1. Physics 6B Electric Field Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3. q2q2 q1q1 x=0 x=0.2mx=-0.3m x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4. q2q2 q1q1 x=0 x=0.2mx=-0.3m The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

5. q2q2 q1q1 x=0 x=0.2mx=-0.3m This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 gives an E-field vector to the right. E1E1 E2E2 x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6. q2q2 q1q1 x=0 x=0.2mx=-0.3m This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. E1E1 E2E2 x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

7. q2q2 q1q1 x=0 x=0.2mx=-0.3m E1E1 E2E2 x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

8. q2q2 q1q1 x=0 x=0.2mx=-0.3m E total (This means 400 N/C in the negative x-direction) x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

9. q2q2 q3q3 q1q1 x=0 x=0.2mx=-0.3m For part b) all we need to do is multiply the E-field from part a) times the new charge q 3. x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. (This means 400 N/C in the negative x-direction) E total Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

10. q2q2 q3q3 q1q1 x=0 x=0.2mx=-0.3m Note that this force is to the right, which is opposite the E-field This is because q 3 is a negative charge: E-fields are always set up as if there are positive charges. x 17.22Two point charges are located on the x-axis as follows: charge q 1 = +4 nC at position x=0.2m and charge q 2 = +5 nC at position x = -0.3m. a)Find the magnitude and direction of the net electric field produced by q 1 and q 2 at the origin. b)Find the net electric force on a charge q 3 =-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q 1 will produce an E-field vector that points left, and q 2 gives an E-field vector to the right. This is how we can put the +/- signs on the E-fields when we add them up. (This means 400 N/C in the negative x-direction) For part b) all we need to do is multiply the E-field from part a) times the new charge q 3. E total F on3 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

11. 17.28Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

12. 17.28Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? The formula for electric force between 2 charges is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

13. 17.28Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? The formula for electric force between 2 charges is If both charges are doubled, we will have Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

14. 17.28Two unequal charges repel each other with a force F. If both charges are doubled in magnitude, what will be the new force in terms of F? The formula for electric force between 2 charges is If both charges are doubled, we will have So the new force is 4 times as large. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

15. 17.29Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

16. 17.29Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? The formula for electric force between 2 charges is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

17. 17.29Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? The formula for electric force between 2 charges is We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

18. 17.29Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? The formula for electric force between 2 charges is Canceling and cross-multiplying, we get We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

19. 17.29Two unequal charges attract each other with a force F when they are a distance D apart. How far apart (in terms of D) must they be for the force to be 3 times as strong as F? The formula for electric force between 2 charges is Canceling and cross-multiplying, we get We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. Square-roots of both sides gives us the answer: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

20. 17.30When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

21. 17.30When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2 nd law says that F net = ma. So this is really a problem about the force on the heavier charge. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

22. 17.30When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2 nd law says that F net = ma. So this is really a problem about the force on the heavier charge. The formula for electric force between 2 charges is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

23. 17.30When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2 nd law says that F net = ma. So this is really a problem about the force on the heavier charge. The formula for electric force between 2 charges is If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

24. 17.30When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2 nd law says that F net = ma. So this is really a problem about the force on the heavier charge. The formula for electric force between 2 charges is If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: We cancel common terms and cross-multiply to get Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

25. 17.30When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to 1/5 of this value, how far (in terms of d) should the charges be released? Recall that Newton's 2 nd law says that F net = ma. So this is really a problem about the force on the heavier charge. The formula for electric force between 2 charges is If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: We cancel common terms and cross-multiply to get Square-root of both sides: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

26. 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

27. 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB -4nC x=0x=0.8m +6nC x

28. -4nC x=0x=0.8m +6nC x The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

29. a For part a) which direction do the E-field vectors point? 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm -4nC x=0x=0.8m +6nC x The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0x=0.8m +6nC x Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

30. a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

31. a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

32. a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right b E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

33. a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right b E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

34. a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right b E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points right and E 2 points left c E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

35. a For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E1E1 E2E2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.41A point charge of -4 nC is at the origin, and a second point charge of +6 nC is placed on the x-axis at x=0.8m. Find the magnitude and direction of the electric field at the following points on the x-axis: a) x=20 cm; b) x=1.20m; c) x= -20cm The electric field near a single point charge is given by the formula: This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0x=0.8m +6nC x Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points left and E 2 points right b E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC For part b) E 1 points right and E 2 points left c E1E1 E2E2 Q 1 = -4nC x=0x=0.8m x Q 2 = +6nC

36. 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

37. x y Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM 12 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

38. x y Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E total = 0 12 E1E1 E2E2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

39. x y x y 12 21 Part b): both vectors point away from their charge. E1E1 E2E2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) E1E1 E2E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E total = 0

40. x y x y 12 21 Positive x-direction Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) E1E1 E2E2 Part b): both vectors point away from their charge. Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E total = 0 Positive x-direction E1E1 E2E2

41. x y x y 12 21 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) E1E1 E2E2 Part b): both vectors point away from their charge. Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E total = 0 Positive x-direction E1E1 E2E2

42. x y Part c): both vectors point away from their charge. We will need to use vector components to add them together. 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)

43. x y 12 E 1,y (0.15,- 0.4) (0.15,0)(- 0.15,0) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together.

44. x y 12 E 1,y (0.15,- 0.4) (0.15,0)(- 0.15,0) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together.

45. x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) E2E2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. E 1,y

46. x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m 0.3m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) E2E2 E 1,y Part c): both vectors point away from their charge. We will need to use vector components to add them together.

47. x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m 0.3m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) E 1,y Part c): both vectors point away from their charge. We will need to use vector components to add them together. E 2,x E 2,y

48. x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m 0.3m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) E 1,y Part c): both vectors point away from their charge. We will need to use vector components to add them together. E 2,x E 2,y

49. x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m 0.3m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) E 1,y Part c): both vectors point away from their charge. We will need to use vector components to add them together. E 2,x E 2,y Add together the x-components and the y-components separately:

50. x y 12 (0.15,- 0.4) (0.15,0)(- 0.15,0) The 0.5m in this formula for E 2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. Add together the x-components and the y-components separately: Now find the magnitude and the angle using right triangle rules: 75.7º E total

51. x y Part d): TRY THIS ONE ON YOUR OWN FIRST... 12 (0.15,0)(- 0.15,0) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m) (0,0.2)

52. x y Part d): both vectors point away from their charge. We will need to use vector components to add them together. 12 E1E1 E2E2 (0,0.2) (0.15,0)(- 0.15,0) The 0.25m in this formula is the distance to each charge using the Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. From symmetry, we can see that E 2 will have the same components, except for +/- signs. Now we can add the components (the x-component should cancel out) The final answer should be 1382.4 N/C in the positive y-direction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 17.42A point charge of q=+6 nC is at the point (x=0.15m,y=0m), and an identical point charge is placed at (-0.15m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.15m,-0.4m); d) (0m,0.2m)