Electrostatics - finding velocity of an electron

[milkyway11]

The problem asks an electron is brought to rest by a potential difference of 1kV. what was the initial velocity of the electron?

To begin this problem, I first find the change in potential energy by \DeltaU=q\DeltaV. Since electron has a negative charge, I would have a negative \DeltaU. with \DeltaU=-\DeltaKE, I can find the initial velocity. But here is my problem, I would end up with a negative velocity. Did I do something wrong?

Another question A proton experiences a force of 10mN as it travels between teh plates of a parallel plate capacitor, parallel to the plates. If the capacitor holds 1mC of charge and has a potential of 10V, what is the separation between the plates?

I solved this one by using F=qE. Then I can use E to find distance via E=V/d. However, I am not sure which charge to use for the force equation. I would assume it's the charge of proton since the force is experienced by the proton. But I am not sure what the given 1mC of capacitor will do?

 

[gneill]

The problem didn't state what the polarity of the potential difference was. It only claimed that it brought the electron to rest.

 

[milkyway11]

So what exactly does that mean?

The problem didn't state what the polarity of the potential difference was. It only claimed that it brought the electron to rest.

 

[gneill]

So what exactly does that mean?

You were worried about the sign of q∆V leading to a negative velocity in your calculations for the speed of the electron. You can pick whichever sign you wish for ∆V, because only the magnitude of the change was specified.

 

[milkyway11]

Thank you! Can you also take a look at my second question?

You were worried about the sign of q∆V leading to a negative velocity in your calculations for the speed of the electron. You can pick whichever sign you wish for ∆V, because only the magnitude of the change was specified.

 

[gneill]

For your second question, I suspect that the charge is just throwaway information. There is enough information without it to determine the field strength, and hence the separation.