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Two positive point charges are placed on the x-axis. One, of magnitude 4Q, is placed at the origin. The other, of magnitude Q i...Asked by Steve
Two positive point charges are placed on the x-axis. One, of magnitude 4Q, is placed at the origin. The other, of magnitude Q is placed at x=3 m. Neither charge is able to move. Where on the x-axis in meters can I place a third positive point charge such that the magnitude of the net force on the third charge is zero?
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Lets assume a charge of q is placed at a point (x,o) where 0<x<3.
Your net force on that middle charge will be the sum of two electrostatic forces. Your plan is to use a distance of x for the force calculation between the leftmost charges and a distance of 3-x for the force calculation between the rightmost charges. These two forces should be equal to achieve equilibrium.
Basically, thanks to Coulomb's Law, I have 4Q/(x2) = Q/((3-x)2) after simplification. This yields
x^2-8x+12=0
(x-2)(x-6)=0
x=2 or x=6
Now just note that x has to be between 0 and 3 meters because that's the only way the field directions will oppose each other.
Anyways, this is quite a nice application of Coulomb's law. However, next time please refrain from posting live brilliant problems.
Your net force on that middle charge will be the sum of two electrostatic forces. Your plan is to use a distance of x for the force calculation between the leftmost charges and a distance of 3-x for the force calculation between the rightmost charges. These two forces should be equal to achieve equilibrium.
Basically, thanks to Coulomb's Law, I have 4Q/(x2) = Q/((3-x)2) after simplification. This yields
x^2-8x+12=0
(x-2)(x-6)=0
x=2 or x=6
Now just note that x has to be between 0 and 3 meters because that's the only way the field directions will oppose each other.
Anyways, this is quite a nice application of Coulomb's law. However, next time please refrain from posting live brilliant problems.
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