Three spiders are resting on the vertices of a triangular web. The sides of the

Question

Three spiders are resting on the vertices of a triangular web. The sides of the triangular web have a length of a = 0.81 m, as depicted in the figure. Two of the spiders (S_1 and S_3) have +5.6 Mu C charge, while the other (S_2) has -5.6 Mu C charge. What are the magnitude and direction of the net force on the third spider (S_3)? Suppose the third spider (S_3) moves to the origin. Would the net force on the third spider (S_3) be greater than, less than, or equal to the magnitude found in part (a)? greater than in part (a) less than in part (a) equal to the part (a) What are the magnitude and direction of the net force on the third spider (S_3) when it is resting at the origin?

Explanation / Answer

As given in the question,

q1 = 5.6 C = 5.6*10^-6 C, q2 = - 5.6 C = - 5.6*10^-6 C,

q3 =  5.6 C = 5.6*10^-6 C and a = 0.81 m

(a)   For the net force on the spider S3,

F(13) = k*q1*q3 / a^2 = (8.9*10^9)*(5.6*10^-6)*(5.6*10^-6) / (0.81)^2 = 0.425 N

F(23)  = k*q2*q3 / a^2 = (8.9*10^9)*(-5.6*10^-6)*(5.6*10^-6) / (0.81)^2 = -0.425 N

Here the forces F(13) and F(23) are at an angle of 120°,

So, the magnitude of the resultant force: F = |F(13)| = |F(23)| = 0.425 N

The required angle of the resultant force F:    = 270°

(b)   The correct answer will be:   Option (a) greater than in part (a)

(c) For the net force on the spider S3 for the new position,

The distance between charges in the new position: r = a/2 = 0.81/2 = 0.405 m

F(13)  = k*q1*q3 / r^2 = (8.9*10^9)*(5.6*10^-6)*(5.6*10^-6) / (0.405)^2 = 0.106 N

F(23)  = k*q2*q3 / a^2 = (8.9*10^9)*(-5.6*10^-6)*(5.6*10^-6) / (0.81)^2 = -0.106 N

The magnitude of the resultant force,

F = |F(13)| + |F(23)| = 0.212 N

The required angle of the resultant force F:    = 270°

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