
#1
Aug1811, 06:23 PM

P: 7

1. The problem statement, all variables and given/known data
Three point charges are arranged as shown in the figure below. (Take q1 = 5.70 nC, q2 = 4.60 nC, and q3 = 2.79 nC.) (a) Find the magnitude of the electric force on the particle at the origin. (b) Find the direction of the electric force on the particle at the origin. 2° (counterclockwise from the +xaxis) 2. Relevant equations Fab= k q1q2/r2 3. The attempt at a solution I am attaching a photo of my last attempt to do this problem but I really think that I don't know what I am doing. I think I am using the wrong length and so I can't get the magnitude right. I can't even get to the direction part but I can't do that either because I don't know if I'm supposed to use sin or cos. 



#2
Aug1811, 06:53 PM

HW Helper
P: 2,316

You won't use pythagorus on the triangle of distances, you will use it on the triangle of force, force and net force. whether you use sin, cos or tan will depend which two forces you chose to work with: are the opposite, adjacent or hypotenuse of the triangle. 



#3
Aug1811, 07:02 PM

P: 461

I didn't study your work in detail because the first thing I noticed was that in your calculations you were somehow using the distance between q1 and q3. That distance does not directly affect the force on q2 and is not needed for a simple solution. Also, the distances on your work paper appear to be ten times greater than those shown on the problem.
The force on a particle is the sum of the forces contributed by all other particles. So to solve this, separately calculate the force between q1 and q2 (which, conveniently, has only an X component), and the force between q3 and q2 (which, conveniently, has only a Y component) and add those two vectors together to get the force vector on q2. 



#4
Aug1811, 07:41 PM

P: 7

Magnitude and direction of the net force on a particle. Please help!
Thank you both so much for the help. I feel like such a dummy about the distance. I think I have just been looking at these problems for too long.
Ok so I made the corrections you indicated and I'm still 10% wrong but idk where the problem is. I'm attaching my new work. 



#5
Aug1811, 08:07 PM

Mentor
P: 11,415

In what directions are each of the forces directed? (Pay attention to both the relative positions of the charges and their signs!). How do you add forces directed in different directions to find the net force and its magnitude?




#6
Aug1811, 08:08 PM

HW Helper
P: 2,316

You need to draw triangle of forces, and use pythagorus to find the resultant force. eg 2N west plus 1N South = 2.236N "southwestish". Trig can find the real direction. 



#7
Aug1811, 08:20 PM

P: 461

I didn't do the arithmetic so I'll assume its right, but you must still add those two vectors to get the sum of the forces on q2. Then convert to polar coords (magnitude and direction) because the question wants that. It looks like you entered in the 'magnitude' box only the magnitude of one of those vectors.
EDIT: Yep, what Peter said. 



#8
Aug1811, 08:21 PM

P: 7

Would that be sq rt of (9.15X10^7)^2 + (1.38X10^6)^2 I don't want to enter the wrong answer again cuz it only gives me so many chances. 



#9
Aug1811, 08:38 PM

HW Helper
P: 2,316





#10
Aug1811, 08:42 PM

P: 7

It should have been 2.622X10^6 and 1.155X10^5 I was just looking at the wrong page when I typed that. 



#11
Aug1811, 08:51 PM

HW Helper
P: 2,316





#12
Aug1811, 09:17 PM

P: 7

YAY!!! Thank you so much! Now I just have to figure out how to do the second half and I will be done. I just have no idea where to even start. It has been so long since I did any of this and I don't remember much. Will you please just help me get started?




#13
Aug1811, 09:47 PM

Mentor
P: 11,415

Try sketching the force vectors on your diagram; place arrowed line segments with their "feet" attached to the object on which the force is acting and the pointy ends pointing in the direction that the force acts. Between charge pairs this will inevitably lie along a line joining the centers of the charge pair. After you've done this, then it's time to look at the vectors' orientation in the coordinate system and decide whether the x and y components of the vectors are positive or negative. 



#14
Aug1811, 09:58 PM

P: 7





#15
Aug1811, 10:00 PM

Mentor
P: 11,415

You should be able to identify the x and y components of your resultant force vector (after all, you calculated the magnitude of the resultant from them!). 



#16
Aug1811, 10:26 PM

P: 7

Thanks again for your help everyone!



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