23.4 The Electric Field

23.4 The Electric Field

Proton

+

+A positive test charge

would be repelled by

the field

23.4 The Electric 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?

- +Q1 = -25 mC Q2 = +50 mC

P

2 cm 8 cm

- +

Q1 = -25 mC Q2 = +50 mC

P2 cm 8 cm

E2E1

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.

-+

Q1 = 7.00 mC Q2 = -5.00 mC

0.300 m

0.400 m

P

-+

Q1 = 7.00 mC Q2 = -5.00 mC

0.300 m

0.400 m

P

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

c = 0.500 m

q

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

-+

P

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)

q

E1

E2

E2x

E2y

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

Pf

Electric Field: Example 5

Calculate the electric field at point A, as shown in the diagram

-+

Q1 = +50.0 mC Q2 = -50.0 mC

52 cm

30 cm

A

Ans: E = 4.5 X 106 N/C at an angle of 76o

Electric Field: Example 6

Calculate the electric field at point B, as shown in the diagram.

-+

Q1 = +50.0 mC Q2 = -50.0 mC

26 cm

30 cm

B

26 cm

Ans: E = 3.6 X 106 N/C along the +x direction