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Physics 6B Electric Field Examples

Physics 6B Electric Field Examples. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB. 17.22 Two 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.

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Physics 6B Electric Field Examples

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

  2. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  3. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. q2 q1 x x=0 x=-0.3m x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  4. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-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. q2 q1 x x=0 x=-0.3m x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  5. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: E2 E1 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 gives an E-field vector to the right. q2 q1 x x=0 x=-0.3m x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  6. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: E2 E1 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 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. q2 q1 x x=0 x=-0.3m x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  7. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: E1 E2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 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. q2 q1 x x=0 x=-0.3m x=0.2m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  8. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Etotal This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 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. q2 q1 x x=0 x=-0.3m x=0.2m (This means 400 N/C in the negative x-direction) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  9. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Etotal This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 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. q2 q3 q1 x x=0 x=-0.3m x=0.2m (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 q3. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  10. 17.22 Two point charges are located on the x-axis as follows: charge q1 = +4 nC at position x=0.2m and charge q2 = +5 nC at position x = -0.3m. a) Find the magnitude and direction of the net electric field produced by q1 and q2 at the origin. b) Find the net electric force on a charge q3=-0.6nC placed at the origin. The electric field near a single point charge is given by the formula: Etotal Fon3 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q1 will produce an E-field vector that points left, and q2 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. q2 q3 q1 x x=0 x=-0.3m x=0.2m (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 q3. Note that this force is to the right, which is opposite the E-field This is because q3 is a negative charge: E-fields are always set up as if there are positive charges. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  11. 17.28 Two 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. 17.28 Two 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. 17.28 Two 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. 17.28 Two 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. 17.29 Two 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. 17.29 Two 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. 17.29 Two 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. 17.29 Two 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. Canceling and cross-multiplying, we get Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  19. 17.29 Two 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. Canceling and cross-multiplying, we get Square-roots of both sides gives us the answer: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  20. 17.30 When 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. 17.30 When 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 2nd law says that Fnet = 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. 17.30 When 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 2nd law says that Fnet = 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. 17.30 When 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 2nd law says that Fnet = 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. 17.30 When 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 2nd law says that Fnet = 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. 17.30 When 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 2nd law says that Fnet = 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. 17.41 A 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. 17.41 A 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 +6nC x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  28. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  29. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m For part a) which direction do the E-field vectors point? -4nC +6nC a x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  30. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 E1 Q1 = -4nC Q2 = +6nC a x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  31. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 E1 Q1 = -4nC Q2 = +6nC a x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  32. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 E1 Q1 = -4nC Q2 = +6nC a x x=0 x=0.8m E2 Q1 = -4nC Q2 = +6nC E1 For part b) E1 points left and E2 points right b x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  33. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 E1 Q1 = -4nC Q2 = +6nC a x x=0 x=0.8m E2 Q1 = -4nC Q2 = +6nC E1 For part b) E1 points left and E2 points right b x x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  34. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 E1 Q1 = -4nC Q2 = +6nC a x x=0 x=0.8m E2 Q1 = -4nC Q2 = +6nC E1 For part b) E1 points left and E2 points right b x x=0 x=0.8m E2 E1 Q1 = -4nC Q2 = +6nC c x For part b) E1 points right and E2 points left x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  35. 17.41 A 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: -4nC +6nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x x=0 x=0.8m For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 E2 E1 Q1 = -4nC Q2 = +6nC a x x=0 x=0.8m E2 Q1 = -4nC Q2 = +6nC E1 For part b) E1 points left and E2 points right b x x=0 x=0.8m E2 E1 Q1 = -4nC Q2 = +6nC c x For part b) E1 points right and E2 points left x=0 x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  36. 17.42 A 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. 17.42 A 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) y Part a): TRY DRAWING THE E-FIELD VECTORS ON THE DIAGRAM 2 1 x Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  38. 17.42 A 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) y E1 E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. Etotal = 0 2 1 x Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  39. 17.42 A 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) y E1 E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. Etotal = 0 2 1 x y Part b): both vectors point away from their charge. E1 E2 2 1 x Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  40. 17.42 A 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) y E1 E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. Etotal = 0 2 1 x y Part b): both vectors point away from their charge. Positive x-direction E1 E2 2 1 x Positive x-direction Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  41. 17.42 A 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) y E1 E2 Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. Etotal = 0 2 1 x y Part b): both vectors point away from their charge. Positive x-direction E1 E2 2 1 x Positive x-direction Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  42. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x (0.15,- 0.4) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  43. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x (0.15,- 0.4) E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  44. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x (0.15,- 0.4) E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  45. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x (0.15,- 0.4) E2 E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  46. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m (0.15,- 0.4) 0.3m E2 E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  47. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m (0.15,- 0.4) E2,x 0.3m E2,y E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  48. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m (0.15,- 0.4) E2,x 0.3m E2,y E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  49. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. 0.4m (0.15,- 0.4) E2,x 0.3m E2,y E1,y Add together the x-components and the y-components separately: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  50. 17.42 A 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. y (- 0.15,0) (0.15,0) 2 1 x The 0.5m in this formula for E2 is the distance to charge 2, using Pythagorean theorem or from recognizing a 3-4-5 right triangle when you see it. (0.15,- 0.4) 75.7º Etotal Add together the x-components and the y-components separately: Now find the magnitude and the angle using right triangle rules: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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