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Finding Magnitude and Direction of Force on a Charge 
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#1
Aug2711, 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 F_{CB} pointing to the left and F_{AB} 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. 


#2
Aug2711, 11:42 AM

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P: 41,579




#3
Aug2711, 11:48 AM

P: 92

F_{CB}=[(9x10^{9}(6x10^{9})(5x10^{9})) / (.3^{2})][cos 180][itex]\widehat{x}[/itex] + [(9x10^{9}(6x10^{9})(5x10^{9})) / (.3^{2})][sin 180][itex]\widehat{y}[/itex]
F_{AB}=[(9x10^{9}(3x10^{9})(5x10^{9})) / (.1^{2})][cos 270][itex]\widehat{x}[/itex] +[(9x10^{9}(3x10^{9})(5x10^{9})) / (.1^{2})][sin 270][itex]\widehat{y}[/itex] 


#4
Aug2711, 11:56 AM

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P: 41,579

Finding Magnitude and Direction of Force on a Charge



#5
Aug2711, 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. 


#6
Aug2711, 12:05 PM

Mentor
P: 41,579

Give me your values for sin(180), cos(180), sin(270), and cos(270).



#7
Aug2711, 01:40 PM

P: 92

sin(180) = 0
cos(180) = 1 sin(270) = 1 cos(270)=0 


#9
Aug2711, 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.



#10
Aug2711, 08:38 PM

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P: 41,579




#11
Aug2811, 12:35 AM

P: 53

C and B, sorry. They'll produce an opposite charge.



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