General Physics II Spring Electric Forces and Fields

Transcription

1 General Physics II Spring 2008 Electric Forces and Fields 1

2 Coulomb s Law 2

3 The direction of the electric force is always along the line joining the two charges. Charges of the same sign repel; charges of opposite sign attract. Coulomb s Law 3

4 Checking Understanding Two uniformly charged spheres are firmly fastened to and electrically insulated from frictionless pucks on an air table. The charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces? 4

5 Superposition of Forces Electric forces obey the superposition principle: If a charge interacts with several other charges, then the net force on this charge is the vector sum of the forces that the other charges would exert on it individually (all charges are at rest). F = F + F + F +... net,1 2 on 1 3 on 1 4 on 1 Electric forces between ordinary charged objects are quite small in magnitude (fraction of a newton) because the net charge on such objects is small. These amounts of charge are usually of the order of microcoulombs (1 μc = 10-6 C) or nanocoulombs (1 nc = 10-9 C). 5

6 Checking Understanding All charges in the diagrams below are of equal magnitude. In each case, a small, positive charge is placed at the black dot. In which case is the magnitude of the force on the small, positive charge the smallest? 6

7 Chapter 20, Problem 47 F F 3 on 1 2 on

8 Chapter 20, Problem 47 y (F 3 on 1 ) y (F 2 on 1 ) y F F 3 on 1 2 on 1 60 (F 3 on 1 ) x 60 (F 2 on 1 ) x x Force-vector Diagram 8

9 Chapter 20, Problem 47 P Prepare: The charges are point charges. Please refer to Figure P The electric force on charge q 1 is the vector sum of the forces F 2 on 1 and F 3 on1, where q 1 is the 1 nc charge, q 2 is the left 2 nc charge, and q 3 is the right 2 nc charge. Solve: We have F K q q =, away from q r (1 10 m) on ( Nm /C )(1 10 C)(2 10 C) = 2 2, away from q 2 4 = ( N, away from q2) 4 4 F2 on 1 x = = 4 4 ( F2 on 1) y = ( N)(sin60 = ( ) ( N)(cos60 ) ( N) ) ( N) K q q F =, away from q = ( N, away from q ) on r2 4 4 ( F3 on 1) x = ( N)(cos60 ) = ( N) 4 4 ( F3 on1) y = ( N)(sin60 ) = ( N) ( Fon 1) x= ( F2 on 1) x+ ( F3 on 1) x= 0 4 ( F ) = ( F ) + ( F ) = N on 1 y 2 on 1 y 3 on 1 y So the force on the 1 nc charge is N directed upward. Assess: The magnitude and symmetry of q 2 and q 3 ensure that their x-component of the net force is zero. 9

10 The Electric Field How can two charges exert forces on each other across empty space? Answer: Every charge modifies the space around it, creating an electric field. When another charge enters this field, it experiences an electric force. Thus, the electric field is the intermediary for the exertion of the electric force. Consider a test charge (or probe charge) q' placed at some location in space where an electric field is present. The electric field at the location of the test charge is defined to be the electric force per unit charge at that point in space: E = F on q q. 10

11 The Electric Field Note that the electric field, which is created by the source charges, exists independently of the test charge. Any point charge q' placed in an existing electric field will experience an electric force F = qe. If the charge is positive, the force is in the same direction as the field. If the charge is negative, the force is in the opposite direction to that of the field. Thus, the direction of the electric field at any point is the direction of the force on a positive test charge placed at that point. 11

12 Visualizing the Electric Field 12

13 The Electric Field of a Single Point Charge Let the source of the electric field be the source charge q. A test charge q is placed at a distance r from the source charge. By Coulomb s law, the magnitude of the force on the test charge is q q Fon q = K r2. The magnitude of the electric field at the location of the test charge is therefore F E = on q q E = K r 2, i.e.,. (point charge) The direction of the electric field is away from the source charge if it is positive and toward it if it is negative. q 13

14 Electric Field of a Point Charge 14

15 Workbook: Chapter 20, Question 23 15

16 Uniform Electric Fields A uniform electric field has the same value (magnitude and direction) everywhere in a region of space. The field field is very nearly uniform between the plates of a parallel-plate capacitor. 16

17 Superposition of Electric Fields Because the electric force exerted on a point charge obeys the Principle of Superposition, so does the electric field at the position of the charge. Thus, the total electric field at a point is simply the vector sum of the individual electric field values at that point: E= E + E + E Problem-solving involving superposition of electric fields due to several point charges is very similar to problems involving the superposition of forces. Remember that: The electric field due to a single positive charge points away from it. The electric field due to a single negative charge points toward it. 17

18 Checking Understanding All charges in the diagram below are of equal magnitude. In each of the four cases below, two charges lie along a line, and we consider the electric field due to these two charges at a point along this line represented by the black dot. In which of the cases below is the field to the right? 18

19 Chapter 20 Problem 46 y 1 (E 2 ) x θ E 1 x E 2 (E 2 ) y 2 19

20 Chapter 20 Problem 46 P Prepare: The electric field is that of the two 1 nc charges located on the y-axis. Please refer to Figure P We denote the top 1 nc charge by q 1 and the bottom 1 nc charge by q 2. The electric field E 1 of the positive charge is directed away from the charge and the electric field E 2 due to the negative charge is directed toward it. With vector addition, they yield the net electric field E net at the point P indicated by the dot. Solve: The electric fields from q 1 and q 2 are q1 ( Nm /C )(1 10 C) E1 = K, along + x-axis, along x-axis 2 = + 2 r1 (0.05 m) = (3600 N/C, along + x-axis) q2 E2 = K 2, θ below x -axis = (720 N/C, θ below x-axis) r2 Because tan θ = 10 cm/5 cm, θ = tan 1 (2) = We will now calculate the components of these electric fields. The electric field due to q 1 is away from q 1 along + x and that due to q 2 is toward q 2 in the third quadrant. Their components are The x and y components of the net electric field are Thus, the strength of the electric field at P is E 1x = E 1 E 1y = 0 E 2 x = E 2 cos63.45 E 2 y = E 2 sin63.45 (E ne t ) x = E 1x + E 2 x = E 1 E 2 cos63.45 = 3278 N/C (E ne t ) y = E 1y + E 2 y = 0 E 2 sin63.45 = 644 N/C E net = (3278 N/C) 2 + ( 644 N/C) 2 = 3300 N/C To find the angle this net vector makes with the horizontal, we calculate tan φ = (E ) net y (E net x ) = 644N/C φ = N/C Thus, the strength of the net electric field at P is 3300 N/C and E net makes an angle of 11 below the +x-axis. Assess: Because of the inverse square dependence on distance, E 2 < E 1. Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis. 20