Three spiders are resting on the vertices of a triangular web. The sides of the triangular web have a length of a = 0.85 m, as depicted in the figure. Two of the spiders (S1 and S3) have +7.7 µC charge, while the other (S2) has

Question

−7.7 µC
charge.
a)What are the magnitude and direction of the net force on the third spider
(S3)
c)What are the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin?

cant seem to get this one, what are the steps to do this?

F = k Q1 Q2/r^2

If all three sides are the same length a, we have an equilateral triangle with 60 degree interior angles
Q1=Q3 = Q
Q2 = -Q
find force on Q3
F due to Q1 in direction from Q1 through Q2 (repulsive)
F= k Q^2/a^2
F from Q2 on Q3 (attractive)
F= k Q^2 /a^2 same magnitude

so if x axis is along direction from Q1 to Q3 then
Fx = F - F cos 60
and if y is perpendiculat to x
Fy = F sin 60

Now move Q3 to the origin and do part c

so magnitude is (8.99E9*-7.7)/(.85)^2 ?

im still confused on distance?

Answer 1

To calculate the net force on the third spider (S3), we need to find the magnitudes of the forces between each pair of spiders and their respective directions.

First, let's find the force between S1 and S3, both charged with +7.7 µC:
F1 = k * Q1 * Q3 / r^2
F1 = (8.99E9) * (7.7E-6) * (7.7E-6) / (0.85)^2
F1 = 46.20 N

Next, let's find the force between S2 and S3, charged with -7.7 µC and +7.7 µC respectively:
F2 = k * Q2 * Q3 / r^2
F2 = (8.99E9) * (-7.7E-6) * (7.7E-6) / (0.85)^2
F2 = -46.20 N

Now, we need to find the directions of these forces. In this equilateral triangle, we can use 60° angles between the sides.

For F1 (between S1 and S3), the force is repulsive and directed radially outwards along the line between S1 and S3.

For F2 (between S2 and S3), the force is attractive and directed radially inwards along the line between S2 and S3.

To find the net force on S3, we can find the x and y components of the forces.

F1x = 0 (as it's along the line between S1 and S3)
F1y = F1 (as it's perpendicular to x-axis)

F2x = F2 * cos(60°) = -46.20 * 0.5 = -23.10 N
F2y = F2 * sin(60°) = -46.20 * 0.87 = -40.18 N

Now, we find the net force components:

Fx = F1x + F2x = 0 + (-23.10) = -23.10 N
Fy = F1y + F2y = 46.20 + (-40.18) = 6.02 N

Now, to find the magnitude and direction of the net force on S3:

F_net = sqrt(Fx^2 + Fy^2) = sqrt((-23.10)^2 + (6.02)^2) = 23.98 N

To find the direction, we can use the arctangent function:

θ = arctan(Fy / Fx) = arctan(6.02 / -23.10) = -14.42° (measured counterclockwise from the line connecting S1 and S3)

Now, when S3 is resting at the origin, the net force on S3 will have the same magnitude as the force between S1 and S2:

F_net (origin) = 46.20 N

The direction of the net force when S3 is at the origin will be directly towards the midpoint of the line connecting S1 and S2 (as both forces are attractive):

θ' = 60° (measured counterclockwise from the line connecting S1 and S3)

So for part (a):
Magnitude: 23.98 N
Direction: -14.42°

For part (c):
Magnitude: 46.20 N
Direction: 60°