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mlostrac
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Homework Statement
A charge of +2q is placed at the origin and a second charge of –q is placed at
x = 3.0 cm. Where can a third charge +Q be placed so that it experiences a zero
force?I think I've found the solution, I just need someone to verify why the math is the way it is at the end.
1) The overall force acting on the the new charge must equal zero, therefore I set them equal to each other:
k (2q x Q)/(r^2) = k (q x Q)/(r-0.03)^2
(r - 0.03)^2 = r^2 (2q x Q)/(q x Q)
(r - 0.03)^2 = 2r^2
r - 0.03 = +/- 1.414r
-0.03 = -2.414r
r = .0124
OR
-0.03 = 0.414r
r = -.072So would this mean that the charge would be .012 metres to the right of the negative charge? (Assuming that -q is placed 3 cm to the right of +2q)