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I have a charged ring in x-y plane.a positive charge is kept at a distance z along z axis.It is prvided a certain velocity,so that it overcomes the repulsive force.plz explain the motion of charged particle,particularly about its speed and positions as it reaches near the ring.

I have a charged ring in x-y plane.a  positive charge is kept at a distance z along z axis.It is prvided a certain velocity,so that it overcomes the repulsive force.plz explain the motion of charged particle,particularly about its speed and positions as it reaches near the ring.

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1 Answers

Chetan Mandayam Nayakar
312 Points
9 years ago

let 'r' be radius of ring, Q its charge, and charge along z-axis be q, difference in potential energy between (0,0,z) and (0,0,0)

=(Qq/4πε0)((1/r) -(1/√((r^2)+(z^2))))≡W,let velocity provided to q be such that its initial kinetic energy is U. If U<W, the charged particle stops for an instant before reaching the centre of the ring, and the motion is reversed. If U=W,then the particle comes to rest at the centre of the ring. If U>W, the particle slows down as it reaches the centre, assumes a minimum speed at the centre,and then accelerates away from the centre, travelling in the same direction as the initial velocity.

Electric field E= ( Qz k)/( 4pi(epsilon-zero)√((r^2+z^2)^3)), near the ring z is very small compared to r, then E=                 Qz k/(4pi(epsilon-zero)r^3), the electric field, and as a consequence of it, the force is directly proportional to displacement from equilibrium position, which is the centre of the ring. But the constant of proportionality is positive. If it was negative, there would have been simple harmonic oscillation.   

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