HW7Due: 11:59pm on Tuesday, September 15, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] Parallel Lines of Charge A very long uniform line of charge has charge per unit length 4.90 of charge has charge per unit length -2.30 Part A What is the magnitude of the net electric field at point ANSWER: 5 = 6.42×10 Correct and lies along the x-axis. A second long uniform line = 0.392 . and is parallel to the x-axis at = 0.218 on the y-axis? Part B What is the direction of the net electric field at point ANSWER: -y-axis +y-axis Correct = 0.218 on the y-axis? Part C What is the magnitude of the net electric field at point ANSWER: 4 = 4.28×10 Correct = 0.614 on the y-axis? Part D What is the direction of the net electric field at point ANSWER: -y-axis +y-axis Correct = 0.614 on the y-axis? Part E What is the size of the attractive force, by the positively-charged line, on a one-meter length of the negatively-charged line? ANSWER: 0.517 Correct Problem 22.48 A very long, solid cylinder with radius . Part A Derive the expression for the electric field inside the volume at a distance density . ANSWER: Correct from the axis of the cylinder in terms of the charge has positive charge uniformly distributed throughout it, with charge per unit volume Part B What is the electric field at a point outside the volume in terms of the charge per unit length ANSWER: Correct in the cylinder? The Electric Field inside and outside a Charged Insulator A slab of insulating material of uniform thickness , lying between are constant across the entire surface. ANSWER: = 1. The electric field is zero in the middle of the slab. .5 Compute the electric flux Hint not displayed Express your answer in terms of ANSWER: = Correct . as shown in the figure.3 A Gaussian surface for this problem Hint not displayed Hint C. the magnitude of the electric field inside the slab as a function of How to approach the problem ? . which is perpendicular to the x direction? Express your answer in radians. Part D What is Hint D. the magnitude of the electric field outside the slab? is not given as a function of . extends infinitely in the y and z directions. Both the magnitude and the direction of The direction of varies across the surface but its magnitude is constant. Hint C. Both the magnitude and the direction of Correct Part B What is the angle that the field makes with the surface of the slab. not just As implied by the fact that at the surface. at .4 Calculate the enclosed charge Hint not displayed Hint C. vary across the surface. and .to along the x axis. The slab has a uniform charge density . Part A Which of the following statements is true of the electric field ANSWER: at the surface of one side of the slab? The direction of is constant but its magnitude varies across the surface. this magnitude is constant everywhere outside the slab.1 How to approach the problem Hint not displayed Hint C.57 Correct Part C What is .2 Gauss's law Hint not displayed Hint C. in terms of .1 . 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. Before application of the electric field.e. However. Such a setup (usually called a PN junction) can be used as an electric "one-way street.2 A Gaussian surface for this problem Hint not displayed Hint D. i.. The rod is positioned along the x axis.e. and that it remains so for this discussion. these atoms were distributed evenly throughout the rod. you can check that the electric field would be zero in the regions where there is no charge. The atoms in the rod are composed of positive nuclei (indicated by plus signs) and negative electrons (indicated by minus signs). To illustrate the behavior of charge inside conductors. consider a long conducting rod that is suspended by insulating strings (see the figure).. The nuclei move to the right and the electrons move to the left through equal distances.2 Masses and charges of nuclei and electrons Hint not displayed ANSWER: Both electrons and nuclei move to the right. if a slab with negative charge were added behind this slab. The nuclei experience a force to the right and the electrons experience a force to the left. Assume that the rod is initially electrically neutral. from to . and severely inhibits the flow of current in the opposite direction. 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.) Hint A. i. 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." since it supports the flow of positive charge only in the positive direction.4 Compute the flux Hint not displayed ANSWER: = Correct Basic models of diodes and transistors (which are components used in more complex circuits. like those on computer chips) treat regions inside them as slabs of charge.3 Calculate the enclosed charge Hint not displayed Hint D. The electrons experience a force to the left but the nuclei experience no force.Hint not displayed Hint D. The electrons move to the left and the nuclei are almost stationary. Correct . Correct Part B What is the motion of the negative electrons and positive atomic nuclei caused by the external field? Hint B. The Electric Field inside a Conductor Learning Goal: To understand how the charges within a conductor respond to an externally applied electric field. because the fields due to the positive and negative charges cancel.1 How to approch this part Hint not displayed Hint B. and that the electric field in the regions where there is charge is always in the positive x direction. In this example you found that the electric field points in opposite directions on the two sides of . The electrons are almost stationary and the nuclei move to the right. The electrons experience no force but the nuclei experience a force to the right. along the electric field. 1 How to approach this part Remember that the rod as a whole must remain electrically neutral even if the charges are redistributed. What is the electric field 1. what will the net charge on the right and left ends of the rod become? Hint C. a rod (or wire) can conduct current indefinitely. leaving an excess of stationary nuclei at the right end. moves to the right.19 Part A How many excess electrons must be added to an isolated spherical conductor of diameer 34. In an insulator.584 = 171 Correct ANSWER: Part C At a distance of 0.0 of 1125 ANSWER: just outside the surface? 10 = 2. Exercise 22. For this reason. the electric field is 485 .26×10 Correct to produce an electric field Exercise 22. the electric field is .22 . The motion of the electrons due to the external electric field constitutes an electric current.2 Correct Part B At a distance of 0. In what direction will this field point? Hint D. Correct An electric field that exists in an isolated conductor will cause a current flow. Part C Imagine that the rightward current flows in the rod for a short time. an isolated conductor will have no static electric field inside it. This conclusion does not apply to a conductor whose ends are connected to an external circuit. ANSWER: It will point to the right and enhance the initial applied field. halting the motion of the charges on a nanosecond time scale for meter-sized conductors. while the electrons relatively move a lot. why does the right end of the rod become positively charged? The reason is that some electrons have moved to the left end. the current. the electric field is 485 . This flow sets up an electric field that opposes the original electric field. which is defined as the "flow" of positive charge. and at most. It will point to the left and oppose the initial applied field. 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. Part D The charge imbalance that results from this movement of charge will generate an additional electric field near the rod. the charge distribution is displaced slightly. In a circuit.206 485 from the axis of a very long charged conducting cylinder with radius from the axis of the cylinder? .1 Direction of the electric field The electric field point away from positive charges and towards negative ones. What is the electric field 0.594 ANSWER: = 57. the electrons are constrained to stay with their atoms (or molecules).The nuclei of the atoms of a conducting solid remain almost in their places in the crystal lattice.204 from the center of a charged conducting sphere with radius from the center of the sphere? . only redistributes it. This is because applying an electric field does not change the charge on the rod. and will have a reduced electric field near it.22 Part A At a distance of 0. What is the electric field 0.192 from a large uniform sheet of charge. Since the negatively charged electrons are moving to the left. As a result. . You received 44.from the sheet? ANSWER: = 485 Correct Score Summary: Your score on this assignment is 98%.08 out of a possible total of 45 points.