Final Speed of a Proton Accelerated by Potential Difference V: Velocity Formula

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The discussion focuses on calculating the final speed of a proton accelerated by a potential difference V. It highlights that the work done on the proton, represented as energy transferred, is equal to the product of charge and voltage (delta(E) = qV). The kinetic energy gained by the proton, initially at rest, is equated to this energy transfer. The final speed can be derived from the kinetic energy formula, leading to the expression for speed as v = sqrt(2qV/m). The relationship between voltage and the proton's speed is clarified through the energy conversion process.
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A proton of mass m and charge e is acclerated from rest through a potential difference V. The final speed of the proton is: ?

I tried to figure this out.. but I can't find the equation for this.

thank you
 
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A common definition for electric potential (voltage) is the work done per unit charge (or Joules per Coulomb) to move a charge from infinity to a point in an electric field.

With that knowledge, how much energy was transferred to this proton of charge e?

Knowing that the proton of mass m was initially at rest, and has now gained some kinetic energy E, what is it's speed?
 
so then answer is 2Ve/m...?
 
i get 2q/m... how does voltage apply to the answer?
 
A charge of q accelerated across a potential V will have a change in energy of delta(E) = qV. If at rest, the change in energy is equal to the change in kinetic energy, assuming it is non-relativistic, which is a reasonable assumption in the case.
 

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