A uniform thin ring of charge, with radius 3.70 cm and total charge 5.70 µC, is

Question

A uniform thin ring of charge, with radius 3.70 cm and total charge 5.70 µC, is located in the yz-plane and centered on the origin. (In other words, the x-axis passes through the center of the ring, and is perpendicular to the ring.) A particle of mass 12.3 g and charge 4.40 µC is on the negative x-axis, initially very far from the ring (effectively at .) The particle is launched in the positive x direction with some initial velocity, toward the center of the ring. Calculate the minimum initial velocity required for the particle to get past the center of the ring and come out the other side.

Explanation / Answer

Take v=sqrt(2*k*q1*q2/mR) You have q1, q2, m, and R, and you probably have the electostatic constant k in your textbook also. This equation came from solving initial KE = final PE for v (1/2)mv^2 = k*q1*q2/R The potential energy of the charge q1 in the middle of the ring is k*q1*q2/R where q2 is the total charge on the ring and R is the radius of the ring. Rearrange the equation to solve for v and you get the equation you were given. any problems let me know and i will help :)

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