University:
California State University, Northridge
Course:
PHYS 100B | General Physics II
Academic year:

2008
Views:

368

Pages:

25
Author:

winterenpih

Chapter 19 Electric Charges, Forces, and Fields Outline 19-1 Electric Charge 19-2 Insulators and Conductors 19-3 Coulomb’s Law 19-4 The Electric Field 19-5 Electric Field Lines 19-6 Shield and Charging by Induction 19-7 Electric Flux (and Gauss’s Law) 19-4 The Electric Field Since an electric charge apply a “force field” every around it (see Fig 19-9), it is very natural to define a Electric Field. Figure 19-9. A Electric Field JG Definition of Electric Field, E JG If a test charge q0 experiences a force F at a given location, the electric field at that location is JG JG F E= , q0 (19 − 8) SI unit: N/C Note: 1) The electric field is the force per charge at a given location. 2) The electric field is a vector. JG 3) A positive charge experiences a force in the direction of E . 4) A negative charge experiences a force in the opposite direction JG of E . 5) The magnitude of the force acting on the charge q is F = q E From the definition, one also has JG JG F = qE (19 − 9), JG where q is the charge, and F is the force applied on the charge. Example 19-33: Force Field A + 5.0 uC charge experiences a 0.44 N force in the positive y direction. If this charge is replaced with a -2.7 uC charge, what force it will experience? Picture the Problem (force directions) Solution: G 1. Find E using equation 19-8: G G F 0.44 N yˆ 4 E= = = × 8.8 10 N C ) yˆ ( −6 q 5.0 × 10 C 2. Use equation 19-9 to find the new force: G G F = qE = ( −2.7 ×10− 6 C )( 8.8 × 104 N/C ) yˆ = ( −0.24 N ) yˆ The Electric Field of a Point Force The simplest case is that electric field is created by a Point Charge, as shown in Fig. 19-10. Figure 19-10. The Electric Field of a Point . Suppose a positive point charge q is at the original. A positive test charge q0 is placed a distance r from the original, then the magnitude of the force it experiences is F =k q q0 r2 Applying the definition of the electric field, the magnitude of the field is q q0 (k 2 ) q F r E= = =k 2 q0 q0 r The electric field direction is radially outward. Magnitude of the Electric Field due to a Point Force A simplest case is that electric field is created by a Point Charge, as shown in Fig. 19-10. q E=k 2 r (19 − 10) • If q is positive, the electric field points radially outward; • If q is negetive, the electric field points radially inward; Superposition For case of more than two charges, the net electric field vector is JG JG JG JG E net = E1 + E 2 + E 3 + .... The X and Y components of the net electric field are Enet , x = E1, x + E2, x + E3, x + .... Enet , y = E1, y + E2, y + E3, y + .... The magnitude and direction of the net electric field are 2 2 Enet = Enet + E ,x net , y θ = tan −1 ( Enet , y / Enet , x ) Example 19-5: Superposition in the Electric Field Two charges, each equal to +2.90 uC, are placed at two corners of a square 0.500 on side, as shown in the figure. Find the magnitude and direction of the net electric field at a third of the square. Picture the problem. Example 19-5 Superposition in the Field Solution JG 1) Find the magnitude of the E1 E1 = k q ( 2r ) 2 (2.90 ×10−6 C ) = (8.99 ×10 N .m / C ) 2(0.500m) 2 9 2 = 5.21×104 N / C JG 2) Find the magnitude of the E 2 q E2 = k 2 (r ) (2.90 ×10−6 C ) = (8.99 ×10 N .m / C ) (0.500m) 2 9 2 = 1.04 ×105 N / C JG JG 3) Calculate the components of E1 and E 2 JG E1, x = E1 cos(45.0°) = 3.68 × 104 N / C JG E1, y = E1 sin(45.0°) = 3.68 ×104 N / C JG E 2, x = E2 cos(0°) = 1.04 ×105 N / C JG E 2, y = E2 sin(0°) = 0 N /C JG 4) Calculate the components of E net Enet , x = E1, x + E2, x = 3.68 ×105 + 1.04 × 105 = 1.41×105 N / C Enet , y = E1, y + E2, y = 3.68 ×104 + 0 = 3.68 × 104 N / C JG 5) The magnitude and the direction of the net field E net are 2 2 5 2 4 2 5 Enet = Enet + E = (1.41 × 10 ) + (3.68 × 10 ) = 1.46 × 10 N /C ,x net , y 4 3.68 × 10 ) = 14.6° θ = tan −1 ( Enet , y / Enet , x ) = tan −1 ( 5 1.41×10 19-5 Electric Field Lines Figure 19-14 Electric Field Lines for a Point Charge Rules for Drawing Electric Field Lines JG 1. Point in the direction of the electric field vector E at every point; 2. Start at positive (+) charges or at infinity; 3. End at negative (-) charge or at infinite; 4. The number of lines (density) JG is proportional to the magnitude of the field E . Examples of Electric Field of Two charges Figure 19-15a Electric Field Lines for Systems of Charges (+q and –q) Figure 19-15bc Electric Field Lines for Systems of Charges (2q and –q) (+q and +q) Figure 19-17 A parallel-plate capacitor (Electric Capacitor) Example 19-35: An object with a charge of -3.6 uC and a mass of 0.012 kg experience an upward electric force that is equal in magnitude to its weight, due to the uniform electric field created by the parallel-plate capacitor. Find the direction and magnitude of the electric field. Picture the Problem: An electric charge experiences an upward force due to the presence of an electric field. The negative charge experiences an upward electric force, so we can conclude the electric field points downward. Solution: G 1. (a) Set FE = mg G and solve for E mg ( 0.012 kg )) ( 9.81 m/s ) 4 q E = mg ⇒ E = 3.3 10 N/C = = × −6 q 3.6 × 10 C G E = E ( −yˆ ) = ( −3.3 × 104 N/C ) yˆ 2 Summary 1) Electric Field JG JG F E= , q0 (19 − 8) For point charge q E=k 2 r (19 − 10) 2) Superposition of the Electric Field: vector calculation. 3) Electric Field Lines