
#1
Jan1208, 10:55 PM

P: 10

1. The problem statement, all variables and given/known data
Two positive charges = = 2.0 are located at = 0, = 0.30 and = 0, = 0.30 , respectively. Third point charge = 4.0 is located at = 0.40 , = 0.What is the net force ((a)magnitude and (b)direction) on charge exerted by the other two charges? 2. Relevant equations Columbs law (force between two point charges) F = 1/4 piEo times abs value of q1*q2 divided by r^2 3. The attempt at a solution first i have to find the force that Q exerts on q1 (F1 on q1) i use coulumbs law F = k * [q1q2]/[r^2] k = 8.988 e 9 so i plug in q1 and Q and the distance between them and i get.... F1 on q1 = .29 i break that down into components x and y F1x= .29cos(.40/.50) = .287 F1y = .29sin(.30/.50) = .00301 so now i find the force that q2 exerted on q1 (F2 on q1) q2 and q1 are both on the vertical axis so there will only be a y component so i use cuolumbs law again and now i plug in q1 and q2 and r is the distance between them so.. F2= k [2.0 e 6][2.0 e 6] / (.60 ^2) = .1 so now i add the x components and y components up there is only one x component and 2 y components sooo F1x = .287 F1y + F2y = .10301 F = square root of ( .287^2 + .10301^2) which is equal to .30 (2 sig figs) then to find the angle i would use tan(theta) = Fy/Fx which gives me an angle of 19.7 degrees It is supposed to be in the +x direction so i think i am supposed to subtract it from 180 and that should give me my angle im afraid to type the answer in into masterinphysics because i have only a few attempts left...so i want to make sure im right before i do anything! 



#2
Jan1308, 05:29 PM

Mentor
P: 40,905





#3
Jan1308, 10:46 PM

P: 10

opps sorry the question didnt copy right. my fault
Two positive charges q1 = q2 = 2.0 [tex]\mu[/tex]C are located at coordinate (0,.30) meters and (0,.30) meters, respectively. The third point charge Q = 4.0 [tex]\mu[/tex]C and is located at (.40,0) meters. What is the net force magnitude and direction on chare q1 exerted by the other two charges. i made a lil drawing so that it could help give a visual...its attached So first i found the force exerted by Q on q1 which i will call F1 the diff between q1 and Q is .50 meters F1 on q1 = k (2.0 e 6)(4.0 e 6) / (.50^2) = .29 N i then had to break that down into x and y components i need to use cosine for the x component so... = F1 * cos (.40/.50) and then the y component would be = F1 * sin (.30/.50) i hope that makes sense now 



#4
Jan1408, 05:10 AM

Mentor
P: 40,905

Vector addition in electric field
Much better.
Those are the magnitudes of the components. What are their signs? 



#5
Jan1408, 02:05 PM

P: 10

I guess they are positive?




#6
Jan1408, 02:34 PM

Mentor
P: 40,905

Draw yourself a picture of the force that Q exerts on q1. Which way does the vector point?




#7
Jan1408, 04:45 PM

P: 10

ok so the force that Q exerts on q1 is negative
and the force that q2 exerts on q1 is positive 



#8
Jan1408, 04:46 PM

P: 10

so F1 and its components are negative?




#9
Jan1408, 05:19 PM

P: 10

actually no...only the x component is in the negative direction for F1
the y component of F1 is still positive 



#10
Jan1408, 05:20 PM

Mentor
P: 40,905





#11
Jan1408, 08:14 PM

P: 10

its x component points in the  x direction
the y comp. points in the + y direction 



#13
Jan1508, 10:47 AM

P: 10

ok and the force exerted by q2 on q1 (F2) is also in the positve y direction
and there is no x component for that force so now that i know the components for all the forces do i just add them all up...add the y component of F1 and F2 together? then do i just add the final x and y components together? and that would be the magnitude of the force exerted on q1 by the other two charges right? 



#14
Jan1508, 11:38 AM

Mentor
P: 40,905





#15
Jan1508, 05:37 PM

P: 10

F = [tex]\sqrt{}(Fx^2 + Fy^2)[/tex]
ok i think i got it...Thank you very much for your help!!!!! 


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