Understanding the Relationship between Work and Volts

[jeff1evesque]

Question
I was wondering how the concept of work can have units of volts? I mean work has the units of N/m = J =/ J/(A*S) = Volts? Yet I am reading they are equal? Can someone enlighten me?

 

[Pengwuino]

Can you check your units? [E]=[F][L] which is energy = force * distance. You're saying energy/(current*time) = volts on the right side which is dimensionally correct but were you meaning to equate the two equations? Particularly what do you mean by "I mean work has the units of N/m = J =/[/size] J/(A*S) = Volts" ? Do you mean =/= which is not equal to or do you mean possibly 1/J/(A*s) which is equal to A*s/J?

 

[rl.bhat]

Question
I was wondering how the concept of work can have units of volts? I mean work has the units of N/m = J =/ J/(A*S) = Volts? Yet I am reading they are equal? Can someone enlighten me?

This statement is wrong.
Work = Nm

 

[jeff1evesque]

Can you check your units? [E]=[F][L] which is energy = force * distance. You're saying energy/(current*time) = volts on the right side which is dimensionally correct but were you meaning to equate the two equations? Particularly what do you mean by "I mean work has the units of N/m = J =/[/size] J/(A*S) = Volts" ? Do you mean =/= which is not equal to or do you mean possibly 1/J/(A*s) which is equal to A*s/J?

What I meant was: Nm = J =/= J/(A*S) = Volts

 

[Pengwuino]

What I meant was: Nm = J =/= J/(A*S) = Volts

Well that certainly is true... what seems to be the situation? Joules is definitely not going to be equivalent to Joules divided by charge (that is, current * time).

 

[jeff1evesque]

Well that certainly is true... what seems to be the situation? Joules is definitely not going to be equivalent to Joules divided by charge (that is, current * time).

Note: $$\hat{E} = - \frac{V_0}{d}\hat{z}$$ (volts/meter) is the electric field between a capacitor, and V_0 is the potential difference in a power supply for a circuit.

Then the following statement (from my notes) is not correct?

Work

= $$\int_{bottom}^{Top} \frac{\hat{F}}{Q} \bullet \hat{dl} \equiv potential-difference = \int_{bottom}^{Top} \hat{E} \bullet \hat{dl} = \int_{bottom}^{Top} - \frac{V_{0}}{d}(\hat{z} \bullet \hat{dl}) = -\int_{bottom}^{Top} \frac{V_0}{d}dz = -V_0 (Voltz)$$

thanks,Jeff

 

[Phrak]

I may see the source of confusion. Sometimes physicists will express energy in terms of electron volts, eV. One electron volt is the work done on an electron subject to a changing of one volt in potential. It has units [Q][V].

In some discussion one might simply say "volts" used in place of "electron volts." It saves 3 syllables out of 4 in short-hand discourse.

 

[jeff1evesque]

I may see the source of confusion. Sometimes physicists will express energy in terms of electron volts, eV. One electron volt is the work done on an electron subject to a changing of one volt in potential. It has units [Q][V].

In some discussion one might simply say "volts" used in place of "electron volts." It saves 3 syllables out of 4 in short-hand discourse.

Very cool to know. Thanks a lot.

JL

 

[diazona]

Then the following statement (from my notes) is not correct?

Work

= $$\int_{bottom}^{Top} \frac{\hat{F}}{Q} \bullet \hat{dl} \equiv potential-difference = \int_{bottom}^{Top} \hat{E} \bullet \hat{dl} = \int_{bottom}^{Top} - \frac{V_{0}}{d}(\hat{z} \bullet \hat{dl}) = -\int_{bottom}^{Top} \frac{V_0}{d}dz = -V_0 (Voltz)$$

No, it's not correct. Everything you wrote out in LaTeX code (from $\int \hat{F}/Q\cdot\mathrm{d}\hat{l}$ onward) is right, but it's equal to voltage, not work.

 

[merryjman]

I can't speak about your notes, but maybe a conceptual explanation would help? For example, if a particle with charge 1 Coulomb, initially at rest, was released to move through a potential difference of 1 Volt, after that time the particle would have 1 Joule of kinetic energy. The volt is energy per charge. In the same way, if you had to perform 1 Joule of mechanical work to move that 1-Coulomb particle slowly through an electric field, you would have moved it through a potential difference of 1 Volt.

 

[Redbelly98]

Voltage is equivalent to work done per unit charge. This is not the same thing as work.

In SI units, Volts are equivalent to Joules/Coulomb.

 

[jeff1evesque]

I may see the source of confusion. Sometimes physicists will express energy in terms of electron volts, eV. One electron volt is the work done on an electron subject to a changing of one volt in potential. It has units [Q][V].

In some discussion one might simply say "volts" used in place of "electron volts." It saves 3 syllables out of 4 in short-hand discourse.

You were right, I talked to my teacher today, and he said it's basically what he meant, Quolomb*meters.

Thanks,


Jeff

 

[Phrak]

You were right, I talked to my teacher today, and he said it's basically what he meant, Quolomb*meters.

Odd. He left the charge out of Coulomb-meters?