Download Part A What is the force felt by the electrons and the nuclei in the rod

Part A
What is the force felt by the electrons and the nuclei in the rod when the external field
described in the problem introduction is applied? (Ignore internal fields in the rod for the
moment.)
The nuclei experience a force to the right and the electrons experience a force to the left.
Correct
Part B
What is the motion of the negative electrons and positive atomic nuclei caused by the
external field?
The electrons move to the left and the nuclei are almost stationary.
Correct
The nuclei of the atoms of a conducting solid remain almost in their places in the crystal
lattice, while the electrons relatively move a lot. In an insulator, the electrons are
constrained to stay with their atoms (or molecules), and at most, the charge distribution is
displaced slightly.
The motion of the electrons due to the external electric field constitutes an electric
current. Since the negatively charged electrons are moving to the left, the current, which
is defined as the "flow" of positive charge, moves to the right.
Part C
Imagine that the rightward current flows in the rod for a short time. As a result, what will
the net charge on the right and left ends of the rod become?
left end negative and right end positive
Correct
Given that the positively charged nuclei do not move, why does the right end of the rod
become positively charged? The reason is that some electrons have moved to the left end,
leaving an excess of stationary nuclei at the right end.
Part D
The charge imbalance that results from this movement of charge will generate an
additional electric field near the rod. In what direction will this field point?
It will point to the left and oppose the initial applied field.Correct
An electric field that exists in an isolated conductor will cause a current flow. This flow
sets up an electric field that opposes the original electric field, halting the motion of the
charges on a nanosecond time scale for meter-sized conductors. For this reason, an
isolated conductor will have no static electric field inside it, and will have a reduced
electric field near it. This conclusion does not apply to a conductor whose ends are
connected to an external circuit. In a circuit, a rod (or wire) can conduct current
indefinitely.
Charge Distribution on a Conductor with a Cavity
A
Which of the figures best represents the charge distribution on the inner and outer walls
of the conductor?
3Correct
Charge Distribution on a Conducting Shell – 2
Which of the following figures best represents the charge distribution on the inner and
outer walls of the shell?
1Correct
A Test Charge Determines Charge on Insulating and Conducting Balls
Part A
What is the nature of the force between balls A and B?
strongly attractive
Correct
Part B
What is the nature of the force between balls A and C?
weakly attractive
Correct
Recall that ball C is composed of insulating material, which can be polarized in the
presence of an external charged object such as ball A. Once polarized, there will be a
weak attraction between balls A and C, because the positive and negative charges in ball
C are at slightly different average distances from ball A.
Part C
What is the nature of the force between balls A and D?
attractive
Correct
Part D
What is the nature of the force between balls D and C?
no force
Correct
Because the test charge T is neither strongly attracted to nor repelled from ball C, ball C
must have zero net charge. Since ball D also has zero net charge, there will not be any
force between the two.
Coulomb's Law Tutorial
Part A
Consider two positively charged particles, one of charge
origin, and another of charge
force
(particle 1) fixed on the y-axis at
. What is the net
on particle 0 due to particle 1?
Express your answer (a vector) using any or all of ,
=(-k*q_0*q_1*y_unit)/(d_1)^2Correct
Part B
(particle 0) fixed at the
,
,
, , , and .
Now add a third, negatively charged, particle, whose charge is
Particle 2 fixed on the y-axis at position
from particle 1 and particle 2?
(particle 2).
. What is the new net force on particle 0,
Express your answer (a vector) using any or all of ,
,
,
,
,
, , , and .
=(-k*q_0*q_1*y_unit)/(d_1)^2+(k*q_0*q_2*y_unit)/(d_2)^2Correct
Part C
Particle 0 experiences a repulsion from particle 1 and an attraction toward particle 2. For
certain values of
and
, the repulsion and attraction should balance each other,
resulting in no net force. For what ratio
is there no net force on particle 0?
Express your answer in terms of any or all of the following variables: ,
,
,
.
=sqrt(q_1)/sqrt(q_2)Correct
Part D
Now add a fourth charged particle, particle 3, with positive charge
plane at
. What is the net force
, fixed in the yz-
on particle 0 due solely to this charge?
Express your answer (a vector) using ,
caused by particle 3.
,
,
, , , and . Include only the force
=-k*q_0*q_3/(2*d_2^2)*(sqrt(2)/2)*(y_unit + z_unit)Correct
The Trajectory of a Charge in an Electric Field
Part A
Assume that the charge is emitted with velocity in the positive x direction. Between the
origin and the screen, the charge travels through a constant electric field pointing in the
positive y direction. What should the magnitude
to hit the target on the screen?
Express your answer in terms of
, ,
=(2*y_h*m*v_0^2)/(q*L^2)Correct
,
of the electric field be if the charge is
, and .
Part B
Now assume that the charge is emitted with velocity in the positive y direction.
Between the origin and the screen, the charge travels through a constant electric field
pointing in the positive x direction. What should the magnitude
the charge is to hit the target on the screen?
Express your answer in terms of
, ,
,
of the electric field be if
, and .
=2*L*m*v_0^2/(q*y_h^2)Correct
The equations of motion for this part are identical to the equations of motion for the
previous part, with and
interchanged. Thus it is no surprise that the answers to the
two parts are also identical, with and
interchanged.
Dipole Motion in a Uniform Field
Part A
What is
, the magnitude of the dipole's angular velocity when it is pointing along the
y axis?
Express your answer in terms of quantities given in the problem introduction.
=sqrt((-q*D*E*cos(theta_0)+(q*D*E))/(0.5*I))Correct
Thus increases with increasing
use the trigonometric identity
to write
as
, as you would expect. An easier way to see this is to
.
Part B
If
is small, the dipole will exhibit simple harmonic motion after it is released. What is
the period of the dipole's oscillations in this case?
Express your answer in terms of and quantities given in the problem introduction.
=2*pi/(sqrt(q*D*E/I))Correct