- #1
hteezy
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Homework Statement
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?
Homework Equations
Columbs law (force between two point charges)
F = 1/4 piEo times abs value of q1*q2 divided by r^2
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 I am right before i do anything!