6 4 24 10 10 pm chapter 23 electric potential chapter 23 electric pote
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6/4/24, 10:10 PM
Chapter 23: Electric Potential
Chapter 23: Electric Potential
To understand how points are awarded, read the Grading Policy for this assignment.
Exercise 23.1 - Enhanced - with Solution
A point charge with a charge q₁ = 2.20 μC is held stationary at the origin. A second point charge with a charge q2 = -4.50
μC moves from the point x = 0.160 m, y = 0 to the point x = 0.250 m, y = 0.250 m.
For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Conservation of energy with
electric forces.
Part A
How much work is done by the electric force on 92?
Express your answer in joules.
ANSWER:
J
Exercise 23.8 - Enhanced - with Feedback
Three equal 1.10 μC point charges are placed at the corners of an equilateral triangle with sides 0.800 m long.
Part A
What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are
infinitely far apart.)
Express your answer with the appropriate units.
ANSWER:
U =
Exercise 23.4 - Enhanced - with Feedback
Part A
How much work would it take to push two protons very slowly from a separation of 2.00 x 10-10m (a typical atomic
distance) to 3.00 × 10-15 m (a typical nuclear distance)?
Express your answer in joules.
ANSWER:
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W =
Chapter 23: Electric Potential
J
Part B
If the protons are both released from rest at the closer distance in part A, how fast are they moving when they reach their
original separation?
Express your answer in meters per second.
ANSWER:
v =
m/s
Exercise 23.10 - Enhanced - with Feedback
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint.
Part A
How much work is done by the Coulomb force when the alpha particle moves to the midpoint of one of the sides of the
square?
Express your answer in joules.
ANSWER:
W =
J
Exercise 23.12 - Enhanced - with Feedback
An object with charge q = -7.00×10-9 C is placed in a region of uniform electric field and is released from rest at point A.
After the charge has moved to point B, 0.500 m to the right, it has kinetic energy 4.00×10-7 J.
Part A
If the electric potential at point A is +30.0 V, what is the electric potential at point B?
Express your answer with the appropriate units.
ANSWER:
VB =
Part B
What is the magnitude of the electric field?
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Express your answer with the appropriate units.
ANSWER:
E =
Part C
What is the direction of the electric field?
ANSWER:
from point B to point A
from point A to point B
perpendicular to the line AB
Chapter 23: Electric Potential
Exercise 23.16 - Enhanced - with Feedback
Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 cm. An electron is released from
rest at a point midway between the two charges and moves along the line connecting the two charges.
Part A
What is the speed of the electron when it is 10.0 cm from the +3.00 nC charge?
Express your answer with the appropriate units.
ANSWER:
v =
Exercise 23.19 - Enhanced - with Solution
Two point charges q₁ = 2.30 nC and 92 = -6.20 nC are 0.100 m apart. Point A is midway between them; point B is
0.0800 m from 91 and 0.0600 m from 92. Take the electric potential to be zero at infinity.
For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Potential and potential energy.
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Chapter 23: Electric Potential
Part A
Find the potential at point A.
Express your answer in volts.
ANSWER:
VA
=
Part B
Find the potential at point B.
Express your answer in volts.
ANSWER:
VB =
V
V
0.080 m
B
-0.060 m
+0.050 m→<0.050 m→
91
A
Part C
Find the work done by the electric field on a charge of 2.80 nC that travels from point B to point A.
Express your answer in joules.
ANSWER:
W =
Exercise 23.44
J
In a certain region of space the electric potential is given by V =
B = 8.00 V/m³.
=
+Ax²y – Bxy², where A = 5.00 V/m³ and
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92
4/13 6/4/24, 10:10 PM
Part A
Chapter 23: Electric Potential
Calculate the magnitude of the electric field at the point in the region that has cordinates x = 2.50 m, y = 0.400 m, and
z = 0.
Express your answer with the appropriate units.
ANSWER:
E =
Part B
Calculate the direction angle of the electric field at the point in the region that has cordinates x = 2.50 m, y = 0.400 m,
and z = 0.
Express your answer in degrees.
ANSWER:
0 =
measured counterclockwise from the positive x axis in the xy plane
Exercise 23.28 - Enhanced - with Feedback
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 28.0 cm. The
potential is zero at a point at infinity.
Part A
Find the value of the potential at 45.0 cm from the center of the sphere.
Express your answer in volts.
ANSWER:
V =
V
Part B
Find the value of the potential at 28.0 cm from the center of the sphere.
Express your answer in volts.
ANSWER:
V =
V
Part C
Find the value of the potential at 11.0 cm from the center of the sphere.
Express your answer in volts.
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