# Potential difference given velocity of electron at 2 points

## Homework Statement

An electron moves through an electric field, and its speed drops from 2900 {\rm m/s} to 1500 {\rm m/s} .

What's the potential difference between the two points at which the speed was measured?

## Homework Equations

U=qV
K=mv^2/2

q=electron
m=mass of electron

## The Attempt at a Solution

Ki=.5 * 9.109*10^-31 * 2900^2
= 3.83*10^-24J
Kf=.5 * 9.109*10^-31 * 1500^2
=1.0247*10^-24J
deltaK=2.80557*10^-24

p.d= deltaK/q
= 2.80557*10^-24/1.602*10^-19C
=1.8*10^-5V

Is it supposed to be negative 1.602*10^-19C for the charge on the electron and is that what is stuffing me up? Onto my final go of this on the online homework so trying to get it :)

rock.freak667
Homework Helper
I believe your method looks correct. But the negative sign does not really matter, as your change in KE would be negative so your pd would still be positive.

Damn the homework already said no to the answer 1.8*10^-5V and 1.7*10^-5V :'(

Maybe it doesn't like V and wants J/C I dunno :(

tiny-tim
Homework Helper
hi louza8!

(try using the X2 icon just above the Reply box )
Ki=.5 * 9.109*10^-31 * 2900^2
= 3.83*10^-24J

oooh … it's not 24

EDIT: oops! yes it is 24

Last edited:
Perhaps you need a more precise answer like 1.75. If the homework system you are using is something like UTexas, 3 decimals are needed as they only tolerate 2% error.

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vela
Staff Emeritus
Homework Helper
I believe your method looks correct. But the negative sign does not really matter, as your change in KE would be negative so your pd would still be positive.
Energy conservation gives you ΔU=-ΔK, so ΔU>0. ΔV=ΔU/q should be negative.

The OP should also verify he or she has the right number of sig figs.

To verify, the correct number of sig figs are 2, deduced from the question. or are they 3? whoops? confused.

Last edited:
ehild
Homework Helper
The speed drops so ΔK is negative. The work of the field is negative.

ehild

Edit: I left out one sentence before, saying that the potential difference was postive. Work is negative p.d. times charge. It is electron, so the p.d. is negative. -1.751 *10-5 V

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vela
Staff Emeritus
Homework Helper