, z) =
Therefore, the equation of an equipotential surface at a potenial V 0 is given by
V0
.
E
This is the equation of a plane that is parallel to the plates of the capacitor and perpendicular to the electric
field. In particular, the lower plate, which is at zero potential, corresponds to the surface z = 0.
z=
Part F
What is the distance ∆z between two surfaces separated by a potential difference ∆V ?
Express your answer in terms of E and ∆V .
Hint 1. How to approach the problem
Use the equation you found in Part E to find equations that represent the potentials V 1 and V 2 of the planes
located at z 1 and z 2 . Use these expressions to find an equation for ∆z.
ANSWER:
∆z =
Problem 23.73
Description: Electric charge is distributed uniformly along a thin rod of length a, with total charge Q. Take the
potential to be zero at infinity. Find the potential at the following points: (the figure ). (a) Find the potential at the point
P, a distance x to...
Electric charge is distributed uniformly along a thin rod of length a , with total charge Q . Take the potential to be zero at
infinity. Find the potential at the following points: (the figure ).
Part A
Find the potential at the point P, a distance x to the right of the rod.
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
V =
Also accepted:
,
Part B
Find the potential at the point
R , a distance y above the right-hand end of the rod.
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
V =
Also accepted:
,
Part C
In part A, what does your result reduce to as x becomes much larger than a?
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
V =
Also accepted:
,
Part D
In part B, what does your result reduce to as y becomes much larger than a?
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
V =
Also accepted:
,
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