# Electron vector analysis question

3 electrons q1, q2, q3 setup in a right angled triangle

q1
|\
| \
q3-q2

charges are

q1 = +2.5 X 10^-17C
q2 = +3.0 X 10^-17C
q3 = +3.5 X 10^-17C

distances between the charges
q1-q3 = .03 m
q1-q2 = .05 m

setup FBD

\
\ 2F1
\
/
/ 3F1
q1

2F1 = k*q1*q2 / .05^2
= 2.7 X 10^-21N
3F1 = k*q3*q1 / .03^2
= 8.75 X 10^-21N

now I think I use cosine law to figure this out

Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos135)
= 7.1 X 10^-21

with all charges being positive, all qs are experiencing repulsion so the direction is away from q1

the question "calculate the magnitude and direction of force on q1"

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rl.bhat
Homework Helper
To calculate the magnitude and direction of force on q1, find the force due to q3 on q1 and force due to q2 on q1. Find the resultant of these forces. Find the cosine of angle between these forces. It is not cos135. The direction of the resultant is given by angle made by the resultant with one of the forces.

well, I believe I have found the forces acting on q1 (by q2 and q3). I guess my problem is I dont know what to do next. From the example I am given in the text they used cosine rule and a FBD to calculate the resultant of these forces. Im guessing my problem is my Trig is rusty......

how do I determine what theta should be??? (in the book they use an isosceles and use 120degrees in this example its a right angle triangle).

Kurdt
Staff Emeritus
Gold Member
You have a right angled triangle and you know the length of two sides. That is all the information you need to calculate anything about that triangle. Can you remember the trigonometric ratios for a right angled triangle?

sin0 = opp/hyp

sin^-1(.03/.05) = 36.8 = 37 degrees

Kurdt
Staff Emeritus
Gold Member
You don't have the opposite, you have the adjacent.

Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos37)

??

I really need to touch up on my trig before I go back to school

Kurdt
Staff Emeritus
Gold Member
The net force will be the vector sum or resultant of the two vectors.

sorry Im lost here, unfortunately the book provided for this distance ed course is poorly written and the examples provided dont really help with answering of questions.

what in the trig am I messing up??

seeing as this is based on the original distribution of forces, Im asuming Im working with the right angled triangle.

Im trying to figure out theta for the upper left portion of this triangle (it probably has a name but I havnt touched trig in 10+ years). when looking for theta in this region, the triangle is setup as

| = A
\ = H
__ = O

q1 - q3 = A = .03m
q1 - q2 = H = .05m

given this information and the little trig I know, I can use cosine rule to solve for theta

where cos = A/H

cos^-1 (.03/.05) = 53 Degrees

am I then using
Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos53)

Its absolutely absurd that I have nobody through the distance education program to ask questions and I am left to throw my self at the mercy of the internet forums.

Thanks for any guidance

Kurdt
Staff Emeritus
Gold Member
Its just the final part of adding the forces now that you're not doing correctly. You're gonna have to set up a coordinate system and find the components of each vector along each of the axes. The following page may help you with that.

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

so is this where I draw the FBD??

and I guess I will need to know the angles of the other vectors

so
q1 - q2 = 53
q3 - q1 = 90
q2 - q3 = 180 - (53 + 90) = 37

and

2F1 = k*q1*q2 / .05^2
= 2.7 X 10^-21N
3F1 = k*q3*q1 / .03^2
= 8.75 X 10^-21N
2F3 = k*q2*q3 / .04^2
=5.9 X 10^-21 N

So with all this information I should be able to answer the question?????

the reason I keep going back to this
Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos0)

is because thats how they answered the question in the text (solving trignometrically using cosine rule with the FBD)

Kurdt
Staff Emeritus