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 = 0.89 m, as depicted in the figure. Two of the spiders (S1 and S3) have +6.8 pC charge, while the other (S2) has -6.8 HC charge. a/2 a/2 S2 (a) What are the magnitude and direction of the net force on the third spider (S3)? 0.742 How do you find the magnitude of the electric force? How do you find the total net force?N 225 Does the sign of a charge effect the direction of the electric field? counterclockwise from the +x-axis magnitude direction (b) Suppose the third spider (S3) moves to the origin. Would the net force on the third spider (S3) be greater than, less than, or equal to the magnitude found in part (a)? o greater than in part (a) less than in part (a) equal to the part (a) (c) What are the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin? magnitude direction 71920 How do you find the magnitude of the electric force? How do you find the total net force? N 270 counterclockwise from the +x-axis

Explanation / Answer

a)
magnitude of force on s3,

|F13| = |F23| = k*q^2/a^2

= 9*10^9*(6.8*10^-6)^2/0.89^2

= 0.525 N

here the angle between F13 and F23 is 120 degrees

so, Fnet = sqrt(F13^2 + F23^2 + 2*F13*F23*cos(120))

= F13

= 0.525 N

direction : 270 degrees

b) greater than in part a

c) magnitude of force on s3,

|F13| = |F23| = k*q^2/(a/2)^2

= 9*10^9*(6.8*10^-6)^2/(0.89/2)^2

= 2.10 N

Fnet = |F13| + |F23|

= 2.10 + 2.10

= 4.20 N

direction : 270 degrees

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