Transcript Slide 1
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 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 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 x=-0.3m q1 x=0 x x=0.2m 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 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: E kq R2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. q2 x=-0.3m q1 x=0 x x=0.2m 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 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: E kq R2 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. E1 q2 x=-0.3m E2 q1 x=0 x x=0.2m 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 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: E kq R2 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. E1 q2 x=-0.3m E2 q1 x=0 x x=0.2m This is how we can put the +/- signs on the E-fields when we add them up. Etotal E1 E2 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 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: E kq R2 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 x=-0.3m E2 q1 x=0 x x=0.2m This is how we can put the +/- signs on the E-fields when we add them up. Etotal E1 E2 (9 109 Nm )(4 109 C) C2 2 Etotal 2 (0.2m) (9 109 Nm )(5 109 C) C2 2 2 (0.3m) 900 NC 500 NC 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 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: E kq R2 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. q2 x=-0.3m q1 x=0 x x=0.2m This is how we can put the +/- signs on the E-fields when we add them up. Etotal E1 E2 (9 109 Nm )(4 109 C) C2 2 Etotal 2 (0.2m) Etotal 400 NC (9 109 Nm )(5 109 C) C2 2 2 (0.3m) 900 NC 500 NC (This means 400 N/C in the negative x-direction) 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 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: E kq R2 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. q2 x=-0.3m q3 x=0 q1 x x=0.2m This is how we can put the +/- signs on the E-fields when we add them up. Etotal E1 E2 (9 109 Nm )(4 109 C) C2 2 Etotal 2 (0.2m) Etotal 400 NC (9 109 Nm )(5 109 C) C2 2 2 (0.3m) 900 NC 500 NC (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 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: kq E 2 R 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. q2 x=-0.3m q3 x=0 q1 x x=0.2m This is how we can put the +/- signs on the E-fields when we add them up. Etotal E1 E2 (9 109 Nm )(4 109 C) C2 2 Etotal 2 (0.2m) Etotal 400 NC (9 109 Nm )(5 109 C) C2 2 2 (0.3m) 900 NC 500 NC (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. Fonq3 (0.6 109 C)( 400 NC ) 2.4 107N 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 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 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 Felec kq1q2 R2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 Felec kq1q2 R2 If both charges are doubled, we will have Felec k(2q1)(2q2 ) 4 kq1q2 R2 R2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 Felec kq1q2 R2 If both charges are doubled, we will have Felec k(2q1)(2q2 ) 4 kq1q2 R2 R2 So the new force is 4 times as large. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 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 Felec kq1q2 D2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 Felec kq1q2 D2 We want the force to be 3 times as strong, so we can set kq q kq q 3 12 2 21 2 up the force equation and solve for the new distance. D Dnew Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 Felec kq1q2 D2 We want the force to be 3 times as strong, so we can set kq q kq q 3 12 2 21 2 up the force equation and solve for the new distance. D Dnew Canceling and cross-multiplying, we get 2 Dnew 1 3 D2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 Felec kq1q2 D2 We want the force to be 3 times as strong, so we can set kq q kq q 3 12 2 21 2 up the force equation and solve for the new distance. D Dnew Canceling and cross-multiplying, we get 2 Dnew 1 3 D2 Square-roots of both sides gives us the answer: Dnew 1 3 D Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 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 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 F kq1q2 elec d2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 F kq1q2 elec d2 If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: Fnew 1 5 kq1q2 2 dnew Fold 1 5 kq1q2 d2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 F kq1q2 elec d2 If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: Fnew 1 5 kq1q2 2 dnew Fold 1 5 kq1q2 d2 2 2 We cancel common terms and cross-multiply to get dnew 5 d Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 F kq1q2 elec d2 If we want the acceleration to be 1/5 as fast, we need the force to be 1/5 as strong: Fnew 1 5 kq1q2 2 dnew Fold 1 5 kq1q2 d2 2 2 We cancel common terms and cross-multiply to get dnew 5 d Square-root of both sides: d 5d new Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 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 x=0 +6nC x x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC x=0 +6nC x x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4nC +6nC x=0 x x=0.8m For part a) which direction do the E-field vectors point? -4nC x=0 a +6nC x x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 Etotal E1 E2 -4nC +6nC x=0 x x=0.8m E2 Q1 = -4nC x=0 E1 a Q2 = +6nC x x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 Etotal E1 E2 (9 109 Nm )(4 109 C) C2 2 (0.2m) (9 109 Nm )(6 109 C) C2 +6nC x=0 x x=0.8m E2 Q1 = -4nC x=0 2 Etotal -4nC E1 a Q2 = +6nC x x=0.8m 2 2 (0.6m) 1050 NC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 Etotal E1 E2 (9 109 Nm )(4 109 C) C2 2 (0.2m) (9 109 Nm )(6 109 C) C2 x x=0.8m E2 Q1 = -4nC E1 a Q2 = +6nC x x=0.8m 2 2 (0.6m) For part b) E1 points left and E2 points right Etotal E1 E2 +6nC x=0 x=0 2 Etotal -4nC 1050 NC E2 Q1 = -4nC x=0 Q2 = +6nC E1 b x x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 This is only the magnitude. The direction is away from a positive charge, and toward a negative one. For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 Etotal E1 E2 (9 109 Nm )(4 109 C) C2 2 (0.2m) (9 109 Nm )(6 109 C) C2 2 2 2 (1.2m) x=0.8m E2 Q1 = -4nC E1 a Q2 = +6nC x x=0.8m (0.6m) 1050 NC (9 109 Nm )(6 109 C) C2 E2 Q1 = -4nC x=0 Etotal E1 E2 Etotal x 2 For part b) E1 points left and E2 points right (9 109 Nm )(4 109 C) C2 +6nC x=0 x=0 2 Etotal -4nC Q2 = +6nC E1 b x x=0.8m 2 2 (0.4m) 312.5 NC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 -4nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x=0 For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 Q1 = -4nC x=0 2 Etotal 2 (0.2m) (9 109 Nm )(6 109 C) C2 2 2 (1.2m)2 E1 a Q2 = +6nC x x=0.8m (0.6m) 1050 NC E2 Q1 = -4nC x=0 Etotal E1 E2 Etotal x=0.8m 2 For part b) E1 points left and E2 points right (9 109 Nm )(4 109 C) C2 x E2 Etotal E1 E2 (9 109 Nm )(4 109 C) C2 +6nC (9 109 Nm )(6 109 C) C2 2 (0.4m)2 For part b) E1 points right and E2 points left 312.5 NC Q2 = +6nC E1 b x x=0.8m E2 E1 c Q1 = -4nC x=0 Q2 = +6nC x x=0.8m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 E kQ charge is given by the formula: R2 -4nC This is only the magnitude. The direction is away from a positive charge, and toward a negative one. x=0 For part a) both E-field vectors point in the –x direction Call the -4nC charge #1 and the +6nC charge #2 Q1 = -4nC x=0 2 Etotal 2 (0.2m) (9 109 Nm )(6 109 C) C2 2 2 (1.2m)2 (0.6m) x=0 (9 109 Nm )(6 109 C) C2 2 (9 109 Nm )(4 109 C) C2 2 2 (0.2m) a Q2 = +6nC x x=0.8m E2 Q1 = -4nC (0.4m)2 312.5 NC (9 109 Nm )(6 109 C) C2 Q2 = +6nC E1 b x x=0.8m E2 E1 c For part b) E1 points right and E2 points left Etotal E1 E2 Etotal E1 1050 NC Etotal E1 E2 Etotal x=0.8m 2 For part b) E1 points left and E2 points right (9 109 Nm )(4 109 C) C2 x E2 Etotal E1 E2 (9 109 Nm )(4 109 C) C2 +6nC Q1 = -4nC x=0 Q2 = +6nC x x=0.8m 2 2 (1.0m) 846 NC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 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 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): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E1 2 E2 1 x Etotal = 0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E1 E2 2 1 x Etotal = 0 y Part b): both vectors point away from their charge. E1 2 1 x E2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E1 E2 2 1 x Etotal = 0 y Part b): both vectors point away from their charge. (9 109 Nm )(6 109 C) C2 2 E1 E2 2 (0.15m) 9 Nm2 C2 (9 10 2400 NC 2 (0.45m) E1 2 9 )(6 10 C) Positive x-direction 267 NC Positive x-direction 1 x E2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. E1 E2 2 1 x Etotal = 0 y Part b): both vectors point away from their charge. (9 109 Nm )(6 109 C) C2 2 E1 E2 2 (0.15m) 9 Nm2 C2 (9 10 2400 NC 2 (0.45m) E1 2 9 )(6 10 C) Positive x-direction 267 NC Positive x-direction 1 x E2 Etotal 2400 267 2667 NC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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) 2 (0.15,0) 1 x (0.15,- 0.4) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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) 2 (0.15,0) 1 x (0.15,- 0.4) E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. (9 109 Nm )(6 109 C) C2 y 2 E1 E1,x E1,y 2 (0.4m) 0 NC 337.5 NC 337.5 NC (- 0.15,0) 2 (0.15,0) 1 x (0.15,- 0.4) E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. (9 109 Nm )(6 109 C) C2 y 2 E1 E1,x E1,y 2 (0.4m) 0 NC 337.5 NC 337.5 NC (- 0.15,0) 2 (0.15,0) 1 x (0.15,- 0.4) E2 E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. (9 109 Nm )(6 109 C) C2 y 2 E1 E1,x E1,y E2 2 (0.4m) 0 NC 337.5 NC 9 Nm2 C2 (9 10 337.5 NC (- 0.15,0) 2 9 )(6 10 C) 2 (0.5m) 216 NC 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) 1 x 0.4m (0.15,- 0.4) 0.3m E2 E1,y Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. (9 109 Nm )(6 109 C) C2 y 2 E1 E1,x E1,y E2 2 (0.4m) 0 NC 337.5 NC 9 Nm2 C2 (9 10 337.5 NC (- 0.15,0) 2 9 )(6 10 C) 2 (0.5m) 216 NC 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) 1 x 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 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. (9 109 Nm )(6 109 C) C2 y 2 E1 E1,x E1,y E2 2 (0.4m) 0 NC 337.5 NC 9 Nm2 C2 (9 10 337.5 NC (- 0.15,0) 2 9 )(6 10 C) 2 (0.5m) 216 NC E2,x ( 216 N ) ( 3 ) 129.6 N C 5 C ) ( 4 ) 172.8 N E2,y ( 216 N C 5 C 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) 1 x 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 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. (9 109 Nm )(6 109 C) C2 y 2 E1 E1,x E1,y E2 2 (0.4m) 0 NC 337.5 NC 9 Nm2 C2 (9 10 337.5 NC (- 0.15,0) 2 9 )(6 10 C) 2 (0.5m) 216 NC E2,x ( 216 N ) ( 3 ) 129.6 N C 5 C ) ( 4 ) 172.8 N E2,y ( 216 N C 5 C 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) 1 x 0.4m Add together the x-components and the y-components separately: Etotal,x 0 NC 129.6 NC 129.6 NC (0.15,- 0.4) E2,x 0.3m E2,y E1,y Etotal,y 337.5 NC 172.8 NC 510.3 NC Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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. (9 109 Nm )(6 109 C) C2 y 2 E1 E1,x E1,y E2 2 (0.4m) 0 NC 337.5 NC 9 Nm2 C2 (9 10 337.5 NC (- 0.15,0) 2 9 )(6 10 C) 2 (0.5m) 216 NC E2,x ( 216 N ) ( 3 ) 129.6 N C 5 C ) ( 4 ) 172.8 N E2,y ( 216 N C 5 C 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. Add together the x-components and the y-components separately: Etotal,x 0 NC 129.6 NC 129.6 NC (0.15,0) 1 x (0.15,- 0.4) 75.7º Etotal Etotal,y 337.5 NC 172.8 NC 510.3 NC Now find the magnitude and the angle using right triangle rules: Etotal (129.6)2 (510.3)2 526.5 NC tan() 510.3 75.7 below x axis 129.6 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 d): TRY THIS ONE ON YOUR OWN FIRST... y (0,0.2) (- 0.15,0) 2 (0.15,0) 1 x Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 d): both vectors point away from their charge. We will need to use vector components to add them together. E1 9 Nm2 C2 (9 10 9 )(6 10 C) 2 (0.25m) 864 NC ) 518.4 NC E1,x (864 NC )( 00..15 25 ) 691.2 NC E1,y (864 NC )( 00..20 25 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. y E1 E2 (0,0.2) (- 0.15,0) 2 (0.15,0) 1 x From symmetry, we can see that E2 will have the same components, except for +/- signs. E2,x (864 N )( 0.15 ) 518.4 N C 0.25 C N )( 0.20 ) 691.2 N E ( 864 C 0.25 C 2,y Now we can add the components (the x-component should cancel out) Etotal,x 518.4 NC 518.4 NC 0 NC Etotal,y 691.2 NC 691.2 NC 1382.4 NC 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