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Finding Magnitude and Direction of Force on a Charge

  • Thread starter cheerspens
  • Start date
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1. Homework Statement
Three charges are arranged as shown in the picture I attached. Find the magnitude and direction of the electrostatic force on the charge at the origin.


2. Homework Equations
I know that Coulomb's Law must be applied.


3. The Attempt at a Solution
I've drawn a force diagram with FCB pointing to the left and FAB pointing down. I've also broke the vectors into components using Coulomb's Law and added them together. By doing this, I got 2.7 x 10-5 [itex]\hat{x}[/itex] + 0 [itex]\hat{y}[/itex]. From here what do I do to get an answer?

I know the answer is supposed to be 1.38 x 10-5 at 77.5 degrees so I need help with how to get there.
 

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Doc Al

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I've drawn a force diagram with FCB pointing to the left and FAB pointing down. I've also broke the vectors into components using Coulomb's Law and added them together. By doing this, I got 2.7 x 10-5 [itex]\hat{x}[/itex] + 0 [itex]\hat{y}[/itex].
You must have made a mistake, since the resultant must have a component in the y direction. Show how you computed FCB and FAB.
 
92
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FCB=[(9x109|(6x10-9)(5x10-9)|) / (.32)][cos 180][itex]\widehat{x}[/itex] + [(9x109|(6x10-9)(5x10-9)|) / (.32)][sin 180][itex]\widehat{y}[/itex]

FAB=[(9x109|(-3x10-9)(5x10-9)|) / (.12)][cos 270][itex]\widehat{x}[/itex] +[(9x109|(-3x10-9)(5x10-9)|) / (.12)][sin 270][itex]\widehat{y}[/itex]
 

Doc Al

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FCB=[(9x109|(6x10-9)(5x10-9)|) / (.32)][cos 180][itex]\widehat{x}[/itex] + [(9x109|(6x10-9)(5x10-9)|) / (.32)][sin 180][itex]\widehat{y}[/itex]

FAB=[(9x109|(-3x10-9)(5x10-9)|) / (.12)][cos 270][itex]\widehat{x}[/itex] +[(9x109|(-3x10-9)(5x10-9)|) / (.12)][sin 270][itex]\widehat{y}[/itex]
OK, now simplify these results by evaluating those trig functions.
 
92
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I did and thats how I got 2.7x10-9[itex]\hat{x}[/itex] + 0[itex]\hat{y}[/itex]
Was my set up right and I went wrong somewhere in the math? I was always thinking I set it up wrong or wasn't using the right values.
 

Doc Al

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Give me your values for sin(180), cos(180), sin(270), and cos(270).
 
92
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sin(180) = 0
cos(180) = -1
sin(270) = -1
cos(270)=0
 

Doc Al

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What's the point of incorporating Trig functions? If you compute them individually then you get two separate forces, one along the y axis and one along the x axis. Remember also that the force between charges A and B is repulsive, so the force you drew downwards is meant to be upwards.
 
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Doc Al

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What's the point of incorporating Trig functions? If you compute them individually then you get two separate forces, one along the y axis and one along the x axis.
While you don't need to use trig functions, there's nothing wrong with using them.
Remember also that the force between charges A and B is repulsive, so the force you drew downwards is meant to be upwards.
Charges A and B are oppositely charged, thus the force is attractive.
 
53
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C and B, sorry. They'll produce an opposite charge.
 

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