Physics - Electrical Potential help

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SUMMARY

The discussion centers on the relationship between the final speed of a proton and the potential difference (V) it accelerates through. It is established that the final speed of the proton is proportional to the square root of the potential difference, specifically V^(1/2). The kinetic energy gained by the proton, represented as (1/2)mv², is equated to the energy gained from the potential difference, which is q*V joules. This relationship confirms that the correct answer is B: V^(1/2).

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  • Knowledge of electrical potential and units, specifically volts
  • Basic principles of charge and energy conversion in electric fields
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aaronmilk3
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A proton is released from rest and accelerates through a potential difference V. The final speed of the proton is proportional to the potential difference in which of the following ways?

A. V
B. V^1/2
C. V^-1/2
D. V^-1

I think it is V^(1/2) (just√V) but I am no sure exactly how to find this. Any help would be great. I was thinking that I had to take kinetic energy + potential energy = 0 then kinetic energy= -potential energy. But if this is right what would I do? (1/2)mv²=what? No idea and any help is great. Thanks!
 
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The Volt unit is shorthand for Joules per Coulomb. So a charge q traversing a potential difference V gains (or loses) q*V joules of energy. Equate with kinetic energy and you're there.
 

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