Electric Potential related to velocity

AI Thread Summary
A +3μC charge with a mass of 2 x 10^-4 kg is initially moving at 5 m/s in a 1000 V electric potential and needs to be evaluated at 200 V. The conservation of energy principle indicates that the total energy, comprising kinetic and potential energy, remains constant. The initial kinetic energy must be considered alongside the change in electric potential energy to determine the final velocity. The correct approach involves calculating the initial kinetic energy and adjusting it based on the change in potential energy to find the new speed. Ultimately, the final speed is determined to be 7 m/s, highlighting the importance of accounting for both initial conditions and energy conservation.
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=Vot+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=Vot+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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