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Finding Magnitude and Direction of Force on a Charge

by cheerspens
Tags: charge, direction, force, magnitude
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cheerspens
#1
Aug27-11, 11:38 AM
P: 92
1. The problem statement, all variables and given/known data
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. Relevant 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
#2
Aug27-11, 11:42 AM
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Quote Quote by cheerspens View Post
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.
cheerspens
#3
Aug27-11, 11:48 AM
P: 92
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
#4
Aug27-11, 11:56 AM
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Finding Magnitude and Direction of Force on a Charge

Quote Quote by cheerspens View Post
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.
cheerspens
#5
Aug27-11, 12:00 PM
P: 92
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
#6
Aug27-11, 12:05 PM
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Give me your values for sin(180), cos(180), sin(270), and cos(270).
cheerspens
#7
Aug27-11, 01:40 PM
P: 92
sin(180) = 0
cos(180) = -1
sin(270) = -1
cos(270)=0
Doc Al
#8
Aug27-11, 01:43 PM
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Quote Quote by cheerspens View Post
sin(180) = 0
cos(180) = -1
sin(270) = -1
cos(270)=0
Good. Now simplify your equations in post #3 accordingly.
Lobezno
#9
Aug27-11, 07:27 PM
P: 53
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.
Doc Al
#10
Aug27-11, 08:38 PM
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Quote Quote by Lobezno View Post
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.
Lobezno
#11
Aug28-11, 12:35 AM
P: 53
C and B, sorry. They'll produce an opposite charge.


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