Particles of charge Q 1 = +66·µC, Q 2 = +50·µC, and Q 3 = -80·µC are placed in a
ID: 3280334 • Letter: P
Question
Particles of charge Q1 = +66·µC, Q2 = +50·µC, and Q3 = -80·µC are placed in a line. The center charge is 0.35·m from each of the others, as shown if the figure below.
(a) Find the force exterted on Q1 by the other two charges. Take right as the positive direction.
N
(b) Charge Q3 is moved left or right until the net force on it (due to the other charges) is zero.
Where will Q3 be? Choose one answer only.
to the left of Q1
to the right of Q2
between Q1 and Q2
How far will it be from charge Q1? m
Explanation / Answer
Let the x axis lie along the charges. We will calculate the x component of the forces and the other components vanish
Apply Coulomb's law to find the forces
F1x = F12x + F13x
F1x = -k | Q1Q2 | / (0.35 m)2 + k | Q1Q3 | / (0.70 m)2
Pluging given values
F1x = [(9.0 x 109 Nm2/C2) / (0.70 m)2] x (66 µC) [-4 (50µC) + 80 µC]
F1x = -145.46 N (force is to the left, indicated by negative value).
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Force on Q2
F2x = F21x + F23x
F2x = k | Q1Q2 | / (0.35 m)2 + k | Q2Q3 | / (0.35 m)2
F2x = [(9.0 x 109 Nm2/C2) / (0.35 m)2] x (50 µC) [ (66µC) + 80 µC]
F2x = 536.32 N (force is to the right, indicated by positive value)
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Force on Q3:
F3x = F31x + F32x = -k | Q1Q3 | / (0.70 m)2 - k | Q2Q3 | / (0.35 m)2
F3x = [(9.0 x 109 Nm2/C2) / (0.70 m)2] x (80 µC) [ (-66µC) - 50 µC x 4]
F3x = 390.85 N (force is to the left, indicated by positive value)