1MileCrash
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Homework Statement
These are really frustrating because they seem so simple, I work them all out without a hitch only to find that I'm wrong half the time and left wondering why.
In fig 21-26a, particles 1 and 2 have charge 20 microcoulombs each and are held at separation distance d = 1.5 meters.
(1 is located above 2)
a.) What is the magnitude of the electrostatic force on 1 due to 2?
b.) Particle 3 is also of charge 20 microcoulombs and is placed to the right-center of 1 and 2, forming an equilateral triangle with 1 and 2. Find the net electrostatic force on 1 due to 2 and 3.
Homework Equations
The Attempt at a Solution
The first part, I have correct.
[itex]F = kq_{1}q_{2}r^{-2} = (8.99x10^{9})(20x10^{6})(20x10^{6})(1.5)^{2} = 1.6 N[/itex]
The second part, I do not have correct.
Since the magnitude of the force from 2 is 1.6 N and it is located below 1, the vector of the force is just 1.6j.
Since the particles make an equilateral triangle, every angle is 60*. Since 3 is to the right, the positive angle to 1 from 3 is 120*.
Therefore the unit vector that points from 3 to 1 is cos(120)i + sin(120)j.
Since the magnitude of force from 3 to 1 is also 1.6, the vector of the force due to 3 is:
[itex]1.6cos(120)i + 1.6sin(120)j = -.8i + 1.38j[/itex]
Adding the j components, the vector of all force on 1 is:
-.8i + 2.98j which has a magnitude of roughly 3.09 Newtons.
The book answer is 2.77. I don't even understand how their result makes sense. 2 and 3 are purely additive, they are both below 1 and 2 has no x component while 3 does.
Help? TIA
