Magnitude and direction of electric force (file attached)

AI Thread Summary
The discussion centers on a confusion regarding the calculation of the radius for the charge on q3 due to q1, specifically that it should be r^2 = 2a^2, not r = 2a^2. The Pythagorean theorem is applied to find the hypotenuse of a right triangle with both sides measuring "a". Participants clarify that the negative sign in the electric field indicates the direction of the force and acceleration, which are both downward for a positive charge. Additionally, there is a question about the disappearance of the cosine term in a trigonometric context, which is explained by a change in the angle to 2θ. Overall, the conversation aims to clarify these fundamental concepts in electric force calculations.
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Homework Statement


I was kind of confused on how the radius for the charge on q3 by q1 is 2a^2, i attached the file for my notes regarding this. The question is at the bottom of the pdf scan.


Homework Equations





The Attempt at a Solution

 

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The needed distance is the hypotenuse of the right triangle whose sides are both "a". Hint: Pythagorean theorem.
 
c^2=a^2+b^2

so

c^2=a^2+a^2
c^2=2a^2

what about the ^2 for the c, is it just left there? because it seems like it since r=2a^2
 
Careful: it's r^2 = 2a^2, not r = 2a^2.
 
On this other page of notes which has to do with the introduction of acceleration in electric fields, there are three parts of the problem that I am not quite following, i figured it would be easier if i just wrote it down by hand on the actual note sheet than typing it all out.:smile:

For the cosign question, i don't undestand what happened to it because the sin is still there in the next step but not the cos.
 

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If what I am asking isn't really quite clear, i can write it out again more specifically and scan it.
 
(Be sure to include the full problem statement next time.)

(1) Why negative? The field points down (thus is negative), the charge is positive: so the force and acceleration are both downward, thus negative.

(2) They are solving for the value of time that makes y = 0. Set that factor equal to zero and solve for t.

(3) Regarding what happened to the cosine: note that \theta changed to 2\theta. (Review your trig identities for \sin\theta \cos\theta.)
 
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