Electrostatic Force of a Triangle

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Homework Statement



Suppose that the magnitude of the net electrostatic force exerted on the point charge q2 in the figure is 0.65 N . (Figure 1) http://imgur.com/4lZliPq

Find the distance D.


q1 = 2.1 micro C
q2 = 6.3 micro C
q3 = -.89 micro C


Homework Equations



F = k |q1| |q2| / r^2
k = 8.99 E9

The Attempt at a Solution



Normally for this problem I'd try to find the different forces that affect each other but for this problem there is 2 unknowns and I'm not sure how to use the net force of q2 to find the distance.

Fx 1 on 2 = K|q1||q2|(cos 60) / d^2 =

Fy 1 on 2 = K|q1||q2|sin60 / d^2=

Not really sure where to go with this problem :( any help would be appreciated
 

Answers and Replies

  • #2
Simon Bridge
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Force is a vector - you know how to find the relation for the magnitude of each force individually, and you know what they have to add up to, and you know how to do a vector sum.

You are doing the trig a little early - just add them head-to-tail.
(Remember about triangles and parallelograms?)
 
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  • #3
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Force is a vector - you know how to find the relation for the magnitude of each force individually, and you know what they have to add up to, and you know how to do a vector sum.

You are doing the trig a little early - just add them head-to-tail.
(Remember about triangles and parallelograms?)
Hmmmm....Not quite following you, so in a way we should work backwards?

F2= .65 N

.65 = F 1 on 2 + F 3 on 2

Is that what you're saying?
 
  • #4
Simon Bridge
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I'm saying $$|\vec{F}_2|=|\vec{F}_{21}+\vec{F}_{23}|=0.65 \text{N}$$... remember, forces are vectors.
The net force on 2, is the vector sum of the forces due to 3 and 1 - and the magnitude of the resultant force is 0.65N.

Try this - draw a free-body diagram for q2.

Sketch arrows for the directions of the forces (one of them will be a lot bigger than the other, and be careful about the directions since the other two charges have opposite signs.) Concentrate on getting the angles right - use a protractor.

Draw the parallelogram of vectors to get the resultant.
This will give you two triangles and some angles - you also know the size of the resultant vector.
Use your knowledge of triangles - you know, all that geometry you did in math class?
sum of angles, the sine rule, the cosine rule, that stuff.
 
  • #5
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I'm saying $$|\vec{F}_2|=|\vec{F}_{21}+\vec{F}_{23}|=0.65 \text{N}$$... remember, forces are vectors.
The net force on 2, is the vector sum of the forces due to 3 and 1 - and the magnitude of the resultant force is 0.65N.

Try this - draw a free-body diagram for q2.

Sketch arrows for the directions of the forces (one of them will be a lot bigger than the other, and be careful about the directions since the other two charges have opposite signs.) Concentrate on getting the angles right - use a protractor.

Draw the parallelogram of vectors to get the resultant.
This will give you two triangles and some angles - you also know the size of the resultant vector.
Use your knowledge of triangles - you know, all that geometry you did in math class?
sum of angles, the sine rule, the cosine rule, that stuff.
So this is what I have so far...

.65N=F12 + F32
.65N = (Kq2 / R^2)(q1cos60 + q1sin60 + q3cos60 + q3sin60)

Am I on the right track?
 
  • #6
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Anyone have any ideas?
 
  • #7
Simon Bridge
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65N = (Kq2 / R^2)(q1cos60 + q1sin60 + q3cos60 + q3sin60)
This does not make sense. Where does all this trig come from? It looks to me that you don't know how to add vectors geometrically. OK then - define an x and y direction and resolve the vectors against them.
 

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