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Resultant Force on Charge in the middle of square 
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#1
Jan811, 01:29 PM

P: 18

1. The problem statement, all variables and given/known data
(a) Find the resultant force on a negative charge of 3q, placed at the center of a square of side b, which has charges on the 4 corners, respectively, q, 2q, 4q, 2q in clockwise order. (Magnitude and direction) 2. Relevant equations F= k*q1*q*(1/ r^2) Ftotal= F_{15} + F_{25} + F_{35} + F_{45} 3. The attempt at a solution I drew the figure, and see that r = square root 2 *b* (1/2) (pythagorean theorem) One question is that F25 and F45 are equal and opposite. Do they cancel eachother out? If so, I wouldn't have to calculate the forces for those, yes? I'm just a little confused. Sorry if it's super easy. 


#2
Jan811, 02:38 PM

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[tex]r = \sqrt{2}b/2 = .707b[/tex] AM 


#3
Jan811, 03:46 PM

P: 18

Ok awesome!
So now I have F_{15}= (2ke_{3}q^{2})/(b^{2}) N and F_{35}=(2ke_{4}q^{2})/(b^{2}) N But how do I find the x and y components without numbers? Do I just put in F_{15x}= F_{15} sin(135) F_{15y}= F_{15} cos(135) and F_{35x}= F_{35} sin(135) F_{35y}= F_{35} cos(135) The fact that theres no value for q shouldn't be throwing me off this much, but it is. 


#4
Jan811, 08:24 PM

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Resultant Force on Charge in the middle of square
AM 


#5
Jan811, 09:12 PM

P: 18

45 degrees?
I think I was confused because I made the middle charge the origin and thought the angle came from that. . . And I don't really understand why we split the 4q because if one if +q and one is 5q we would get a net result of 4q anyway. . . I assume the direction is towards the first corner q. Thanks so much again, sorry if something is going over my head. . . 


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