Rate of work done by F on charge carrier with average velocity v

[zenterix]

TL;DR Summary

I am reading the book Electricity and Magnetism by Purcell and Morin. In Chapter 4, entitled "Electric Currents", there is a small section 4.8 called "Energy Dissipation in Current Flow". I'd like to understand a certain snippet from this section.

In the book Electricity and Magnetism by Purcell and Morin, there is the following snippet

The flow of current in a resistor involves the dissipation of energy. If it takes a force $\vec{F}$ to push a charge carrier along with average velocity $\vec{v}$, any agency that accomplishes this must do work $F\cdot v$. If an electric field E is driving the ion of charge $q$, then $\vec{F}=q\vec{E}$, and the rate at which work is done is $q\vec{E}\cdot\vec{v}$.

My question is how to arrive at $\vec{F}\cdot\vec{v}$ as the rate that work is done?

 

[zenterix]

After writing out the question I realized that

$$W=\int_C \vec{F}\cdot d\vec{r}$$

$$=\int_{t_0}^{t} \vec{F}(\vec{r}(t))\cdot \vec{v}(t)dt$$

and so

$$\frac{dW}{dt}=\vec{F}(\vec{r}(t))\cdot \vec{v}(t)$$

 

[Gavran]

After writing out the question I realized that

$$W=\int_C \vec{F}\cdot d\vec{r}$$

$$=\int_{t_0}^{t} \vec{F}(\vec{r}(t))\cdot \vec{v}(t)dt$$

and so

$$\frac{dW}{dt}=\vec{F}(\vec{r}(t))\cdot \vec{v}(t)$$

This is correct.

There is a simpler and more understandable approach.
The rate at which work is done is power by definition, so:

$P = \frac {dW} {dt} = \frac {d} {dt} (\vec F \cdot \vec r) = \vec F \cdot \frac {d\vec r} {dt} = \vec F \cdot \vec v$

.

 

[weirdoguy]

There is a simpler and more understandable approach.

Which is not general since you used a very narrow, high-school definition of work.

 

[Gavran]

Which is not general since you used a very narrow, high-school definition of work.

Power is the rate at which work is done is definition of power, not of work. This definition was used here to make the answer to the question how to arrive at $\vec F \cdot \vec v$ as the rate that work is done more complete.

 

[weirdoguy]

Power is the rate at which work is done is definition of power, not of work.

I know, but you used high-school definition of work: $\vec{F}\cdot\vec{r}$. Thus your approach is not general. @zenterix approach is the most general.

 

[weirdoguy]

is definition of power

Also, the more general definition is the rate at which energy is transferred. Work is not the only way to transfer energy, there is also heat.