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
−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?
so magnitude is (8.99E9*-7.7)/(.85)^2 ?
im still confused on distance?
Well, I may not be an electrician, but I can certainly give it a shot!
a) To find the magnitude and direction of the net force on the third spider (S3), we need to calculate the electrical forces between the spiders. The formula for the electrical force between two charged objects is given by Coulomb's Law:
F = k * (q1 * q2) / r^2
Where F is the force, k is the Coulomb's constant, q1 and q2 are the charges of the spiders, and r is the distance between them.
First, we need to calculate the distance between the spider S3 and the other two spiders, S1 and S2. Since the sides of the triangular web have a length of a = 0.85 m, the distance between the spiders is also a.
Now, let's calculate the force between S3 and S1:
F1 = k * (q1 * q3) / a^2
F1 = k * (7.7 * 10^-6 C * 7.7 * 10^-6 C) / (0.85 m)^2
Likewise, let's calculate the force between S3 and S2:
F2 = k * (q2 * q3) / a^2
F2 = k * (-7.7 * 10^-6 C * 7.7 * 10^-6 C) / (0.85 m)^2
Now, to find the net force on S3, we need to calculate the vector sum of F1 and F2. Since F2 is directed towards S2, which is opposite in sign to S1 and S3, we can subtract F2 from F1:
Net Force = F1 - F2
And there you have it! You now have the magnitude and direction of the net force on the third spider (S3).
c) To find the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin, you would need to calculate the forces between the spiders for this particular scenario. Since the spider S3 is now at the origin, the distance between S3 and S1 (or S2) would be the magnitude of the vector connecting the origin to the position of spider S1 (or S2), which is still a.
Using the same formula as before, calculate the forces F1 and F2 with a distance of a = 0.85 m. Then, find the net force by subtracting F2 from F1.
I hope this helps, and remember, when it comes to electric spider webs... it's all about the shocking details!
To find the net force on the third spider (S3), we can use Coulomb's Law which states that the force between two point charges is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.
Step 1: Calculate the magnitude of the net force:
1. Calculate the force between S1 and S3 using Coulomb's Law:
F1 = k * (q1 * q3) / r^2
where F1 is the force between S1 and S3, q1 and q3 are the charges of S1 and S3 respectively, r is the distance between them, and k is the electrostatic constant.
F1 = (9 * 10^9 N * m^2 / C^2) * (7.7 * 10^-6 C) * (7.7 * 10^-6 C) / (0.85 m)^2
2. Calculate the force between S2 and S3 using Coulomb's Law:
F2 = k * (q2 * q3) / r^2
where F2 is the force between S2 and S3, q2 is the charge of S2, and r is the distance between them.
F2 = (9 * 10^9 N * m^2 / C^2) * (-7.7 * 10^-6 C) * (7.7 * 10^-6 C) / (0.85 m)^2
3. Calculate the net force on S3:
Net force = F1 + F2
Step 2: Determine the direction of the net force:
To find the direction of the net force, we need to consider the vector nature of forces.
1. Draw a diagram depicting the triangular web and indicate the direction of each force (F1 and F2) on S3.
2. Use vector addition to find the resultant force, which is the net force on S3.
Add the forces F1 and F2 as vectors using the parallelogram law of vector addition.
The resultant force will give both magnitude and direction.
For part c), when the third spider (S3) is resting at the origin, the process is the same, but you need to calculate the forces between S1 and S3 and between S2 and S3 using the distance from S1 and S2 to the origin instead of the side length of the triangle.
To find the magnitude and direction of the net force on the third spider (S3), we can use the concept of electrostatic forces.
a) When S3 is at the vertex of the triangular web, we can calculate the force between each pair of spiders using Coulomb's Law:
F = (k * q1 * q2) / r^2
where F is the force between the spiders, k is the electrostatic constant (9 * 10^9 Nm^2/C^2), q1 and q2 are the charges of the spiders, and r is the distance between them.
Since S1 and S3 have the same charge (+7.7 µC), their forces cancel out due to being opposite in direction (they are at the ends of the base of the triangle). Therefore, we only need to consider the forces between S2 and S3.
1. Calculate the distance between S2 and S3:
- Since the triangle is equilateral, all sides are equal to 0.85 m.
- Using the Law of Cosines, we can find the distance between S2 and S3:
r = √(s^2 + s^2 - 2s * s * cos(60°))
= √(0.85^2 + 0.85^2 - 2 * 0.85 * 0.85 * cos(60°))
2. Calculate the force between S2 and S3 using Coulomb's Law.
b) To find the direction of the net force on S3, we need to consider the direction of each individual force and how they add up.
c) When S3 is resting at the origin, we need to calculate the forces between each spider and the origin (S1, S2, and S3 with the origin as O). The steps are similar to those in part a), using the distance from each spider to the origin.
Remember to consider the sign of the charge when calculating the forces, as it will affect the direction of the forces.
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