11:30 PM Sun Jan 21 < 88 0 10... x Exam-1-Key-2... A+B: 8=-30 C+D: =6 Lesson...

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11:30 PM Sun Jan 21 < 88 0 10... x Exam-1-Key-2... A+B: 8=-30 C+D: =6 Lesson 02 Problem Set 1Completed 3 Lesson 02 Pro... SG-1-Due_Ja... T 49% SG-1-Due_Ja... Lesson 02... 5. If a positive test charge is placed in a uniform field, the test charge will move in the same direction as the field. If a negative test charged is placed in a uniform field, the test charge will move against the field. Explain why this happens (3 points). This occurs because a uniform field means that the strength is the same throughout the field, and the test charges have the same potential for force. When a positive charge is placed, the test charge will move in the same direction because they are attracted by the negative in the field that are also moving in that direction. When a negative charge is placed, repulsion occurs from the negative charges and it moves against the field because it is experiencing force in the opposite direction. 6. Particle A is placed at position (3, 3) m, particle B is placed at (-3, 3) m, particle C is placed at (-3, -3) m, and particle D is placed at (3, -3) m. Particles A and B have a charge of -q and particles C and D have a charge of +2q. 9=-3 A+B=3C a) Draw a properly labeled coordinate plane with correctly placed and labeled charges (3 points). B(-313) A (3,3) q 0 (31-3) 29 29 ( -3,-3) b) Draw and label a vector diagram showing the electric field vectors at position (0, 0) m (3. points). EA points). Xdirections cancel out = +6C c) Solve for the magnitude and direction of the net electric field strength at position (0, 0) m (7 r =3 +3 Ec=ED EA = EB = 1/2 Ec E = K 12 24.24 m = (ax109 Nm (c) (6x10-5) EA = 1501.SN = EB (4-24m)2 Ec = 3003 N/C ED = E= {Ey = (1501.5 NIC + 15.01.5N/C + 3003.74 NIC + 3003.74 NIC) (sinus) ZE = 6.37x103 NIC @ 90 9=640 29 =12C 11:30 PM Sun Jan 21 < 88 0 10... x Exam-1-Key-2... A+B: 8=-30 C+D: =6 Lesson 02 Problem Set 1Completed 3 Lesson 02 Pro... SG-1-Due_Ja... T 49% SG-1-Due_Ja... Lesson 02... 5. If a positive test charge is placed in a uniform field, the test charge will move in the same direction as the field. If a negative test charged is placed in a uniform field, the test charge will move against the field. Explain why this happens (3 points). This occurs because a uniform field means that the strength is the same throughout the field, and the test charges have the same potential for force. When a positive charge is placed, the test charge will move in the same direction because they are attracted by the negative in the field that are also moving in that direction. When a negative charge is placed, repulsion occurs from the negative charges and it moves against the field because it is experiencing force in the opposite direction. 6. Particle A is placed at position (3, 3) m, particle B is placed at (-3, 3) m, particle C is placed at (-3, -3) m, and particle D is placed at (3, -3) m. Particles A and B have a charge of -q and particles C and D have a charge of +2q. 9=-3 A+B=3C a) Draw a properly labeled coordinate plane with correctly placed and labeled charges (3 points). B(-313) A (3,3) q 0 (31-3) 29 29 ( -3,-3) b) Draw and label a vector diagram showing the electric field vectors at position (0, 0) m (3. points). EA points). Xdirections cancel out = +6C c) Solve for the magnitude and direction of the net electric field strength at position (0, 0) m (7 r =3 +3 Ec=ED EA = EB = 1/2 Ec E = K 12 24.24 m = (ax109 Nm (c) (6x10-5) EA = 1501.SN = EB (4-24m)2 Ec = 3003 N/C ED = E= {Ey = (1501.5 NIC + 15.01.5N/C + 3003.74 NIC + 3003.74 NIC) (sinus) ZE = 6.37x103 NIC @ 90 9=640 29 =12C