the electric field inside a conductor

7/29/2019 The Electric Field Inside a Conductor

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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 themoment.)

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 crystallattice, 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 electriccurrent. 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 willthe 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 rodbecome 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 anadditional 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
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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 areconnected 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 attractiveCorrect

Part B

What is the nature of the force between balls A and C?
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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 ballC are at slightly different average distances from ball A.

Part C

What is the nature of the force between balls A and D?

attractiveCorrect

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 anyforce between the two.

Coulomb's Law Tutorial

Part A

Consider two positively charged particles, one of charge (particle 0) fixed at the

origin, and another of charge (particle 1) fixed on the y-axis at . What is the net

force on particle 0 due to particle 1?

Express your answer (a vector) using any or all of , , , , , , and .

=(-k*q_0*q_1*y_unit)/(d_1)^2Correct

Part B
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Now add a third, negatively charged, particle, whose charge is (particle 2).

Particle 2 fixed on the y-axis at position . What is the new net force on particle 0,

from particle 1 and particle 2?

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 CParticle 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 , fixed in the yz-

plane at . What is the net force on particle 0 due solely to this charge?

Express your answer (a vector) using , , , , , , and . Include only the forcecaused by particle 3.

=-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 of the electric field be if the charge isto hit the target on the screen?

Express your answer in terms of , , , , and .

=(2*y_h*m*v_0^2)/(q*L^2)Correct

Part B
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Now assume that the charge is emitted with velocity in the positivey 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 of the electric field be if

the charge is to hit the target on the screen?

Express your answer in terms of , , , , 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 , as you would expect. An easier way to see this is touse the trigonometric identity

to write as .

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
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the electric field inside a conductor

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