Question

In: Physics

Three charged particles form a triangle: particle 1 with charge Q1 = 100.0 nC is at...

Three charged particles form a triangle: particle 1 with charge Q₁ = 100.0 nC is at xy coordinates (0, 5.70 mm), particle 2 with charge Q₂ is at (0, -5.70 mm), and particle 3 with charge q = 50.0 nC is at (6.90 mm, 0). What are (a) the x component and (b) the y component of electrostatic force on particle 3 due to the other two particles if Q₂ is equal to 100.0 nC? What are (c) the x component and (d) the y component of electrostatic force on particle 3 due to the other two particles if Q₂ is equal to -100.0 nC, ?

Solutions

Expert Solution

Force on particle three due to Q₁

$F_{A} = \frac{kQ_{1}Q_{3}}{AC^{2}}$

Force on particle three due to Q₂

Where

$AC = BC = \sqrt{(5.70)^{2}+(6.90)^{2}}= 8.95 mm$

When Q₂ = 100 nC

$F_{A}= F_{B} = \frac{9\times 10^{9}\times 100\times 10^{-9}\times 50\times 10^{-9}}{(8.95\times 10^{-3})^{2}}= 0.562 N$

Angle $\angle ABC = \angle BAC = \tan^{-1}\left ( \frac{6.90mm}{5.70 mm} \right )= 50.44^{0}$

Now

Part A) the x component

$F_{x} = F_{AX}+F_{BX}$

$F_{x} = F_{A}\sin 50.44^{0}+F_{B}\sin 50.44^{0}$

$F_{x} = 0.562 N\sin 50.44^{0}+0.562 N\sin 50.44^{0} = 0.867 N$

Part B) The Y component

$F_{y} = F_{Ay}-F_{By}$

$F_{y} = F_{A}\cos 50.44^{0}-F_{B}\cos50.44^{0}$

$F_{y} = 0.562 N\cos50.44^{0} -0.562 N\cos50.44^{0} = 0 N$

Part C) X component when Q₂ = -100 nC

$F_{x} = F_{AX}-F_{BX}$

$F_{x} = F_{A}\sin 50.44^{0}-F_{B}\sin 50.44^{0}$

$F_{x} = 0.562 N\sin 50.44^{0}+0.562 N\sin 50.44^{0} = 0 N$

Part B) The Y component when Q₂ = 100 nC

$F_{y} = F_{Ay}-F_{By}$

$F_{y} = F_{A}\cos 50.44^{0}-F_{B}\cos50.44^{0}$

$F_{y} = 0.562 N\cos50.44^{0} -0.562 N\cos50.44^{0} = 0 N$

genius_generous answered 2 years ago

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