- #1
Punchlinegirl
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Two point charges lie on the y-axis. A third charge q is placed somewhere in space such that the resultant force on it is zero. What distance from the origin must this third charge be placed such that the resultant force on it is zero? Answer in units of m.
The first charge is 8[tex] \mu C[/tex] and is located at 0.5 m.
The second charge is -37 [tex] \mu C[/tex] and is located at -1 m
The third charge is 8 [tex] \mu C [/tex].
I know that this third charge needs to be on the y-axis and for the force to be zero, the forces have to cancel each other out.
so I found the F on 1 & 2 by k(8e-6 *37e-6)/ (1.5^2)
and got this to be 1.184
Then I found F from 1 & 3 by k(8e-6*8e-6)/r^2
solving for r gave me .697, which I added to .5, b/c the third charge is above the 1st one. This gave me 1.2 m,which isn't right.. can someone please help?
The first charge is 8[tex] \mu C[/tex] and is located at 0.5 m.
The second charge is -37 [tex] \mu C[/tex] and is located at -1 m
The third charge is 8 [tex] \mu C [/tex].
I know that this third charge needs to be on the y-axis and for the force to be zero, the forces have to cancel each other out.
so I found the F on 1 & 2 by k(8e-6 *37e-6)/ (1.5^2)
and got this to be 1.184
Then I found F from 1 & 3 by k(8e-6*8e-6)/r^2
solving for r gave me .697, which I added to .5, b/c the third charge is above the 1st one. This gave me 1.2 m,which isn't right.. can someone please help?