Physics Electric Field Problem

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Homework Help Overview

The problem involves a particle with mass m and charge Q being accelerated through a potential difference V, resulting in an increase in kinetic energy K. A second particle with mass m/5 and charge 2Q is also accelerated through the same potential, prompting a question about the kinetic energy gained by this second particle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between mass, charge, and kinetic energy, questioning whether mass needs to be considered in the context of the potential difference and kinetic energy gained.

Discussion Status

There are various interpretations regarding the relationship between the kinetic energies of the two particles. Some participants suggest that the kinetic energy gained by the second particle may be directly related to its charge, while others seek clarification on the implications of the equations presented.

Contextual Notes

Participants note the use of voltage in the original problem statement and discuss the implications of charge on kinetic energy without reaching a consensus on the final relationship between the kinetic energies of the two particles.

asheik234
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Homework Statement


A particle with a mass m and a charge Q is accelerated through a potential difference V, causing its kinetic energy to increase by an amount K. Another particle of mass m/5 and charge 2Q is accelerated through the same potential. How much kinetic energy does this second particle gain?


Homework Equations


Vq = U

(1/2)mv^2 = U;


The Attempt at a Solution


I can't find an equation that relates mass, charge and kinetic energy. Any help would be appreciated.
 
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It seems that you don't need to involve mass.

Remember, 1 volt = 1 joule of energy per coulomb
 
If Q*C = K;
then 2Q*C = 2K'

So the increase in kinetic energy is K, correct me if I'm wrong.
 
asheik234 said:
If Q*C = K;
then 2Q*C = 2K'
I think you mean 2Q*C = K'
 
haruspex said:
I think you mean 2Q*C = K'

Then the kinetic energy doesn't change?
 
asheik234 said:
Then the kinetic energy doesn't change?
K' stands for the KE of the 2Q charge, right? (Just noticed the OP uses V, not C, so switching back to that here...)
If Q*V = K and 2Q*V = K' then what is the relationship between K and K'?
 
haruspex said:
K' stands for the KE of the 2Q charge, right? (Just noticed the OP uses V, not C, so switching back to that here...)
If Q*V = K and 2Q*V = K' then what is the relationship between K and K'?

So K' = 2K, right, because Q is proportional to KE.
 
asheik234 said:
So K' = 2K, right, because Q is proportional to KE.
Right, because KE is proportional to Q here.

Remember, 1 volt = 1 joule per coulomb
 

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