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Vector addition in electric field

by hteezy
Tags: addition, electric, field, vector
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hteezy
#1
Jan12-08, 10:55 PM
P: 10
1. The problem statement, all variables and given/known data

Two positive charges = = 2.0 are located at = 0, = 0.30 and = 0, = -0.30 , respectively. Third point charge = 4.0 is located at = 0.40 , = 0.What is the net force ((a)magnitude and (b)direction) on charge exerted by the other two charges?

2. Relevant equations


Columbs law (force between two point charges)
F = 1/4 piEo times abs value of q1*q2 divided by r^2

3. The attempt at a solution

first i have to find the force that Q exerts on q1 (F1 on q1)
i use coulumbs law
F = k * [q1q2]/[r^2]
k = 8.988 e 9

so i plug in q1 and Q and the distance between them and i get....

F1 on q1 = .29
i break that down into components x and y

F1x= .29cos(.40/.50) = .287
F1y = .29sin(.30/.50) = .00301

so now i find the force that q2 exerted on q1 (F2 on q1)

q2 and q1 are both on the vertical axis so there will only be a y component

so i use cuolumbs law again and now i plug in q1 and q2 and r is the distance between them so..

F2= k [2.0 e -6][2.0 e -6] / (.60 ^2) = .1

so now i add the x components and y components up

there is only one x component and 2 y components sooo

F1x = .287
F1y + F2y = .10301

F = square root of ( .287^2 + .10301^2)

which is equal to .30 (2 sig figs)

then to find the angle i would use tan(theta) = Fy/Fx
which gives me an angle of 19.7 degrees
It is supposed to be in the +x direction so i think i am supposed to subtract it from 180 and that should give me my angle


im afraid to type the answer in into masterinphysics because i have only a few attempts left...so i want to make sure im right before i do anything!
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Doc Al
#2
Jan13-08, 05:29 PM
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Quote Quote by hteezy View Post
1. The problem statement, all variables and given/known data

Two positive charges = = 2.0 are located at = 0, = 0.30 and = 0, = -0.30 , respectively. Third point charge = 4.0 is located at = 0.40 , = 0.What is the net force ((a)magnitude and (b)direction) on charge exerted by the other two charges?
It's hard to understand what you've written (when I compare it to what follows). Please fill in the gaps. What are the charges? Where are they? (I assume those are x and y coordinates?) You need the force on which charge?

2. Relevant equations


Columbs law (force between two point charges)
F = 1/4 piEo times abs value of q1*q2 divided by r^2

3. The attempt at a solution

first i have to find the force that Q exerts on q1 (F1 on q1)
i use coulumbs law
F = k * [q1q2]/[r^2]
k = 8.988 e 9
OK.

so i plug in q1 and Q and the distance between them and i get....

F1 on q1 = .29
What are the charges?
i break that down into components x and y

F1x= .29cos(.40/.50) = .287
F1y = .29sin(.30/.50) = .00301
Redo these calculations. Also: Direction (and thus signs) counts.
hteezy
#3
Jan13-08, 10:46 PM
P: 10
opps sorry the question didnt copy right. my fault

Two positive charges q1 = q2 = 2.0 [tex]\mu[/tex]C are located at coordinate (0,.30) meters and (0,-.30) meters, respectively. The third point charge Q = 4.0 [tex]\mu[/tex]C and is located at (.40,0) meters. What is the net force magnitude and direction on chare q1 exerted by the other two charges.

i made a lil drawing so that it could help give a visual...its attached

So first i found the force exerted by Q on q1 which i will call F1
the diff between q1 and Q is .50 meters

F1 on q1 = k (2.0 e -6)(4.0 e -6) / (.50^2) = .29 N

i then had to break that down into x and y components

i need to use cosine for the x component so... = F1 * cos (.40/.50)
and then the y component would be = F1 * sin (.30/.50)

i hope that makes sense now
Attached Thumbnails
untitled.JPG  

Doc Al
#4
Jan14-08, 05:10 AM
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Vector addition in electric field

Much better.
Quote Quote by hteezy View Post
So first i found the force exerted by Q on q1 which i will call F1
the diff between q1 and Q is .50 meters

F1 on q1 = k (2.0 e -6)(4.0 e -6) / (.50^2) = .29 N
Good. That's the magnitude of the force.

i then had to break that down into x and y components

i need to use cosine for the x component so... = F1 * cos (.40/.50)
F1*cos(theta) = F1*(.40/.50)
and then the y component would be = F1 * sin (.30/.50)
F1*sin(theta) = F1*(.30/.50)

Those are the magnitudes of the components. What are their signs?
hteezy
#5
Jan14-08, 02:05 PM
P: 10
I guess they are positive?
Doc Al
#6
Jan14-08, 02:34 PM
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Draw yourself a picture of the force that Q exerts on q1. Which way does the vector point?
hteezy
#7
Jan14-08, 04:45 PM
P: 10
ok so the force that Q exerts on q1 is negative
and the force that q2 exerts on q1 is positive
hteezy
#8
Jan14-08, 04:46 PM
P: 10
so F1 and its components are negative?
hteezy
#9
Jan14-08, 05:19 PM
P: 10
actually no...only the x component is in the negative direction for F1

the y component of F1 is still positive
Doc Al
#10
Jan14-08, 05:20 PM
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Quote Quote by hteezy View Post
ok so the force that Q exerts on q1 is negative
and the force that q2 exerts on q1 is positive
I don't understand what you mean by saying that the force is negative or positive.

Quote Quote by hteezy View Post
so F1 and its components are negative?
F1 is shown on your diagram in post #3. Which way does its x-component point? Its y-component?
hteezy
#11
Jan14-08, 08:14 PM
P: 10
its x component points in the - x direction
the y comp. points in the + y direction
Doc Al
#12
Jan15-08, 04:40 AM
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Quote Quote by hteezy View Post
its x component points in the - x direction
the y comp. points in the + y direction
Right!
hteezy
#13
Jan15-08, 10:47 AM
P: 10
ok and the force exerted by q2 on q1 (F2) is also in the positve y direction
and there is no x component for that force

so now that i know the components for all the forces do i just add them all up...add the y component of F1 and F2 together?

then do i just add the final x and y components together?

and that would be the magnitude of the force exerted on q1 by the other two charges right?
Doc Al
#14
Jan15-08, 11:38 AM
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Quote Quote by hteezy View Post
ok and the force exerted by q2 on q1 (F2) is also in the positve y direction
and there is no x component for that force
Good.
so now that i know the components for all the forces do i just add them all up...add the y component of F1 and F2 together?
Yes.
then do i just add the final x and y components together?
No. The x and y components are perpendicular to each other--so you can't just add them like numbers. How do you find the magnitude (and direction) of a vector, given its components?
hteezy
#15
Jan15-08, 05:37 PM
P: 10
F = [tex]\sqrt{}(Fx^2 + Fy^2)[/tex]

ok i think i got it...Thank you very much for your help!!!!!


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