This statement is wrong.jeff1evesque said: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 said: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?
jeff1evesque said:What I meant was: Nm = J =/= J/(A*S) = Volts
Pengwuino said: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).
Phrak said: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.
No, it's not correct. Everything you wrote out in LaTeX code (from [itex]\int \hat{F}/Q\cdot\mathrm{d}\hat{l}[/itex] onward) is right, but it's equal to voltage, not work.jeff1evesque said:Then the following statement (from my notes) is not correct?
Work
= [tex] \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)[/tex]
Phrak said: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 said:You were right, I talked to my teacher today, and he said it's basically what he meant, Quolomb*meters.
The relationship between work and volts is that work is the product of the electric potential difference (volts) and the amount of charge (coulombs) that moves through a circuit. In other words, work is equal to the product of volts and coulombs.
Voltage plays a crucial role in determining the amount of work done in an electrical circuit. As voltage increases, the amount of work done also increases, assuming the same amount of charge is moving through the circuit. This is because voltage is a measure of the electric potential difference, which is the force that drives the flow of charge and ultimately determines the amount of work done.
No, work cannot be done without voltage. Voltage is necessary to create an electric potential difference, which is the driving force for the flow of charge and the basis for the concept of work in electrical systems. Without voltage, there would be no energy transfer and therefore no work done.
Work does not directly affect voltage in a circuit. Instead, it is voltage that affects the amount of work done. This is because voltage is a measure of the electric potential difference, which is the force that drives the flow of charge and ultimately determines the amount of work done.
Yes, there is a limit to the amount of work that can be done with a specific voltage. This limit is determined by the amount of charge that can flow through the circuit. Once all of the available charge has been used to do work, the voltage will no longer have an effect on the amount of work done.