23.4 The Electric Field

23.4 The Electric Field

+ Proton 23.4 The Electric Field A positive test charge would be repelled by the field +

Electron A positive test charge would be attracted by the field + -

Opposite charges attract

23.4 The Electric Field This equation gives us the force on a charged particle placed in an electric field. If q is positive, the force is in the same direction as the field. If q is negative, the force and the field are in opposite directions.

23.4 The Electric Field

Example 23.5 Electric Field Due to Two Charges

Electric Field: Example 2 Calculate the magnitude and direction of an electric field at a point 30 cm from a source charge of Q = -3.0 X 10-6 C. E = kQ r2 E = (9.0 X 109 N-m2/C2)(3.0 X 10-6 C) (0.30 m)2 E = 3.05 X 105 N/C towards the charge

Electric Field: Example 3 Two point charges are separated by a distance of 10.0 cm. What is the magnitude and direction of the electric field at point P, 2.0 cm from the negative charge? 2 cm 8 cm + P - Q1 = -25 mC Q2 = +50 mC

E2 E1 2 cm 8 cm P - + Q1 = -25 mC Q2 = +50 mC

E = E1 + E2 (both point to the left) E = kQ r2 E1 = (9.0 X 109 N-m2/C2)(25 X 10-6 C) (0.020 m)2 E1 = 5.625 X 108 N/C E2 = (9.0 X 109 N-m2/C2)(50 X 10-6 C) (0.080 m)2 E2 = 7.031 X 107 N/C E = E1 + E2 = 6.3 X 108 N/C

Electric Field: Example 4 Charge Q1 = 7.00 mC is placed at the origin. Charge Q2 = -5.00 mC is placed 0.300 m to the right. Calculate the electric field at point P, 0.400 m above the origin. P 0.400 m 0.300 m + - Q1 = 7.00 mC Q2 = -5.00 mC

P 0.400 m c = 0.500 m q 0.300 m + - Q1 = 7.00 mC Q2 = -5.00 mC c2 = a2 + b2 c2 = (0.400 m)2 + (0.300 m)2 c = 0.500 m tan q = opp/adj = 0.400/0.300 q = 53.1o

E = kQ r2 E1 = (9.0 X 109 N-m2/C2)(7.00 X 10-6 C) (0.400 m)2 E1 = 3.94 X 105 N/C E2 = (9.0 X 109 N-m2/C2)(5.00 X 10-6 C) (0.500 m)2 E2 = 1.80 X 105 N/C

E2x = E2cosq = (1.80 X 105 N/C)(cos 53.1o) P E2 E2y q + - E2x = E2cosq = (1.80 X 105 N/C)(cos 53.1o) E2x = 1.08 X 105 N/C (to the right) E2y = E2sinq = (1.80 X 105 N/C)(sin 53.1o) E2y = -1.44 X 105 N/C (down)

Ex = E2x Ex = 1.08 X 105 N/C (to the right) Ey = E1 + E2y Ey = 3.94 X 105 N/C + -1.44 X 105 N/C Ey = 2.49 X 105 N/C ER2 = (1.08 X 105 N/C )2 + (2.49 X 105 N/C)2 ER = 2.72 X 105 N/C tan f = Ey/Ex = 2.49 X 105/ 1.08 X 105 f = 66.6o f P

Electric Field: Example 5 Calculate the electric field at point A, as shown in the diagram Ans: E = 4.5 X 106 N/C at an angle of 76o A 30 cm 52 cm + - Q1 = +50.0 mC Q2 = -50.0 mC

Electric Field: Example 6 Calculate the electric field at point B, as shown in the diagram. Ans: E = 3.6 X 106 N/C along the +x direction B 30 cm 26 cm 26 cm + - Q1 = +50.0 mC Q2 = -50.0 mC