TI-Nspire™ Activities: Physics
March 4, 2008 through March 11, 2008
Electric Fields Created by Point Charges
| Problem 1 – The electric field created by a point charge | |
| Step 1: Students should open the file PhysWeek06_ElectricFields.tns, read the first three pages, and then move to page 1.4, which shows two point charges. One of the charges is positive, and the other is negative. A circle surrounds each point charge, and a vector on that circle shows the magnitude and direction of the electric field at a given point. Students can vary the position of the test charge and the magnitude of the point charge (Q) that creates the electric field, and observe the changes in the direction and magnitude of the electric field. This exploration is for qualitative purpose only. Students should use the motion control buttons to start and pause the animations. To switch between the two animations, students should press /e. After students have explored the animation, they should answer questions 1–3. Note: The moving test charges in the two animations do not represent actual moving charges. The animations simply allow the students to visualize the electric field lines at each point on the circle. Make sure students understand that this is not an animation of a moving particle. Q1. Describe the direction of the electric fields created by a positively charged particle and a negatively charged particle. A. The electric field near a positively charged particle is directed radially away from the particle. The electric field near a negatively charged particle is directed radially toward the particle. Q2. What happens to the magnitude of the electric field when the distance from the charge that creates the electric field changes? A. When the distance from the charge that creates the electric field increases, the magnitude of the electric field decreases. When the distance decreases, the magnitude increases. The increases and decreases are proportional to the inverse square of the distance, as described by Coulomb's law. Q3. What happens to the magnitude of the electric field when the magnitude of the charge that creates the electric field changes? A. As the magnitude of the charge increases, the magnitude of the field increases. As the magnitude of the charge decreases, the magnitude of the field decreases. |
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| Problem 2 – The electric field created by two point charges | |
| Step 1: Students should read the text on pages 2.1 and 2.2. Then, they should move to page 2.3, which shows two positive point charges, q1 and q2, with fixed positions. The vector at point A shows the magnitude and direction of the net electric field at that point. The vectors E1 and E2 show the electric force on point A due to q1 and q2, respectively. The magnitude of the field is shown at the bottom of the screen (variable E), and the angle between the vector and a line parallel to the x-axis is also given. Students should vary the position of point A and the magnitudes of q1 and q2 and observe the effects on the net electric field. Note: The simulation will not give accurate results if the point charges are given negative signs. |
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| Step 2: Next, students should move to page 2.4, which shows the same system as page 2.3; however, in this case, the point charge q2 has a negative sign. Again, students should vary the position of point A and the magnitudes of charges q1 and q2 and observe the effects on the net electric field. Then, students should answer questions 4–10. Q4. For both simulations, describe how the direction and magnitude of the net electric field changes under each of three conditions: as point A is moved vertically from the x-axis, when point A is on the x-axis between q1 and q2, and when point A is on the x-axis on either side of the charges. A. In both cases, when point A moves away from both charges (i.e., vertically), the magnitude of the electric field decreases. In terms of direction, in the first case, because both charges are positive, they counteract each other in horizontal direction when point A is in between the charges and reinforce each other when point A is outside that region. They always reinforce each other in the vertical direction. In the second case, when charges are opposite, the forces reinforce each other in the horizontal direction when point A is in between two charges, but they counteract each other outside that area, and they always counteract each other in the vertical direction. Q5. Describe how changing the magnitudes of charges q1 and q2 affected the direction and magnitude of the net electric field in both cases. A. When the magnitudes of the charges increase, the magnitude of the electric field also increases. If the charges are not equal, the components of the electric field do not counteract as much, and that affects the direction of the net electric field. The charge with the larger absolute value (magnitude) dominates in determining the direction of the net vector. Q6. Describe the symmetry in the electric field created by two point charges of equal positive charge. A. There are two axes of symmetry, one that goes through the two charges and one that is the perpendicular bisector to the segment connecting the point charges. When point A is on the x-axis, the electric field vector is always horizontal. When point A is on the y-axis, the electric field vector is always vertical. At the origin, the field is undefined. The symmetry is for both the direction and magnitude of the electric field. Q7. Describe the symmetry of the electric field created by two point charges with equal magnitudes but opposite signs. Is it different from the symmetry of the system when both charges are positive? If so, how? A. The axes of symmetry are the same, although the electric field direction is different at some points on the plane. It is still horizontal when point A is on the x-axis, but when point A is on the y-axis, it remains horizontal. This is because the direction of the electric field created by charge q2 is now opposite to what it was in the previous case. Q8. Consider the electric field at three points, (–10, –2), (0, 0), and (10, 10). For the situation on page 2.3 with q2 = q1, list these points in order of decreasing field magnitude. Will this be the same order as in the system on page 2.4 with q2 = –q1? Why? Base your explanations on physics principles, not visual observations. A. For the situation with two positive charges, the largest magnitude of the electric field is at (–10, –2), and the smallest magnitude is at (0, 0). The magnitude of the electric field is proportional to the inverse square of the distance, and point (–10, –2) is very close to one of the charges, so the field is very large at that point. At point (0, 0), the charges counteract and cancel each other, so the magnitude of the electric field is zero. In the case of opposite charges, however, the situation is different. The magnitude of the electric field is still largest at (–10, –2), but at (0, 0), the electric fields add up instead of canceling, and this point is closer to the charges than is point (10, 10), so the smallest magnitude electric field is at (10, 10). Q9. A particle with charge +0.1 μC is located at point (–10, 0). Another particle with charge +0.02 μC is located at point (10, 0). Calculate the magnitude and direction of the net electric field at point (8, 4). Confirm your calculations by setting this problem up on page 2.3. A. The magnitude of the electric field E produced by a point charge q at a distance r from the point charge is given by  , wherek = 9.0 × 109 N•m2/C2. The magnitude of the x-component of the net electric field, Ex, is equal to the x-component of the electric field due to charge q1, Ex1, minus the x¬-component of the electric field due to charge q2, Ex2. The minus sign appears because the forces act in opposite directions.  The distance r1 is the distance between charge q1 and point (8, 4). In other words, it is the hypotenuse of a triangle with base 18 and height 4. The angle θ1 is the angle between the hypotenuse of this triangle and the x-axis. Therefore, from the Pythagorean theorem, r2 = 182 + 42, and from trigonometry,  . Therefore,  . By similar reasoning,  . Therefore,  Ey is calculated in a similar way, as shown on the next page:  The magnitude of the electric field, E, is given by  . The angle θ between the net electric field vector and the x-axis is given by  Note that, in order to use the simulation to check their answers, students will need to first mark the coordinates of point A using the Coordinates and Equations tool (Menu > Actions > Coordinates and Equations) and then manually change the coordinates of the point to (8, 4). Q10. Two point charges, q1 = +0.3 μC and q2 = –0.01 μC are separated by a distance of 20 m. The electric field at a certain point P is zero. Calculate how far point P is from q1. Confirm your answer by setting this problem up on page 2.4. A. Point P must be at the location at which the electric force from the field produced by charge q1 exactly cancels that produced by charge q2. Assume that charge q1 is located at (0, 0) and charge q2 is located at (20, 0). Between charges q1 and q2 (i.e., 0 < x < 20), the electric fields from the two charges reinforce one another, so x cannot be between 0 and 20. For x < 0, the force from charge q1 is always larger than the force from q2, so x cannot be less than zero. The net electric field at point P, EP, is equal to the field at P due to charge q1, EP1, minus the field at P due to q2, EP2. Therefore, EP is given by the following equation:  The net field at point P is zero, so the equation above becomes a quadratic equation in x. The equation has two solutions, x = 17 and x = 24.46. From the reasoning above, we know that x cannot be less than 20, so the answer must be x = 24.46 m. If you wish, you may have the students use the simulation on page 2.4 to confirm their answers. Note: It is suggested that students share their solutions to questions 4–10 with the class and that you lead a class discussion about the solutions. It is particularly important that students discuss and understand the symmetry of the electric field in each case before moving on to the next four problems. If you wish, you may assign each of the next four problems (problems 3–6) to a different group of students, and then have students share their results with the class. |
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| Problem 3 – The electric field along a line connecting two positive charges of equal magnitude | |
| Step 1: Next, students should read page 3.1. Then, they should move to page 3.2, which shows a point A constrained to move between equal charges q1 and q2, both of which are positively charged. Students should press the play button (¢) to begin the animation. When point A has moved across the entire length of the displayed x-axis once, students should press pause (||) to stop the animation, and then move to page 3.3. |
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| Step 2: Page 3.3 shows a Lists & Spreadsheet application on the left and a Data & Statistics application on the right. The Lists & Spreadsheet application is set up to automatically capture data on the position of point A and the magnitude of the electric field at that point as the animation on page 3.2 proceeds. The Data & Statistics application shows a scatter plot of the magnitude of the electric field (efield) vs. the position of point A (distance) at that point. Students may need to change the window settings (Menu > Window > Zoom-Data or Window Settings) in order to analyze specific regions of the scatter plot (they can press /e to move between the two applications). Students should examine the scatter plot and then answer questions 11–13. Q11. What happens to the magnitude of the electric field when point A approaches the points where the charges q1 and q2 are located? A. The magnitude of the electric field approaches infinity. Q12. What happens to the magnitude of the electric field at the point that is exactly midway between the charges q1 and q2? A. The net electric field is zero because the vectors from q1 and q2 are opposite to each other in direction and equal in magnitude, so they cancel each other. Q13. Find an equation for the electric field E(x) created by two positive charges each of magnitude q located at a distance a from the origin, as a function of distance x from the origin. A.  |
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| Problem 4 – The electric field along a line connecting a positive charge and a negative charge | |
| Step 1: Next, students should read page 4.1 and then move on to page 4.2. Page 4.2 shows a situation similar to that shown on page 3.2, but in this case, the two charges have equal magnitudes but opposite signs. Students should again use the play and pause buttons to observe how the electric field at point A changes as point A move along the x-axis. |
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| Step 2: Once point A has moved along the full length of the displayed x-axis, students should move to page 4.3, which contains a display similar to that on page 3.3. Students should again examine the scatter plot of electric field strength vs. position and then answer questions 14–16. Q14. What happens to the magnitude of the electric field when point A approaches the points where the charges q1 and q2 are located? A. The magnitude of the electric field approaches infinity. Q15. What happens to the magnitude of the electric field at the point that is exactly midway between the two charges? A. The electric field vectors due to the charges are the same in direction and magnitude, so they add constructively. Q16. Find an equation for the electric field E(x) created by two charges of equal magnitude q but with opposite charges located at the distance a from the origin, as a function of distance x from the origin. A.  |
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| Problem 5 – The electric field along the perpendicular bisector of a segment connecting two positive charges | |
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| Step 2: Once point A has moved along the full length of the displayed y-axis, students should move to page 5.3, which contains a display similar to that on page 3.3. Students should again examine the scatter plot of electric field strength vs. position and then answer questions 17–19. Q17. Analyze the scatter plot of the electric field as point A moves away from the origin, and explain the resulting changes in the magnitude of the electric field. A. As point A moves away from the origin, the electric field first increases from zero, reaches its maximum value, and then decreases, approaching zero at infinity. At the origin, the vectors of the electric field created by the charges cancel each other. When you start moving away along the y-axis in either direction, the x-components of the vectors still cancel each other, but the y-components add up. As the angle increases, the y-components increase, but at the same time, the distance is increasing as well. These two factors play against each other, at some point reaching the optimal condition for the magnitude of the net electric field. Q18. Find the equation for the electric field E created by two positive charges of equal magnitude q located at a distance a from the origin along the x-axis, as a function of distance from the origin along the y-axis. A. The equation is given below:  Because the horizontal portions of the electric fields created by each particle cancel, the total field is vertical. Both charges contribute equally to the field, so the equation describing the field is equal to twice the magnitude of the field due to either charge. Q19. On the next page, plot the function you derived in question 18, along with the data you collected on page 5.3. Use the values of q and a from page 5.2. Do your data match your derived equation? Explain. A. The function matches the scatter plot for y > 0. Because the electric field is negative when y < 0, the scatter plot shows the absolute value of the function for the region where y < 0. Note: Remind students to substitute x for y when they plot the function. |
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| Problem 6 – The electric field along the perpendicular bisector of a line connecting a positive charge and a negative charge | |
| Step 1: Next, students should read page 6.1 and then move on to page
6.2. Page 6.2 shows a situation similar to that shown on page 5.2, but
in this case, the charges q1 and q2 have opposite signs. Students
should again use the play and pause buttons to observe how the electric
field at point A changes as point A move along the y-axis. |
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| Step 2: Once point A has moved along the full length of the displayed
y-axis, students should move to page 6.3, which contains a display
similar to that on page 5.3. Students should again examine the scatter
plot of electric field strength vs. position and then answer questions
20–23. Q20. Analyze the scatter plot of the electric field as point A moves away from the origin, and explain any changes in the magnitude of the electric field. A. As point A moves away from the origin, the electric field magnitude decreases to zero. The direction of the electric field does not change. Q21. Describe the properties of the function E(y). A. It is an even, positive-valued function with a maximum at y = 0 and a horizontal asymptote at y = 0. Q22. Find the equation for the electric field E created by two equal but opposite charges of magnitude q located at a distance a from the origin along the x-axis, as a function of distance from the origin along the y-axis. A. The equation is given below:  Because the vertical portions of the electric fields created by each particle cancel, the total field is horizontal. Both charges contribute equally to the field, so the equation describing the field is equal to twice the magnitude of the field due to either charge. Q23. On the next page, plot the function you derived in question 22, along with the data you collected on page 6.3. Use the values of q and a from page 6.2. Do your data match your derived equation? Explain. A. The function perfectly matched to the scatter plot. Since this is a positive-valued function, the absolute value of the function is the function itself, so the magnitude of the electric field is equal to the electric field at all points on the y-axis. |
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