Problem 7)
The expression for the electric force
F = k q1q2 / r2
The net force about charge -aq
F = F12 – F32
calculate the distance
F12 = k q1 q2 / r122
r12 = 8 -3 = 5 cm = 5 10-2 m
r32 = 11-8 = 3 cm = 3 10-2 m
calculate the force
Data
q = 5 μC = 5 10-6 C
a = 7.50
F12 = k q1 q2 / r122
F12 = 8.987 109 5 10-6 (7.50 5 10-6 )/ (5 10-2)2
F12 = 6.74 102 N
F32 = k q3 q2 / r322
F32 = 8.987 109 5 10-6 (7.50 5 10-6 )/ (3 10-2)2
F32 = 1.87 103 N
The net force is
F = 6.74 102 N - 1.87 103 N
F = -1198 N = -1.20 N
The negative sign indicates that the force is directed to the left
Problem 8
The expression for the electric field is
E = k q / r2
r = 25 cm = 0.25 m
Eb = 8.987 109 10 10-6 /0.252
Eb = 1.44 106 N/C
Ec= 8.987 109 5 10-6 / 0.252
Ec = 7.19 105 N/C
These field are in the line of the triangle we use trigonometry to find components
Cos 30 = Eby/Eb
Eby = Eb Cos 30
Eby = 1.44 106 Cos 30
Eby = 1.247 106 N/C
Sin 30 = Ebx /Eb
Ebx = Eb Sin 30
Ebx = 1.44 106 Sin 30
Ebx = 0.72 106 N/C
Ecy = Ec Cos 30
Ecy = 6.227 105 N/C
Ecx = Ec Sin 30
Ecx = 3.595 105 N/C
The net field is the sum of each axis
Ex = Ebx + Ecx
Ex = 0.72 106 + 3.595 105
Ex =10.795 105 N/C
Ey = Eby – Ecy
Ey = 12.47 105 - 6.227 105
Ey = 6.243 105 N/C
E = sqrt (Ex2 + Ey2)
E = 105 sqrt (10.7952 + 6.2432 )
E = 12.47 105 N/C
Direction
Tan θ = Ey/Ex
Tan θ = 6.243/10.795
θ = 30
Part b)
F = q E
F = 1.5 10-9 12.47 105
F = 1.87 10-3 N
As the charge is positive the direction does not change
θ = 30
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