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olga2289 [7]
3 years ago
12

Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot

al force on the particle with charge q is

Physics
1 answer:
Gelneren [198K]3 years ago
5 0

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = \frac{kQq}{r^{2} }

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

N_{x} = N cosine 60° + (-N cosine 60°) = 0

for the y-component

N_{y} = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero (N_{x} = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

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(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

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from the question we are given the following values:

mass (m) = 1.6 kg

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(A) initial potential energy of the object = final potential energy of the spring

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              mg(1.05 + l)  = 0.5 x k x l^{2}

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mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

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         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

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      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

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        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

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