Electric Potential related to velocity

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SUMMARY

The discussion centers on calculating the final velocity of a +3μC charge with a mass of 2 x 10^-4 kg as it moves from an electric potential of 1000 V to 200 V, assuming energy conservation. The initial velocity is 5 m/s, leading to an initial kinetic energy of 1/2 * (2 x 10^-4 kg) * (5 m/s)^2. The correct final velocity, after accounting for the change in electric potential energy, is determined to be 7 m/s, contrasting with the incorrect calculation of 24 m/s. The key takeaway is that the change in kinetic energy must be calculated from the initial kinetic energy, not just the change in potential energy.

PREREQUISITES
  • Understanding of electric potential energy and its relation to charge (V = electric potential energy / q)
  • Knowledge of kinetic energy formula (KE = 1/2 mv^2)
  • Familiarity with the conservation of energy principle in physics
  • Basic algebra for solving equations involving kinetic and potential energy
NEXT STEPS
  • Review the conservation of energy principle in electrostatics
  • Learn how to calculate changes in kinetic energy due to electric potential differences
  • Study the relationship between electric potential and kinetic energy in charged particles
  • Explore examples of energy conservation in different physical systems
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and energy conservation principles, as well as educators seeking to clarify concepts related to electric potential and kinetic energy.

jeandempsey
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Homework Statement


A +3μC charge with a mass of 2 x 10 ^-4 kg is moving with a speed of 5 m/s at a position where the electric potential is 1000 V. How fast will it be moving when it gets to a position where the electric potential is 200 V? Assume energy is conserved.


Homework Equations


V=electric potential energy / q
D=volt+1/2at2
KE=1/2mv^2

The Attempt at a Solution


I got the change in electric potential energy to be .0024. And then I set that equal to the kinetic energy equation. I got my answer to be 24m/s, but the answer is 7 m/s. I think I might be doing this totally wrong.
 
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jeandempsey said:

Homework Statement


A +3μC charge with a mass of 2 x 10 ^-4 kg is moving with a speed of 5 m/s at a position where the electric potential is 1000 V. How fast will it be moving when it gets to a position where the electric potential is 200 V? Assume energy is conserved.


Homework Equations


V=electric potential energy / q
D=volt+1/2at2
KE=1/2mv^2

The Attempt at a Solution


I got the change in electric potential energy to be .0024. And then I set that equal to the kinetic energy equation. I got my answer to be 24m/s, but the answer is 7 m/s. I think I might be doing this totally wrong.

In what way did you "set that equal to the kinetic energy equation"? Did you account for the kinetic energy due to the initial velocity? Perhaps you should show the calculation that you performed.
 
I did this: my electric potential energy was -2.4x10^-3
So- -2.4x10^-3 = 1/2(2x10^-4)v^2
 
jeandempsey said:
I did this: my electric potential energy was -2.4x10^-3
So- -2.4x10^-3 = 1/2(2x10^-4)v^2

There was an initial velocity, so an initial kinetic energy. The change in potential results in a change in the kinetic energy, not the total.
 
Thank you, but I'm still kind of confused as to how you would get the velocity from that
 
jeandempsey said:
Thank you, but I'm still kind of confused as to how you would get the velocity from that

What is the relationship between kinetic energy and speed? What, then, is the initial kinetic energy of the charged particle to begin with? Then, when it has moved across the 1000V to
200V potential difference its kinetic energy will have CHANGED from that initial value.

If you want to look at the situation as a single formula, then using the law of conservation of energy you can write that the sum of the kinetic energy and potential energy for the system is constant. So that

KE_1 + PE_1 = KE_2 + PE_2
 

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