# Potential Difference between two points

## 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?

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?

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:

Dick
Homework Helper
Looks fine to me. Well done.

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

Dick
Homework Helper

## 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?

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?

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?

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?

Dick
Homework Helper
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!

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?

Dick
Homework Helper
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.

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.

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