1. 23.4 The Electric Field

2. 23.4 The Electric Field

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

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

5. Opposite charges attract

6. 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.

7. 23.4 The Electric Field

9. Example 23.5 Electric Field Due to Two Charges

11. 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

12. 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

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

14. 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 = X 108 N/C
E2 = (9.0 X 109 N-m2/C2)(50 X 10-6 C) (0.080 m)2
E2 = X 107 N/C
E = E1 + E2 = 6.3 X 108 N/C

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

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

17. 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

18. 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 = X 105 N/C (down)

19. Ex = E2x Ex = 1.08 X 105 N/C (to the right) Ey = E1 + E2y
Ey = 3.94 X 105 N/C 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

20. 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 = mC
Q2 = mC

21. 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 = mC
Q2 = mC