What Is the Force Felt by the Electrons and the Nuclei in the Rod When the External Field

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.)

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The nuclei experience a force to the right and the electrons experience a force to the left.
Correct

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Part B

What is the motion of the negative electrons and positive atomic nuclei caused by the external field?

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The electrons move to the left and the nuclei are almost stationary.
Correct

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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?

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left end negative and right end positive
Correct

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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?

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It will point to the left and oppose the initial applied field.Correct

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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

Which of the figures best represents the charge distribution on the inner and outer walls of the conductor?

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3Correct

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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?

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1Correct

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A Test Charge Determines Charge on Insulating and Conducting Balls

Part A

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

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strongly attractive
Correct

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Part B

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

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weakly attractive
Correct

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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?

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attractive
Correct

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Part D

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

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no force
Correct

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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 (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 .

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

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Part B

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 .

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=(-k*q_0*q_1*y_unit)/(d_1)^2+(k*q_0*q_2*y_unit)/(d_2)^2Correct

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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: , , , .

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=sqrt(q_1)/sqrt(q_2)Correct

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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 force caused by particle 3.

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=-k*q_0*q_3/(2*d_2^2)*(sqrt(2)/2)*(y_unit + z_unit)Correct

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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 is to hit the target on the screen?

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

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

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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 of the electric field be if the charge is to hit the target on the screen?

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

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

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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.

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=sqrt((-q*D*E*cos(theta_0)+(q*D*E))/(0.5*I))Correct

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Thus increases with increasing , as you would expect. An easier way to see this is to use 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.

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=2*pi/(sqrt(q*D*E/I))Correct

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