Potential difference given velocity of electron at 2 points

AI Thread Summary
An electron's speed decreases from 2900 m/s to 1500 m/s while moving through an electric field, leading to a calculated change in kinetic energy. The potential difference (p.d.) is derived from the change in kinetic energy divided by the charge of the electron, resulting in a value around 1.75 x 10^-5 V. The discussion highlights confusion regarding the sign of the potential difference, clarifying that it should be negative due to the electron slowing down, indicating it moves to a lower potential. Additionally, there is a debate about the appropriate number of significant figures for the answer, with some suggesting that the homework system may require three figures despite the problem providing only two. Ultimately, the consensus is that the potential difference is negative, reflecting the work done against the electric field.
louza8
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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.
 
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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 don't know :(
 
hi louza8! :smile:

(try using the X2 icon just above the Reply box :wink:)
louza8 said:
Ki=.5 * 9.109*10^-31 * 2900^2
= 3.83*10^-24J

oooh … it's not 24 :redface:

EDIT: oops! yes it is 24 :rolleyes:
 
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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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rock.freak667 said:
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.
 
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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. :blushing: 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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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.
 
  • #10
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?
 
  • #11
No. You are not missing anything. When carrying out sig figs, it really should be two but many systems use three automatically.
 
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