Magnitude and direction of electric force (file attached)

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Homework Help Overview

The discussion revolves around understanding the electric force and the geometry involved in calculating distances between charges, specifically focusing on the radius related to charge q3 and q1. The subject area includes concepts from electrostatics and geometry, particularly the Pythagorean theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of the distance as the hypotenuse of a right triangle formed by the charges. Questions arise regarding the interpretation of the radius and its squared value, as well as the role of trigonometric functions in the context of acceleration in electric fields.

Discussion Status

The discussion is active with participants providing hints and clarifications. Some guidance has been offered regarding the correct interpretation of the radius and the application of trigonometric identities. Multiple interpretations of the problem are being explored, particularly concerning the use of cosine and sine in the calculations.

Contextual Notes

There is a noted need for clearer problem statements and additional context, as some participants express confusion over specific aspects of the problem and the notation used in their notes.

trah22
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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 [itex]\theta[/itex] changed to [itex]2\theta[/itex]. (Review your trig identities for [itex]\sin\theta \cos\theta[/itex].)
 

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