# Homework Help: SI units

1. Jul 12, 2009

### 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?

2. Jul 12, 2009

### 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 =/ 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?

3. Jul 12, 2009

### rl.bhat

This statement is wrong.
Work = Nm

4. Jul 12, 2009

### jeff1evesque

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

5. Jul 13, 2009

### Pengwuino

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).

6. Jul 13, 2009

### jeff1evesque

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

Last edited: Jul 13, 2009
7. Jul 13, 2009

### 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.

Last edited: Jul 13, 2009
8. Jul 13, 2009

### jeff1evesque

Very cool to know. Thanks a lot.

JL

9. Jul 13, 2009

### diazona

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.

10. Jul 13, 2009

### 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.

Last edited: Jul 13, 2009
11. Jul 13, 2009

### Redbelly98

Staff Emeritus
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.

12. Jul 14, 2009

### jeff1evesque

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

Thanks,

Jeff

13. Jul 14, 2009

### Phrak

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