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- HW 2 Electric Fields.pdf

Sahaja L.

10/19/12 HW02: Electric Fields 1/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« HW02: Electric Fields Due: 11:59pm on Friday, September 7, 2012 Note: To understand how points are awarded, read your instructor's Grading Policy. PhET Tutorial: Charges and Electric Fields Learning Goal: To understand the spatial distribution of the electric field for a variety of simple charge configurations. For this problem, use the PhET simulation Charges and Fields. This simulation allows you to place multiple positive and negative point-charges in any configuration and look at the resulting electric field. Start the simulation. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If you select Show E-field in the green menu, red arrows will appear, showing the direction of the electric field. Faint red arrows indicate that the electric field is weaker than at locations where the arrows are brighter (this simulation does not use arrow length as a measure of field magnitude). Feel free to play around with the simulation. When you are done, click Clear All before beginning Part A. Part A Select Show E-field and grid in the green menu. Drag one positive charge and place it near the middle of the screen, right on top of two intersecting bold grid lines. You should see something similar to the figure below. HW02: Electric Fields 2/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« ANSWER: Correct This means that another positive charge, if placed near the original charge, would experience a force directed radially away from the original charge. Part B Now, letâ?s look at how the distance from the charge affects the magnitude of the electric field. Select Show numbers on the green menu, and then click and drag one of the orange E-Field Sensors. You will see the magnitude of the electric field given in units of V/m (volts per meter, which is the same as newtons per coulomb). Place the E-Field Sensor 1 away from the positive charge (1 is two bold grid lines away if going in a horizontal or vertical direction), and look at the resulting field strength. Consider the locations to the right, left, above, and below the positive charge, all 1 away. ANSWER: The electric field produced by the positive charge wraps circularly around the positive charge. is directed radially away from the charge at all locations near the charge. is directed radially toward the charge at all locations near the charge. 10/19/12 HW02: Electric Fields 3/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge. Mathematically, we say the electric field is spherically symmetric. Part C What is the magnitude of the electric field 1 m away from the positive charge compared to the magnitude of the electric field 2 m away? Hint 1. How to approach the problem Use an E-Field Sensor to determine the field strength both at 1 away and at 2 away from the charge. Then, take the ratio of the two field strengths. ANSWER: For these four locations, the magnitude of the electric field is greatest to the left of the charge. greatest to the right of the charge. greatest above the charge. the same. greatest below the charge. The magnitude of the electric field 1 away from the positive charge is equal to two times one-quarter four times one-half the magnitude of the electric field 2 away. 10/19/12 4/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared ( , where is the distance from the charge). You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulombâ?s law, which states that the magnitude of the force between two charged particles is . Part D If the field strength is = 9 a distance of 1 from the charge, what is the field strength a distance of 3 from the charge? Hint 1. How to approach the problem The magnitude of the electric field is inversely proportional to distance squared ( ). So if the distance is increased by a factor of three, the field strength must decrease by a factor of three squared. You could use the simulation to make a measurement (you might have to drag the charge away from the center so you have enough room to get 3 away). ANSWER: Correct Correct. Since , if the distance is increased by a factor of three, the electric field is decreased by a factor of nine. Part E Remove the positive charge by dragging it back to the basket, and drag a negative charge (blue) toward the middle of the screen. Determine how the electric field is different from that of the positive charge. Which statement best describes the differences in the electric field due to a negative charge as compared to a positive charge? ANSWER: = 1 10/19/12 HW02: Electric Fields 5/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct The electric field is now directed toward the negative charge, but the field strength doesnâ?t change. The electric field of a point charge is given by . Because of the sign of the charge, the field produced by a negative charge is directed opposite to that of a positive charge but the magnitude of the field is the same. Part F Now, remove the negative charge, and drag two positive charges, placing them 1 apart, as shown below. Letâ?s look at the resulting electric field due to both charges. Recall that the electric field is a vector, so the net electric field is the vector sum of the electric fields due to each of the two charges. Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)? ANSWER: The electric field changes direction (now points radially inward), and the magnitude of the electric field decreases at all locations. The electric field changes direction (now points radially inward), but the electric field strength does not change. Nothing changes; the electric field remains directed radially outward, and the electric field strength doesnâ?t change. HW02: Electric Fields 6/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct Directly between the two charges, the electric fields produced by each charge are equal in magnitude and point in opposite directions, so the two vectors add up to zero. Part G Consider a point 0.5 above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 . What is the magnitude of the total electric field due to both charges at this location? ANSWER: Correct Notice that this number is less than twice the magnitude of the field due to each charge. This occurs because the horizontal components of the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together. Part H Make an electric dipole by replacing one of the positive charges with a negative charge, so the final configuration looks like the figure shown below. The electric field is nonzero everywhere on the screen. The electric field is roughly zero near the midpoint of the two charges. The electric field is zero at any location along a vertical line going through the point directly between the two charges. 25 zero 36 10/19/12 HW02: Electric Fields 7/28 ANSWER: Correct The electric field due to the positive charge is directed to the right, as is the electric field due to the negative charge. So the net electric field, which is the sum of these two fields, is also to the right. Part I Make a small dipole by bringing the two charges very close to each other, where they are barely touching. The midpoint of the two charges should still be on one of the grid point intersections (see figure below). The electric field at the midpoint is directed to the left. directed to the right. zero. 10/19/12 HW02: Electric Fields 8/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Measure the strength of the electric field 0.5 directly above the midpoint as well as 1 directly above. Does the strength of the electric field decrease as 1 over distance squared ( )? Hint 1. How to approach the problem If the strength of the field is decreasing as , then the ratio of the magnitudes of the electric field measured at two distances, say 0.5 away and 1 away, would be . Compare this value to the value you measure with an E-Field Sensor. ANSWER: No, it decreases less quickly with distance. No, it decreases more quickly with distance. Yes, it does. HW02: Electric Fields 9/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct In fact, it turns out that the strength of the electric field decreases roughly as ! So the field 1 above the midpoint is roughly eight times weaker than at 0.5 above the midpoint. The important lesson here is that, in general, a distribution of charges produces an electric field that is very different from that of a single charge. Part - Make a long line of positive charges, similar to that shown in the figure below. Try to place all of the charges centered along a horizontal grid line. Feel free to look at the electric field, as it is interesting. Measure the strength of the electric field 1 directly above the middle as well as 2 directly above. Does the strength of the electric field decrease as 1 over distance squared ( )? ANSWER: No, it decreases less quickly with distance. Yes, it does. No, it decreases more quickly with distance. Correct In fact, it turns out that the strength of the electric field decreases roughly as . So the field 1 m above the midpoint is roughly half the strength at 0.5 . This is another example showing that a distribution of charges produces an electric field that is very different from that of a single charge. PhET Interactive Simulations University of Colorado http://phet.colorado.edu Magnitude and Direction of Electric Fields A small object A, electrically charged, creates an electric field. At a point P located 0.250 directly north of A, the field has a value of 40.0 directed to the south. Part A What is the charge of object A? Hint 1. How to approach the problem Recall that the electric field at a point P due to a point charge is proportional to the magnitude of the charge and inversely proportional to the square of the distance of P from the charge. Furthermore, the direction of the field is determined by the sign of the charge. Hint 2. Find an expression for the charge Which of the following expressions gives the correct magnitude of charge that produces an electric field of magnitude at a distance from the charge? In the following expressions is a constant that has units of . Hint 1. Magnitude of the electric field of a point charge Given a point charge , the magnitude of the electric field at a distance from the charge is given by , where the constant of proportionality is = 8.99×10 9 . 10/19/12 HW02: Electric Fields 11/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« ANSWER: Hint 3. Find the sign of the charge What is the sign of the charge that produces an electric field that points toward the charge? ANSWER: ANSWER: Correct Part B positive negative 1.11×10 ?9 ?1.11×10 ?9 2.78×10 ?10 ?2.78×10 ?10 5.75×10 12 ?5.75×10 12 10/19/12 HW02: Electric Fields 12/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« 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 1. How to approach the problem Since the electric field is a vector quantity, you need to apply the principle of superposition to find the total field at P. The principle of superposition in terms of electric fields says that the total electric field at any point due to two or more charges is the vector sum of the fields that would be produced at that point by the individual charges. Hint 2. Find the vector sum of the electric fields Which of the following diagrams, where and are the electric fields produced by A and B, respectively, correctly represents the situation described in this problem? ANSWER: Hint 3. Find the electric field produced by B at P What is the magnitude of the electric field produced by the second object B at point P? Express your answer in newtons per coulomb. Hint 1. Magnitude of the electric field of a point charge A B C D 10/19/12 HW02: Electric Fields 13/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Given a point charge , the magnitude of the electric field at a distance from the charge is given by , where the constant of proportionality is = 8.99×10 9 . Hint 2. Find the distance from P to the second object How far ( ) is P from B? Recall that P is located 0.250 north of A and B is located 0.250 south of A. Express your answer in meters. ANSWER: ANSWER: ANSWER: Correct Problem 20.25 A 74 charge experiences a 124 force in a certain electric field. = 0.500 = 10.0 40.0 50.0 30.0 10.0 Part A Find the field strength. Express your answer using two significant figures. ANSWER: Correct Part B Find the force that a 27 charge would experience in the same field. Express your answer using two significant figures. ANSWER: Correct Problem 20.26 A -5.0 charge experiences a 10 electric force in a certain electric field. Part A What force would a proton experience in the same field? Express your answer using two significant figures. ANSWER: Correct = 1.7×10 6 = 45 = ?3.2×10 ?13 10/19/12 HW02: Electric Fields session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Electric Field due to Multiple Point Charges Two point charges are placed on the x axis. The first charge, = 8.00 , is placed a distance 16.0 from the origin along the positive x axis; the second charge, = 6.00 , is placed a distance 9.00 from the origin along the negative x axis. Part A Calculate the electric field at point A, located at coordinates (0 , 12.0 ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Hint 1. How to approach the problem Find the contributions to the electric field at point A separately for and , then add them together (using vector addition) to find the total electric field at that point. You will need to use the Pythagorean theorem to find the distance of each charge from point A. Hint 2. Calculate the distance from each charge to point A Calculate the distance from each charge to point A. Enter the two distances, separated by a comma, in meters to three significant figures. ANSWER: Hint 3. Determine the directions of the electric fields Which of the following describes the directions of the electric fields and created by charges and at point A? ANSWER: , = 20.0 , 15.0 10/19/12 16/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Hint 4. Calculate the components of Calculate the x and y components of the electric field at point A due to charge . Express your answers in newtons per coulomb, separated by a comma, to three significant figures. Hint 1. Calculate the magnitude of the total field Calculate the magnitude of the field at point A due to charge only. Express your answer in newtons per coulomb to three significant figures. ANSWER: Hint 2. How to find the components of the total field Once you have found the magnitude of the field, use trigonometry to determine the x and y components of the field. The electric field of a positive point charge points directly away from the charge, so the direction of the electric field at point A due to charge will be along the line joining the two. Use the position coordinates of and point A to find the angle that the line joining the two makes with the x or y axis. Then use this angle to resolve the electric field vector into components. ANSWER: Hint 5. Calculate the components of Calculate the x and y components of the electric field at point A due to charge . Express your answers in newtons per coulomb, separated by a comma, to three significant points up and left and points up and right. points up and left and points down and left. points down and right and points up and right. points down and right and points down and left. = 0.180 = -0.144 , 0.108 10/19/12 HW02: Electric Fields 17/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« figures. Hint 1. Calculate the magnitude of the total field Calculate the magnitude of the field at point A due to charge only. Express your answer in newtons per coulomb to three significant figures. ANSWER: Hint 2. How to find the components of the total field Once you have found the magnitude of the field, use trigonometry to determine the x and y components of the field. The electric field of a positive point charge points directly away from the charge, so the direction of the electric field at point A due to charge will be along the line joining the two. Use the position coordinates of and point A to find the angle that the line joining the two makes with the x or y axis. Then use this angle to resolve the electric field vector into components. ANSWER: ANSWER: Correct Part B An unknown additional charge is now placed at point B, located at coordinates (0 , 15.0 ). Find the magnitude and sign of needed to make the total electric field at point A equal to zero. Express your answer in nanocoulombs to three significant figures. = 0.240 = 0.144 , 0.192 = 0 , 0.300 10/19/12 HW02: Electric Fields 18/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Hint 1. How to approach the problem You have already calculated the electric field at point A due to and . Now find the charge needed to make an opposite field at point A, so when the two are added together the total field is zero. Hint 2. Determine the sign of the charge Which sign of charge is needed to create an electric field that points in the opposite direction of the total field due to the other two charges, and ? ANSWER: Hint 3. Calculating the magnitude of the new charge Keep in mind that the magnitude of the field due to is , and the field must be equal in magnitude to the field due to charges and . ANSWER: Correct Problem 20.29 The dipole moment of the water molecule is . Part A What would be the separation distance if the molecule consisted of charges ? Note: The effective charge is actually less because electrons are shared by the oxygen and hydrogen atoms. Express your answer using two significant figures. ANSWER: positive negative = 0.300 10/19/12 HW02: Electric Fields 19/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct Problem 20.30 The electric field 20 from a long wire carrying a uniform line charge density is 1.9 . Part A What will be the field strength 38 from the wire? Express your answer using two significant figures. ANSWER: Correct Problem 20.31 Part A What is the line charge density on a long wire if the electric field 49 from the wire has magnitude 270 and points toward the wire? Express your answer using two significant figures. ANSWER: Correct = 3.9×10 ?11 = 1.0 = ?7.4×10 ?6 10/19/12 HW02: Electric Fields 20/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Problem 20.33 In his famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil drops in an electric field. Part A With a field strength of 20 , what mass drop can be suspended when the drop carries a net charge of 10 elementary charges? Express your answer using two significant figures. ANSWER: Correct Problem 20.35 - Copy A proton moving to the right at 3.2×10 5 enters a region where a 60 electric field points to the left. Part A How far will the proton get before its speed reaches zero? Express your answer using two significant figures. ANSWER: Correct Problem 20.47 A 1.0 charge and a 2.3 charge are 15.0 apart. Part A = 3.3×10 ?12 = 8.9×10 ?3 10/19/12 HW02: Electric Fields 21/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Find a point where the electric field is zero. Express your answer using two significant figures. Enter the distance from the required point to the charge 1.0 . ANSWER: Correct Problem 20.55 Two 34 charges are attached to the opposite ends of a spring of spring constant 150 and equilibrium length 50 . Part A By how much does the spring stretch? Express your answer using two significant figures. ANSWER: Correct Problem 20.67 A straight wire 13 long carries 20 distributed uniformly over its length. Part A What is the line charge density on the wire? Express your answer using two significant figures. ANSWER: = 6.0 = 16 = 1.5 10/19/12 HW02: Electric Fields 22/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct Part B Find the electric field strength 11 from the wire axis, not near either end. Express your answer using two significant figures. ANSWER: Correct Part C Find the electric field strength 470 from the wire. Note: make suitable approx imation. Express your answer using two significant figures. ANSWER: Correct Problem 20.64 An electron is at the origin, and an ion with charge 5 is at = 10 . Part A Find a point where the electric field is zero. Express your answer using two significant figures. ANSWER: = 2.5×10 5 = 0.81 10/19/12 HW02: Electric Fields 23/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Correct Dipole Motion in a Uniform Field Consider an electric dipole located in a region with an electric field of magnitude pointing in the positive y direction. The positive and negative ends of the dipole have charges and , respectively, and the two charges are a distance apart. The dipole has moment of inertia about its center of mass. The dipole is released from angle , and it is allowed to rotate freely. 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. Hint 1. How to approach the problem Because there is no dissipation (friction, air resistance, etc.), you can solve this problem using conservation of energy. When the dipole is released from rest, it has potential energy but no kinetic energy. When the dipole is aligned with the y axis, it is rotating, and therefore has both kinetic and potential energy. The sum of potential and kinetic energy will remain constant. Hint 2. Find the potential energy Find the dipole's potential energy due to its interaction with the electric field as a function of the angle that the dipole's positive end makes with the positive y axis. Define the potential energy to be zero when the dipole is oriented perpendicular to the field: . Express your answer in terms of , , , and . = -8.1 10/19/12 HW02: Electric Fields 24/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Hint 1. The formula for the potential energy of a dipole The general formula for the potential energy of an electric dipole with dipole moment in the presence of a uniform electric field is . Hint 2. The dipole moment The dipole moment of the electric dipole , when it makes an angle with the positive y axis can be written as . ANSWER: Correct Hint 3. Find the total energy at the moment of release Find , the total energy (kinetic plus potential) at the moment the dipole is released from rest at angle with respect to the y axis. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: . Express your answer in terms of some or al l of the variables , , , and . ANSWER: Answer Requested Hint 4. Find the total energy when Find an expression for , the total energy (kinetic plus potential) at the moment when the dipole is aligned with the y axis. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: . = = 10/19/12 HW02: Electric Fields 25/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« dipole is oriented perpendicular to the field: . Express your answer in terms of quantities given in the problem introduction and . Hint 1. What is kinetic energy as a function of angular velocity? What is the kinetic energy of a body rotating with angular velocity around an axis about which the moment of inertia is ? ANSWER: ANSWER: ANSWER: Answer Requested 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? = = = 10/19/12 HW02: Electric Fields 26/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Express your answer in terms of and quantities given in the problem introduction. Hint 1. How to approach the problem The equation of motion for a simple harmonic oscillator can always be written in the standard form . To solve this problem, you need to write the equation of motion for the dipole in the standard form with replaced by the angular variable . This will allow you to read off the expression for , which has a simple relationship to the period of oscillation. (Note: Here, the variable does not represent the angular velocity of the dipole; rather, it denotes the frequency of the dipole's oscillation.) Start with the angular analogue of Newton's second law: . Recall that , the angular acceleration, is equal to the second derivative of , just as linear acceleration is equal to the second derivative of position. Hint 2. Compute the torque What is the magnitude of the torque that the electric field exerts about the center of mass of the dipole when the dipole is oriented at an angle with respect to the electric field? Express the magnitude of the torque in terms of quantities given in the problem introduction and . Hint 1. Formula for torque on a dipole The torque on a dipole with dipole moment in an electric field is given by . Alternatively, the torque can be related to the potential energy by . Hint 2. The dipole moment When it makes an angle with the positive y axis, the dipole moment of the electric dipole can be written as . ANSWER: Hint 3. The small-angle approximation = 10/19/12 HW02: Electric Fields 27/28session.masteringphy sics.com/my ct/assignmentPrintView?displayMode=studentView assignmentID=« Because is small, you can apply the small-angle approximation to the expression for torque, and take the torque to be . Up to this point we have been interested only in the magnitude of the torque. Now let's think about the direction. After all, torque is a vector quantity. For a system to oscillate, the torque must be a restoring torque; that is, the torque and the (small) angular displacement must be in opposite directions. (Recall that small angular displacements can be treated as vectors, since they obey vector addition, while large angles do not.) If you did the vector algebra carefully, you would find that the correct vector equation is . For future purposes we will write this as , keeping in mind that now represents the component of in the direction, rather than the magnitude of . Hint 4. Find the oscillation frequency Putting together what you have so far yields . Compare this to the standard form for a simple harmonic oscillator to obtain the oscillation frequency for the motion of the dipole. Express your answer in terms of quantities given in the problem introduction. ANSWER: Hint 5. The relationship between (angular) oscillation frequency and period The relationship between , the angular oscillation frequency of the dipole, and the period of oscillation is given by . ANSWER: = = 10/19/12 HW02: Electric Fields 28/28 Correct Score Summary: Your score on this assignment is 126%. You received 23.57 out of a possible total of 25 points, plus 7.92 points of extra credit.

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