Electric Field Examples. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
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- Scott Haynes
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1 Physics 6B lectric Field amples For ampus Learning Assistance Services at USB
2 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. For ampus Learning Assistance Services at USB
3 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. q q -0.3m 0 0.m For ampus Learning Assistance Services at USB
4 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. q q -0.3m 0 0.m For ampus Learning Assistance Services at USB
5 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q will produce an -field vector that points left, and q gives an -field vector to the right. q q -0.3m 0 0.m For ampus Learning Assistance Services at USB
6 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q will produce an -field vector that points left, and q gives an -field vector to the right. q q -0.3m 0 0.m This is how we can put the +/- signs on the -fields when we add them up. + For ampus Learning Assistance Services at USB
7 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q will produce an -field vector that points left, and q gives an -field vector to the right. q q -0.3m 0 0.m This is how we can put the +/- signs on the -fields when we add them up. + 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(5 0 + (0.3m) ) For ampus Learning Assistance Services at USB
8 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q will produce an -field vector that points left, and q gives an -field vector to the right. q q -0.3m 0 0.m This is how we can put the +/- signs on the -fields when we add them up. + 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(5 0 + (0.3m) ) (This means 400 / in the negative -direction) For ampus Learning Assistance Services at USB
9 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q will produce an -field vector that points left, and q gives an -field vector to the right. q q 3 q -0.3m 0 0.m This is how we can put the +/- signs on the -fields when we add them up. + 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(5 0 + (0.3m) ) (This means 400 / in the negative -direction) 400 For part b) all we need to do is multiply the -field from part a) times the new charge q 3. For ampus Learning Assistance Services at USB
10 7. Two point charges are located on the -ais as follows: charge q +4 n at position 0.m and charge q +5 n at position -0.3m. a) Find the magnitude and direction of the net electric field produced by q and q at the origin. b) Find the net electric force on a charge q 3-0.6n placed at the origin. The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. At the origin, q will produce an -field vector that points left, and q gives an -field vector to the right. F on3 q q 3 q -0.3m 0 0.m This is how we can put the +/- signs on the -fields when we add them up. + 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(5 0 + (0.3m) ) (This means 400 / in the negative -direction) 400 For part b) all we need to do is multiply the -field from part a) times the new charge q 3. F onq 3 ( )( 400 ) ote that this force is to the right, which is opposite the -field This is because q 3 is a negative charge: -fields are always set up as if there are positive charges. For ampus Learning Assistance Services at USB
11 7.8 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? For ampus Learning Assistance Services at USB
12 7.8 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? kqq The formula for electric force between charges is F elec R For ampus Learning Assistance Services at USB
13 7.8 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? kqq The formula for electric force between charges is F elec R If both charges are doubled, we will have F elec k(q )(q ) R kq q 4 R For ampus Learning Assistance Services at USB
14 7.8 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? kqq The formula for electric force between charges is F elec R If both charges are doubled, we will have So the new force is 4 times as large. F elec k(q )(q ) R kq q 4 R For ampus Learning Assistance Services at USB
15 7.9 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? For ampus Learning Assistance Services at USB
16 7.9 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? kqq The formula for electric force between charges is F elec D For ampus Learning Assistance Services at USB
17 7.9 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 charges is F elec kq q D We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. kqq 3 D kq q D new For ampus Learning Assistance Services at USB
18 7.9 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 charges is F elec kq q D We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. kqq 3 D kq q D new anceling and cross-multiplying, we get D new 3 D For ampus Learning Assistance Services at USB
19 7.9 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 charges is F elec kq q D We want the force to be 3 times as strong, so we can set up the force equation and solve for the new distance. kqq 3 D kq q D new anceling and cross-multiplying, we get D new 3 D Square-roots roots of both sides gives us the answer: Dnew 3 D For ampus Learning Assistance Services at USB
20 7.30 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 /5 of this value, how far (in terms of d) should the charges be released? For ampus Learning Assistance Services at USB
21 7.30 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 /5 of this value, how far (in terms of d) should the charges be released? Recall that ewton's nd law says that F net ma. So this is really a problem about the force on the heavier charge. For ampus Learning Assistance Services at USB
22 7.30 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 /5 of this value, how far (in terms of d) should the charges be released? Recall that ewton's 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 charges is kq q F elec d For ampus Learning Assistance Services at USB
23 7.30 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 /5 of this value, how far (in terms of d) should the charges be released? Recall that ewton's 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 charges is If we want the acceleration to be /5 as fast, we need the force to be /5 as strong: F new kq q d new 5 F 5 old kq q d kq q F elec d For ampus Learning Assistance Services at USB
24 7.30 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 /5 of this value, how far (in terms of d) should the charges be released? Recall that ewton's 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 charges is If we want the acceleration to be /5 as fast, we need the force to be /5 as strong: F new kq q d new 5 F 5 old kq q d kq q F elec d We cancel common terms and cross-multiply to get d new 5 d For ampus Learning Assistance Services at USB
25 7.30 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 /5 of this value, how far (in terms of d) should the charges be released? Recall that ewton's 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 charges is If we want the acceleration to be /5 as fast, we need the force to be /5 as strong: F new kq q d new 5 F 5 old kq q d kq q F elec d We cancel common terms and cross-multiply to get d new 5 d Square-root of both sides: d new 5 d For ampus Learning Assistance Services at USB
26 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm For ampus Learning Assistance Services at USB
27 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm -4n +6n 0 0.8m For ampus Learning Assistance Services at USB
28 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4n +6n 0 0.8m For ampus Learning Assistance Services at USB
29 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. For part a) which direction do the -field vectors point? -4n +6n 0 0.8m -4n +6n a 0 0.8m For ampus Learning Assistance Services at USB
30 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4n +6n 0 0.8m For part a) both -field vectors point in the direction all the -4n charge # and the +6n charge # Q -4n a Q +6n 0 0.8m For ampus Learning Assistance Services at USB
31 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4n +6n 0 0.8m For part a) both -field vectors point in the direction all the -4n charge # and the +6n charge # Q -4n a Q +6n 0 0.8m 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(6 0 (0.6m) ) 050 For ampus Learning Assistance Services at USB
32 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4n +6n 0 0.8m For part a) both -field vectors point in the direction all the -4n charge # and the +6n charge # Q -4n a Q +6n 0 0.8m 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(6 0 (0.6m) For part b) points left and points right + ) 050 Q -4n Q +6n 0 0.8m b For ampus Learning Assistance Services at USB
33 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4n +6n 0 0.8m For part a) both -field vectors point in the direction all the -4n charge # and the +6n charge # Q -4n a Q +6n 0 0.8m 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(6 0 (0.6m) For part b) points left and points right + ) 050 Q -4n Q +6n 0 0.8m b (9 0 9 m )(4 0 (.m) 9 m ) (9 0 )(6 0 + (0.4m) ) +3.5 For ampus Learning Assistance Services at USB
34 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4n +6n 0 0.8m For part a) both -field vectors point in the direction all the -4n charge # and the +6n charge # Q -4n a Q +6n 0 0.8m 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(6 0 (0.6m) For part b) points left and points right + ) 050 Q -4n Q +6n 0 0.8m b 9 m (9 0 )(4 0 (.m) 9 m ) (9 0 )(6 0 + (0.4m) For part b) points right and points left ) +3.5 Q -4n c Q +6n 0 0.8m For ampus Learning Assistance Services at USB
35 7.4 A point charge of -4 n is at the origin, and a second point charge of +6 n is placed on the -ais at 0.8m. Find the magnitude and direction of the electric field at the following points on the -ais: a) 0 cm; b).0m; c) -0cm The electric field near a single point charge is given by the formula: kq R This is only the magnitude. The direction is away from a positive charge, and toward a negative one. -4n +6n 0 0.8m For part a) both -field vectors point in the direction all the -4n charge # and the +6n charge # Q -4n a Q +6n 0 0.8m 9 m (9 0 )(4 0 (0.m) 9 m ) (9 0 )(6 0 (0.6m) For part b) points left and points right + 9 m (9 0 )(4 0 (.m) 9 m ) (9 0 )(6 0 + (0.4m) For part b) points right and points left + 9 m (9 0 )(4 0 + (0.m) 9 m ) (9 0 )(6 0 (.0m) ) ) ) Q -4n Q -4n c Q +6n 0 0.8m Q +6n 0 0.8m b For ampus Learning Assistance Services at USB
36 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) For ampus Learning Assistance Services at USB
37 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) y Part a): TRY DRAWIG TH -FILD VTORS O TH DIAGRAM For ampus Learning Assistance Services at USB
38 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) y Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. 0 For ampus Learning Assistance Services at USB
39 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) y Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. 0 Part b): both vectors point away from their charge. y For ampus Learning Assistance Services at USB
40 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) y Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. 0 Part b): both vectors point away from their charge. y 9 m (9 0 )(6 0 ) 400 (0.5m) 9 m (9 0 )(6 0 ) 67 (0.45m) Positive -direction Positive -direction For ampus Learning Assistance Services at USB
41 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) y Part a): both vectors point away from their charge. Since the distances and the charges are equal, the vectors cancel out. 0 Part b): both vectors point away from their charge. y 9 m (9 0 )(6 0 ) 400 (0.5m) 9 m (9 0 )(6 0 ) 67 (0.45m) Positive -direction Positive -direction For ampus Learning Assistance Services at USB
42 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (- 0.5,0) (0.5,0) (0.5,- 0.4) For ampus Learning Assistance Services at USB
43 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (- 0.5,0) (0.5,0) (0.5,- 0.4),y For ampus Learning Assistance Services at USB
44 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (9 0 )(6 0 9 m ,,y (0.4m) ) (- 0.5,0) (0.5,0) (0.5,- 0.4),y For ampus Learning Assistance Services at USB
45 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (9 0 )(6 0 9 m ,,y (0.4m) ) (- 0.5,0) (0.5,0) (0.5,- 0.4),y For ampus Learning Assistance Services at USB
46 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (9 0 )(6 0 9 m ,,y (0.4m) ) 9 m (9 0 )(6 0 ) 6 (0.5m) The 0.5m in this formula for is the distance to charge, using Pythagorean theorem or from recognizing a right triangle when you see it. (- 0.5,0) 0.4m (0.5,0) (0.5,- 0.4) 0.3m,y For ampus Learning Assistance Services at USB
47 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (9 0 )(6 0 9 m ,,y (0.4m) ) 9 m (9 0 )(6 0 ) 6 (0.5m) The 0.5m in this formula for is the distance to charge, using Pythagorean theorem or from recognizing a right triangle when you see it. (- 0.5,0) 0.4m 0.3m (0.5,- 0.4) (0.5,0),,y,y For ampus Learning Assistance Services at USB
48 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (9 0 )(6 0 9 m ,,y (0.4m) ) 9 m (9 0 )(6 0 ) 6 (0.5m) + 3 +, ( 6 ) ( ) ,y ( 6 ) ( ) The 0.5m in this formula for is the distance to charge, using Pythagorean theorem or from recognizing a right triangle when you see it. (- 0.5,0) 0.4m 0.3m (0.5,- 0.4),y (0.5,0),,y For ampus Learning Assistance Services at USB
49 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (9 0 )(6 0 9 m ,,y (0.4m) ) 9 m (9 0 )(6 0 ) 6 (0.5m) + 3 +, ( 6 ) ( ) ,y ( 6 ) ( ) The 0.5m in this formula for is the distance to charge, using Pythagorean theorem or from recognizing a right triangle when you see it. (- 0.5,0) 0.4m 0.3m (0.5,- 0.4),y (0.5,0), Add together the -components and the y-components separately:,,y ,y For ampus Learning Assistance Services at USB
50 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part c): both vectors point away from their charge. We will need to use vector components to add them together. y (9 0 )(6 0 9 m ,,y (0.4m) ) 9 m (9 0 )(6 0 ) 6 (0.5m) + 3 +, ( 6 ) ( ) ,y ( 6 ) ( ) The 0.5m in this formula for is the distance to charge, using Pythagorean theorem or from recognizing a right triangle when you see it. (- 0.5,0) (0.5,0) (0.5,- 0.4) 75.7º Add together the -components and the y-components separately:,,y ow find the magnitude and the angle using right triangle rules: (9.6) + (50.3) 56.5 tan( θ) θ 75.7 below+ ais For ampus Learning Assistance Services at USB
51 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part d): TRY THIS O O YOUR OW FIRST... y (0,0.) (- 0.5,0) (0.5,0) For ampus Learning Assistance Services at USB
52 7.4 A point charge of q+6 n is at the point (0.5m,y0m), and an identical point charge is placed at (-0.5m,0m), as shown. Find the magnitude and direction of the net electric field at: a) the origin (0,0); b) (0.3m,0m); c) (0.5m,-0.4m); d) (0m,0.m) Part d): both vectors point away from their charge. We will need to use vector components to add them together. 9 m (9 0 )(6 0 ) 864 (0.5m),,y (864 + (864 )( )( ) 58.4 ) From symmetry, we can see that will have the same components, ecept for +/- signs. +, (864 )( ) ,y (864 )( ) ow we can add the components (the -component should cancel out) The 0.5m in this formula is the distance to each charge using the Pythagorean theorem or from recognizing a right triangle when you see it. y (- 0.5,0) (0,0.) (0.5,0),,y The final answer should be 38.4 / in the positive y-direction. For ampus Learning Assistance Services at USB
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