what potential difference is needed to stop an electron that has an initial velocity v=4.0×105m/s?

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Potential difference to stop an electron question

• Thread starter cwesto
• Beginning appointment Feb 25, 2009

Homework Statement

An electron with an initial speed of four.70×ten⁵m/south is brought to rest by an electric field.
What was the potential difference that stopped the electron?

Unknown:
potential difference or [tex]\Delta[/tex]V

Known:
v_o=initial speed
m=mass of electron
q=charge of electron

Homework Equations

[tex]\Delta[/tex]Vq+[tex]\frac{1}{2}[/tex]*mv^two=[tex]\Delta[/tex]Vq

The Attempt at a Solution

I effigy basically plug and chug but my equation doesn't seem right. I sure it's not. Basically I don't know where to go from here.

Last edited: Feb 25, 2009

Answers and Replies

Welcome to PF!

Two questions. Are you sure this is the whole question? With these values, nosotros could use most any kind of electric field to stop the electron. Where'd you get that equation?

Well, I'll take a stab at helping out...

You demand to end a moving electron, right? Does that hateful you will have an dispatch (or deceleration, aforementioned thing)?

If you have an acceleration, there's a force. And the force in this example is related to the electric field, right?

So you demand an electrical field such that it creates a force capable of stopping your electron.

If yous tin find the electrical field, y'all *should* be able to relate the electric field to the potential energy, relate that to the electrical potential, and and then find a potential departure...

But so, I haven't worked a problem similar this in awhile and I may not exist solving it with the tools that you already take... Have you done line integrals and all that stuff however?

Sonolum, if the given trouble is the whole trouble, even the smallest force is enough to end the electron somewhen. Why go for the trouble of using line integrals?

I don't really see them so much every bit trouble, myself... I was mostly trying to become a feel for the level of this problem and help out.

I also told the electron will move to a region of lower potential. I got the equation from a previous question that is simliar posted on PF last year. That's all I have. The link to the previous question is below.

(https://world wide web.physicsforums.com/showthread.php?t=234536)

Ah, see, in that problem nosotros know how much distance we have to end the electron. Here, nosotros don't know that - do we have to stop the electron in a sure time? In a certain amount of distance?

Kruum's 100% right, I didn't realize it at get-go, just yeah, any electric field that opposes the electron's motion will *eventually* stop it. It kinda depends on where/when you want to stop it to go anything specific.

I think you need to understand that the kinetic energy is to be absorbed by work from some source. In this case measuring the work can best be done by observing that changing the voltage of a accuse tin be determined by:

West = q*ΔV and that would equal ½mv² in order to stop it.

Hence won't the needed ΔV = ½mv²/q ?

And so in other words I need to know something else?

So in other words I need to know something else?

Like what?

I dunno, Pion's method seems a heckuva lot more eloquent and to the bespeak... Just I pretty sure information technology's (Delta)V = - W/q...

Oh crap! I should read the question a bit more carefully! Pitiful for the confusion cwesto!

The distance in which they want the electron to cease.

I dunno, Pion'south method seems a heckuva lot more eloquent and to the bespeak... Merely I pretty sure information technology's (Delta)V = - W/q...

Happily q is (-) electron which makes the ΔV (-).

The distance in which they want to electron to cease.

No distance required.

If the E-Field is 1000 5/m or ten V/m the distance would change of course but not the ΔV required.

I think you need to empathise that the kinetic free energy is to exist captivated by piece of work from some source. In this case measuring the work can all-time be done by observing that irresolute the voltage of a accuse tin can be determined by:

W = q*ΔV and that would equal ½mv² in lodge to stop it.

Hence won't the needed ΔV = ½mv²/q ?

Wouldn't "ΔV = ½mv²/q" give me ΔV while the speed is 4.70×10⁵one thousand/due south?
I want to know the ΔV when the speed is 0.

I'll ask my professor then post. I have class now. Thanks though.

Ah, but see, that'due south why in that location was mention of changing the kinetic energy into piece of work, correct? The free energy had to become*somewhere*, and the energies should be equivalent before and after, right?

Oh yeah!
ΔV = ½mv²/q
Your right. Thanks.

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