A charged particle, with charge q and mass m is traveling with initial speed, v,

Question

A charged particle, with charge q and mass m is traveling with initial speed, v, in a horizontal direction . It enters a region with a uniform electric field. The field causes the charge to slow down. The field is produced by two parallel vertical counducting plates separated by a distance d which is small compared to the size of the plates. A small hole in each plate allows the charge to enter and exit the region between the plates. What is the maximum voltage between the plates that will allow the charge to exit the second plate?

Explanation / Answer

when it enters the electric field ,it slows down and gain potentialenergy . Let the voltage be V . Then electric field = V/d then Force of charge = Field * charge = Vq/d So acceleration = force / mass = Vq / md Also initial velocity = v Final velocity = 0 (A it just reached the other part ) distance = d Using the equation , final velocity 2 -initial velocity 2 =2aS 0 - v2 = -2Vqd / md V = mv2/2q (You can come to the same answer if you consider the energy view) Here loss in kinetic energy = gain in potential energy loss in kinetic energy=mv2/2 gain in potential energy =Vq equating , V = mv2/2q )

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