MasteringPhysics Assignment
15 Pages

MasteringPhysics Assignment

Course Number: PHYSICS 2760, Spring 2011

College/University: Missouri (Mizzou)

Word Count: 2190

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HW 01 Due: 11:00pm on Friday, January 28, 2011 Note: To understand how points are awarded, read your instructor's Grading Policy. [Switch to Standard Assignment View] Electric Fields and Forces Learning Goal: To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force. Coulomb's law gives the electrostatic force acting between two charges. The magnitude of...

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01 Due: HW 11:00pm on Friday, January 28, 2011 Note: To understand how points are awarded, read your instructor's Grading Policy. [Switch to Standard Assignment View] Electric Fields and Forces Learning Goal: To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force. Coulomb's law gives the electrostatic force acting between two charges. The magnitude of the force between two charges and depends on the product of the charges and the square of the distance between the charges: , where . The direction of the force is along the line connecting the two charges. If the charges have the same sign, the force will be repulsive. If the charges have opposite signs, the force will be attractive. In other words, opposite charges attract and like charges repel. Because the charges are not in contact with each other, there must be an intermediate mechanism to cause the force. This mechanism is the electric field. The electric field at any location is equal to the force per unit charge experienced by a charge placed at that location. In other words, if a charge the electric field at that point is experiences a force , . The electric field vector has the same direction as the force vector on a positive charge and the opposite direction to that of the force vector on a negative charge. An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point charge has magnitude . The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a char ge is placed in an electric field created by , will not significantly affect the electric . field if it is small compared to Imagine an isolated positive point charge with a charge (many times larger than the charge on a single electron). Part A There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend? Check all that apply. ANSWER: the distance between the positive charge and the electron the charge on the electron the mass of the electron the charge of the positive charge the mass of the positive charge the radius of the positive charge the radius of the electron Correct According to Coulomb's law, the force between two particles depends on the charge on each of them and the distance between them. Part B For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend? Check all that apply. ANSWER: the distance between the positive charge and the electron the charge on the electron the mass of the electron the charge of the positive charge the mass of the positive charge the radius of the positive charge the radius of the electron Correct The electrostatic force cannot exist unless two charges are present. The electric field, on the other hand, can be created by only one charge. The value of the electric field depends only on the charge producing the electric field and the distance from that charge. Part C If the total positive charge is = 1.62106 , what is the magnitude of the electric = 1.53 from the charge? field caused by this charge at point P, a distance Enter your answer numerically in newtons per coulomb. ANSWER: 6220 = Correct Part D What is the direction of the electric field at point P? Enter the letter of the vector that represents the direction of . ANSWER: G Correct Part E Now find the magnitude of the force on an electron placed at point P. Recall that the charge on an electron has magnitude Hint E.1 Determine how to approach the problem Hint not displayed Enter your answer numerically in newtons. . ANSWER: = 9.951016 Correct Part F What is the direction of the force on an electron placed at point P? Enter the letter of the vector that represents the direction of . Register to View AnswerCorrect Magnitude and Direction of Electric Fields A small object A, electrically charged, creates an electric field. At a point P located 0.250 Part A What is the charge of object A? Hint A.1 How to approach the problem Hint not displayed Hint A.2 Find an expression for the charge Hint not displayed Hint A.3 Find the sign of the charge Hint not displayed ANSWER: 1.11109 1.11109 directly north of A, the field has a value of 40.0 directed to the south. 2.781010 2.781010 5.751012 5.751012 Correct Part B If a second object B with the same charge as A is placed at 0.250 south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P? Hint B.1 How to approach the problem Hint not displayed Hint B.2 Find the vector sum of the electric fields Hint not displayed Hint B.3 Find the electric field produced by B at P Hint not displayed ANSWER: 40.0 50.0 30.0 10.0 Correct The Electric Field inside a Conductor Learning Goal: To understand how the charges within a conductor respond to an externally applied electric field. To illustrate the behavior of charge inside conductors, consider a long conducting rod that is suspended by insulating strings (see the figure). Assume that the rod is init ially electrically neutral, and that it remains so for this discussion. The rod is positioned along the x axis, and an external electric field that points in the positive x direction (to the right) can be applied to the rod and the surrounding region. The atoms in the rod are composed of positive nuclei (indicated by plus signs) and negative electrons (indicated by minus signs). Before application of the electric field, these atoms were distributed evenly throughout 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.) Hint A.1 Formula for the force on a charge in an electric field Hint not displayed ANSWER: Both electrons and nuclei experience a force to the right. The nuclei experience a force to the right and the electrons experience a force to the left. The electrons experience a force to the left but the nuclei experience no force. The electrons experience no force but the nuclei experience a force to the right. Correct Part B What is the motion of the negative and electrons positive atomic nuclei caused by the external field? Hint B.1 How to approch this part Hint not displayed Hint B.2 ANSWER: Masses and charges of nuclei and electrons Hint not displayed Both electrons and nuclei move to the right. The nuclei move to the right and the electrons move to the left through equal distances. The electrons move to the left and the nuclei are almost stationary. The electrons are almost stationary and the nuclei move to the right. 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? Hint C.1 How to approach this part Hint not displayed ANSWER: left end negative and right end positive left end negative and right end negative left end negative and right end nearly neutral left end nearly neutral and right end positive both ends nearly neutral Correct Given that the positively charged nuclei do not move, wh y 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? Hint D.1 Direction of the electric field Hint not displayed ANSWER: It will point to the right and enhance the initial applied field. 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 Conducting Shell - 1 A positive charge is kept (fixed) at the center inside a fixed spherical neutral conducting shell. Part A The positive charge is equal to roughly 16 of the smaller charges shown on the surfaces of the spherical shell. Which of the pictures best represents the charge distribution on the inner and outer walls of the shell? Hint A.1 Effects of symmetry Hint not displayed ANSWER: 1 2 3 4 5 Correct Charge Distribution on a Conducting Shell - 2 A positive charge is kept (fixed) off-center inside a fixed spherical conducting shell that is electrically neutral, and the charges in the shell are allowed to reach electrostatic equilibrium. The large positive charge inside the shell is roughly 16 times that of the smaller charges shown on the inner and outer surfaces of the spherical shell. Part A Which of the following figures best represents the charge distribution on the inner and outer walls of the shell? Hint A.1 Symmetry inside shell Hint not displayed Hint A.2 Symmetry outside shell Hint not displayed ANSWER: 1 2 3 4 5 Correct Exercise 21.20 Two point charges are placed on the x-axis as follows: charge located at Part A What is the magnitude of the total force exerted by these two charges on a negative point charge = -5.96 that is placed at the origin? 0.201 , and charge = 5.00 is at = 4.01 -0.303 . is ANSWER: = 2.40106 Correct Part B What is the direction of the total force exerted by these two charges on a negative point charge ANSWER: = -5.96 that is placed at the origin? to the + to the direction direction -axis perpendicular to the the force is zero Correct Exercise 21.24 Two charges, one of 2.50 and the other of -3.50 , are placed on the x-axis, one at the origin and the other at = 0.600 , as shown in the figure . Part A Find the position on the ANSWER: -axis where the net force on a small charge zero. -3.27 Correct would be Exercise 21.34 Point charge = -5.00 axis at Part A Calculate the electric fields and at point due to the charges and . Express your results in terms of unit vectors (see example 21.6 in the textbook). Express your answer in terms of the unit vectors comma. , . Enter your answers separated by a is at the origin and point charge . Point is on the y-axis at = +3.00 = 4.00 . is on the - = 3.00 ANSWER: = Part B Use the results of part (a) to obtain the resultant field at form. , expressed in unit vector Correct Express your answer in terms of the unit vectors , . ANSWER: = Correct Exercise 21.40 A 8.756.50point charge is glued down on a horizontal frictionless table. It is tied to a point charge by a light, nonconducting wire. A uniform electric field of is directed parallel to the wire, as shown in the figure magnitude . Part A Find the tension in the wire. ANSWER: 382 = Correct Part B What would the tension be if both charges were negative? ANSWER: 2020 = Correct Exercise 21.44 Two particles with positive charges = 0.520 and a distance of 1.27 . Part A At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero? ANSWER: 0.265 from , between the charges = Correct = 7.50 are separated by Exercise 21.63 Point charges 4.80 3.60 and 4.80 are separated by a distance of , forming an electric dipole. Part A Find the magnitude of the electric dipole moment. ANSWER: = 1.731011 Correct Part B Find the direction of the electric dipole moment? ANSWER: from from Correct to to Part C The charges are in a uniform electric field whose direction makes an angle of 36.6 with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.50109 ANSWER: 728 = Correct ? Score Summary: Your score on this assignment is 101.4%. You received 55.79 out of a possible total of 55 points. [ Print ]

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