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Determine the magnitude of the force on each charge

  • Thread starter meme177
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  • #1
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1. The problem

Two negative and two positive point charges (magnitude Q = 3.16mC ) are placed on opposite corners of a square (side lengths a = 0.105m )

#1. Determine the magnitude of the force on each charge.

I GOT..... F1,F2,F3,F4 =7.44x10^6N (which was marked correct)

I am having trouble with this question.....

#2. Determine the direction of the force on each charge.
I know (Fy/Fx)=1
so tan-1(1)=45

Its asking for θ1,θ2,θ3,θ4 and I have no idea how to get the angles :(
 

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  • #2
BvU
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Hello meme, and welcome to PF.

If you know the tangent, you know the angle (plus or minus pi), isn't it ?
However, the "I know (Fy/Fx)=1" applies to which of the charges, exactly ? How about the other three ?

And: have you altready defined what theta is ? Or was it given ?
 
  • #3
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θ1,θ2,θ3,θ4
I have tried these answers (in order)
225,225,225,225=WRONG (but then I was like NO this is impossible)
Then I tried
135,45,315,225=WRONG

I don't know what to do anymore
 
  • #4
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the "I know (Fy/Fx)=1" applies to which of the charges, exactly ?
It can surely not apply to all four of them !
 
  • #5
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Charge #2 where I calculated the y and x component of the force for question #1
 
  • #6
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I think I see it! charge #2 is 45 .......Charge#1 (45+90=135)......charge#4 (180+45=225).....charge#3 (45+90=135) ??
 
  • #7
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Still a small matter of signs: my hunch for e.g. #2 is that two attracting forces from charges closer by overcome one repelling force from #4. From there, with symmetry considerations, all forces are towards the center of the square. Agree ?
 
  • #8
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okay, I agree.
 
  • #9
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with the forces all pointing to the middle, they cancel (no movement)
Now I have θ1= -45 θ2= -135 θ3= 45 θ4=135
Is this correct, now?
 

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  • #10
BvU
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So theta is the angle wrt the positive x-axis.
I think you swapped 3 and 4 (in your post not in the drawing)
Movement doesn't apply here. The charges are positioned in fixed given places,
and there is no mention of how they are kept there...
 
  • #11
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Yes, so θ1= -45, θ2= -135, θ3= 135, θ4=45 ? Am I doing the angles wrt positive x-axis right?
 
  • #12
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Yes and yes, IMHO.
 
  • #13
SammyS
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with the forces all pointing to the middle, they cancel (no movement)
Now I have θ1= -45 θ2= -135 θ3= 45 θ4=135
Is this correct, now?
Do like charges attract, or do they repel?
 
  • #14
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@Sam: Meme is talking about the net forces, I should think. The magnitude was already calculated in part #1 of post #1.
 
  • #15
SammyS
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@Sam: Meme is talking about the net forces, I should think. The magnitude was already calculated in part #1 of post #1.
Okey-Dokey .
 
  • #16
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Mastering Physics is still counting the angles wrong :/
 
  • #17
SammyS
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Mastering Physics is still counting the angles wrong :/
Does M P want radians?

Otherwise M P may want you to specifically say, degrees, or ° .
 
  • #18
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no, the units are already there (degrees) counterclockwise from the +x axis
 
  • #19
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Here is how it looks
 

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  • #20
SammyS
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Maybe they want

θ1, θ2, θ3, θ4 =

315, 225, 135, 45

but I see you have only 2 tries remaining.
 
  • #21
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I got it! ..... thanx guys!
 

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  • #22
SammyS
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I got it! ..... thanx guys!
That's good to hear.

M P really should have taken your answers with the negative angles. They give exactly the same orientation for the vectors.
 
  • #23
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Goes to show programmers and teachers are different, both only human.
Well done, meme and thanks for helping out, Sam.

M P should do something about this. All the physics was already in part A and part B is more like a detective to find out what their limited answer domain is.
 

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