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rum2563
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Homework Statement
Three spheres, each with a negative charge of 4.0 X 10^6 C, are fixed at the vertices of an equilateral triangle whose sides are 0.20 m long. Calculate the magnitude and direction of the net force on each sphere.
Homework Equations
Fe = kq1q2 / r^2
The Attempt at a Solution
I used vector components to try to solve this question.
I think that we only have to find the net force on one of the spheres since the radius and the charges are same throughout the system. Sphere 1 is the "main" sphere from which I want to find out the net force.
Force of Sphere 2 on 1:
F2 = kq1q2 / r^2
= (9 X 10^9)(4.0 X 10^-6)^2 / (0.2)^2
= 3.6 N
For the x-component of force of sphere 2 on 1:
F2x = 3.6 X sin 30°
= 1.8 N
For the y-component of force of sphere 2 on 1:
F2y = 3.6 X cos30°
= 3.12 N
The force that was exerted by sphere 3 on 1 is the same force as 3.6 N since all the values are the same. Only the direction is different.
Here is the vector sum:
x = -3.6 - 1.8 = -5.4
y = 3.12
c^2 = 3.12 ^2 + (-5.4)^2
= 6.2365
= 6.2 N
Therefore, the net force is 6.2 N.
BUT... how do I get the angle Θ, because in the book it says 150° away from each side.
Please anyone help. It's just a little problem. Thanks.