(adsbygoogle = window.adsbygoogle || []).push({}); 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!

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# Vector addition in electric field

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