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hershey7
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Homework Statement
Two point charges Q1 = 69.5 μC and Q2 = 19.9 μC are fixed on the two ends of the diameter of a semi-circle with a radius of r = 21.3 cm as shown in the figure.
[URL]https://s2.lite.msu.edu/enc/67/e803dcaee3e6ac1f52720314edcaf2afee90f8672567f8e6335a2f5d64ae3b0024edbf942db8a9b89877951ee8e774af588c1fa8ef990eabf675fdb990677deb12f8467ceb901b579f6ad4b974ffe72a8d747e43835a8cf7.gif[/URL]
A massless point charge Q3 = 54.6 μC can slide on the semi-circle without any friction. At which value of the angle θ will the point charge Q3 be in equilibrum?
Homework Equations
Coulomb's Law : F = KQQ/R^2
E=F/q
The Attempt at a Solution
My understanding of this was to find the point at which there was a net Electrical field force of zero on Q that is on the semicircle. However, since all of these were positive charges, I was confused as to how this could stay on a semi-circular path. The attempt to cancel out the X-component Force Vectors (parallel to the diameter) I used the law of cosines to find the distance between Q3--Q1 and Q3--Q2. This distance was used to set the Forces equal to each other and hence have an equation with respect to theta.
My final answer was theta = arccos((Q1^2-Q2^2)/(Q1^2+Q2^2). This was wrong.
I am not looking for a direct answer, merely an explanation of the forces that are acting on Q3 and how they cancel to keep it in equilibrium.
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