# Potential Difference between two points

1. Jul 18, 2009

### bishoy

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

I wish I knew

3. 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

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

W=q(Vb-Va)

3. 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: Jul 18, 2009
2. Jul 18, 2009

### Dick

Looks fine to me. Well done.

3. Jul 18, 2009

### bishoy

Thanks Dick :) what about the second problem? any idea how to go about doing that one?

4. Jul 18, 2009

### Dick

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. Jul 18, 2009

### bishoy

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. Jul 18, 2009

### Dick

Depends on the calculation. You certainly can't ignore it in this one!

7. Jul 18, 2009

### bishoy

alright so using this equation:
$$v = \sqrt { 2 \frac {\mbox{KE}} m }$$

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

32463868.12 (assuming my math is correct)

and that would be in m/s?

8. Jul 18, 2009

### Dick

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. Jul 18, 2009

### bishoy

Well, When I plugged it into my calculator I acually used $$v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.109x10^{-31}Kg }$$

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. Jul 18, 2009

### Dick

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. Jul 18, 2009

### bishoy

my apologies, I misunderstood. I thought you meant that J=kg*m^2/s^2.

Your help is much appreciated :)