Potential difference to bring a proton to rest

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SUMMARY

The discussion centers on calculating the potential difference required to bring a proton, initially traveling at 3.5×10^5 m/s, to rest by reducing its speed by half. The relevant equations include kinetic energy (KE = 0.5 * m * v^2) and electric potential energy (EPE = q * V). The correct potential difference was determined to be 320 V after applying conservation of energy principles. The user initially miscalculated the potential difference but ultimately arrived at the correct solution with guidance.

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  • Familiarity with the mass and charge of a proton
  • Knowledge of conservation of energy principles
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Homework Statement


A proton has an initial speed of 3.5×10^5 . What potential difference is required to reduce the initial speed of the proton by a factor of 2?


Homework Equations


KE=.5*m*v^2
EPE=q*V


The Attempt at a Solution


I did KE=.5(1.67x10^-27)(3.5x10^5/2)^2 and then divided that by the charge of a proton to get V, and I got 160 V which is wrong. Then I tried squaring the velocity first, then dividing it by 2 and I got 320 V, which is also not right. I'm stuck.
 
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this is just conservation of energy

electric potential starts at zero plus kinetic energy starts at some KEi

this must be equal to some electric potential energy plus half of KEi
 
SHISHKABOB said:
this is just conservation of energy

electric potential starts at zero plus kinetic energy starts at some KEi

this must be equal to some electric potential energy plus half of KEi

Thank you so much! I didn't even think of that. I got the answer.
 

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