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 = 0.68 m, as depicted in the figure. Two of the spiders (S1 and S3) have +4.7 µC charge, while the other (S2) has 4.7 µC charge.

(a) What are the magnitude and direction of the net force on the third spider (S3)?

magnitude in ===N

Direction= got that it would be 270 degrees

B) i got this answer

C)What are the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin?

Explanation / Answer

here,

charge on s1 , q1 = 4.7 uC

q1 = q3 = 4.7 * 10^-6 C

charge on s2 , q2 = - 4.7 * 10^-6 C

net force on s3 due to s1 , f13 = k * q1 * q3 / a^2

f13 = 9 * 10^9 * 4.7*10^-6 * 4.7 * 10^-6 /0.68^2

f13 = 0.43 N

due to symmetry , force on s2 due to s3 , f23 = f13 = 0.43 N

but the direction is different

the angle between f23 and f13 is theta = 120 degree

magnitude of net force on s3 , fnet = 2 * f * cos(theta/2)

fnet = 0.43 * 2 * cos(60)

fnet = 0.43 N

the magnitude and direction of the net force on the third spider is 0.43 N counterclockwise 270 degree from x- axis

C)

if spider S3 is at origin

due to symmetry,

net force on S3 , Fnet = 2 * k * q1 * q3 /(a/2)^2

Fnet = 9 * 10^9 * 4.7*10^-6 * 4.7 * 10^-6/(0.34)^2

Fnet = 1.72 N

the net force on the spider S3 is 1.72 N and the direction is 270 degree counterclockwise from the + x axis

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