Potential difference given velocity of electron at 2 points

[louza8]

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 :)

Thanks for help in advance.

 

[rock.freak667]

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.

 

[louza8]

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 don't know :(

 

[tiny-tim]

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

 

[inutard]

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.

 

[vela]

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.

 

[louza8]

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

 

[ehild]

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 positive. Work is negative p.d. times charge. It is electron, so the p.d. is negative. -1.751 *10-⁵ V

 

[vela]

The potential difference is not the work per unit charge done by the field; it's the work per unit charge that would have to be done against the field to move it between two points.

Negative charges are a bit confusing because they want to roll up potential hills. But here, the electron is slowing down, so it's ending up at a lower potential (but higher potential energy) than from where it started. The potential difference is therefore negative.

 

[louza8]

thanks guys, the answer given from mastering physics was 1.75*10^-5V. they were asking for the deltaV.

I don't know why they answered in 3 significant figures when only 2 are provided in the question? am I missing something here?

 

[inutard]

No. You are not missing anything. When carrying out sig figs, it really should be two but many systems use three automatically.