Three spiders are resting on the verticals of a triangular web. The sides of the
ID: 250380 • Letter: T
Question
Three spiders are resting on the verticals of a triangular web. The sides of the triangular web have a length of a= 0.40 m, as depicted in the figure. Two of the spiders (S_1 and S_3) have +8.8 mu C charge, while the other (S_2) has -8.8 mu C charge. What are the magnitude and direction of the force on the third spider (S_3)? magnitude N direction counterclockwise from the +x-axis 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 is resting at the origin? magnitude N direction counterclockwise from the +x-axisExplanation / Answer
Determine the directions of the E fields at S3
Due to S1 the field E1 is at -30° or 330°
Due to S2 the field E2 is at 210°
E = kq/r²
r² = 0.4² = 0.16m
E1 = k*8.8e-6/0.16 = E2 = 495000 V/m (Note: V/m = J/Cm = Nm/Cm = N/C)
Sum the horizontal components
E1*cos330 + E2cos*210 = 0
Sum the vertical components
E1*sin330 + E2sin*210 = -495000 V/m
The force on S3 is qE = 8.8e-6*495000 =4.356N at 270°
(b) less than in part (a)
(c) It is easy as they are in a straight line. like repels like and opposite attracts. thus they are going in the same direction, no x component , so just add the forces obtained by coulombs law, but remember that distance r is half what it was originally.
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