Potential Difference between two points

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Homework Statement


A bird stands on a DC electric transmission line carrying 2800A. The line has 2.5 x 10^-5 ohms resistance per meter, and the bird's feet are 4.0 cm apart. What is the potential difference between the bird's feet?


Homework Equations



I wish I knew

The Attempt at a Solution


Alright so since the resistance is 2.5 x 10^-5 per meter, I found that the resistance is .000001 ohms between the bird's feet. Since the current is 2800A, I used the equation V=IR and got .0028V. Is this correct or is the problem asking for something else?

Also, there is another problem I have

Homework Statement



Part A:
An electron in the picture tube of a television is accelerated from rest through a potential difference Vb-Va= 3000v. What is the change in electric Potential energy of the electron?

Part B:
What is the speed of the electron as a result of this acceleration?

Homework Equations



W=q(Vb-Va)

The Attempt at a Solution


I solved part A using the equation stated above and got 4.8 x10^-16
The part that I need help with is Part B. What equation should I use to find the speed?
 
Last edited:

Answers and Replies

  • #2
Dick
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Looks fine to me. Well done.
 
  • #3
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Thanks Dick :) what about the second problem? any idea how to go about doing that one?
 
  • #4
Dick
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Homework Statement


A bird stands on a DC electric transmission line carrying 2800A. The line has 2.5 x 10^-5 ohms resistance per meter, and the bird's feet are 4.0 cm apart. What is the potential difference between the bird's feet?


Homework Equations



I wish I knew

The Attempt at a Solution


Alright so since the resistance is 2.5 x 10^-5 per meter, I found that the resistance is .000001 ohms between the bird's feet. Since the current is 2800A, I used the equation V=IR and got .0028V. Is this correct or is the problem asking for something else?

Also, there is another problem I have

Homework Statement



Part A:
An electron in the picture tube of a television is accelerated from rest through a potential difference Vb-Va= 3000v. What is the change in electric Potential energy of the electron?

Part B:
What is the speed of the electron as a result of this acceleration?

Homework Equations



W=q(Vb-Va)

The Attempt at a Solution


I solved part A using the equation stated above and got 4.8 x10^-16
The part that I need help with is Part B. What equation should I use to find the speed?

If you put units on 4.8x10^(-16) then you've got the kinetic energy of the electron. It's Joules, right? So what's the relation between speed and kinetic energy?
 
  • #5
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So what's the relation between speed and kinetic energy?

Well, KE= 1/2mV^2, right? I have the KE, so all I need to do is put in the mass of the electron and solve for velocity? But I thought the mass of an electron is negligible in calculations?
 
  • #6
Dick
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Well, KE= 1/2mV^2, right? I have the KE, so all I need to do is put in the mass of the electron and solve for velocity? But I thought the mass of an electron is negligible in calculations?

Depends on the calculation. You certainly can't ignore it in this one!
 
  • #7
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alright so using this equation:
[tex]
v = \sqrt { 2 \frac {\mbox{KE}} m }
[/tex]

and 9.10938188 × 10^(-31) kilograms as the mass of an electron i get
[tex]
v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.10938188 × 10^{-31}Kg }
[/tex]

32463868.12 (assuming my math is correct)

and that would be in m/s?
 
  • #8
Dick
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You should round those numbers off to the precision implied by the problem. But yes, that looks ok. You should be able to figure out the units yourself. J=kg*m^2/s^2. What's sqrt(J/kg)? It's a good habit to get into carrying units around and checking the result has the correct dimension. On a more interesting side note, which you may ignore until you study relativity, that's about 10% of the speed of light. If the answer had been much closer to the speed of light or exceeding it, that would have been a clue that (1/2)*mv^2 is not a good enough approximation and you need to use the relativistic form for kinetic energy.
 
  • #9
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Well, When I plugged it into my calculator I acually used [tex]

v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.109x10^{-31}Kg }

[/tex]

and as far as the units go, That was a good call.... it Should have been m^2/s^2. Thank You for that. I really appreciate your help.
 
  • #10
Dick
Science Advisor
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Well, When I plugged it into my calculator I acually used [tex]

v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.109x10^{-31}Kg }

[/tex]

and as far as the units go, That was a good call.... it Should have been m^2/s^2. Thank You for that. I really appreciate your help.

Uh, J/kg is m^2/s^2. Sqrt(J/kg)=m/s. You were right, it's m/s. m^2/s^2 isn't a even a velocity.
 
  • #11
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my apologies, I misunderstood. I thought you meant that J=kg*m^2/s^2.

Your help is much appreciated :)
 

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