23.4 the electric field. proton + + a positive test charge would be repelled by the field 23.4 the...
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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