Question on a step in deriving Poynting Theorem.

[yungman]

This is in page 346 of Griffiths "Introduction to Electrodynamics". This is regarding to work done by electromagnetic forces dW acting on charges in the interval dt.

dW = \vec F \cdot d \vec l =q( \vec E + \vec v X \vec B) \cdot d \vec l = q( \vec E + \vec v X \vec B) \cdot \vec v d t \;\;\hbox { Where }\; \vec v \;\hbox { is velocity, and }\; d \vec l = \vec v dt

\vec v X \vec B \;\hbox { is perpendicular to }\; \vec v \;\;\Rightarrow \; (\vec v X \vec B) \cdot \vec v \;=\; 0.

\hbox { Therefore }\; dW = \vec F \cdot d \vec l = q \vec E \cdot \vec v dt = \vec E \cdot \vec J d\tau d t

\hbox { Where }\; q=\rho_v d\tau, \;\hbox { and } \; \vec J = \rho_v \vec v

Here is where I have problem. From above:

\frac { d W}{dt} = (\vec E \cdot \vec J) d\tau

But the book gave:

\frac { d W}{dt} = \int_v (\vec E \cdot \vec J) d\tau

What is wrong with my derivation?

 

[Matterwave]

Your equation q=\rho_v d\tau is wrong. It should be dq, so that actually,

q=\int_V \rho d\tau

so now the integral is there.

 

[yungman]

Your equation q=\rho_v d\tau is wrong. It should be dq, so that actually,

q=\int_V \rho d\tau

so now the integral is there.

Thanks.

 

[yungman]

I have another question

dW = \vec F \cdot d \vec l =q( \vec E + \vec v X \vec B) \cdot d \vec l = q( \vec E + \vec v X \vec B) \cdot \vec v d t \;\;\hbox { Where }\; \vec v

dq=\rho_v d\tau, \;\hbox { and } \; \vec J = \rho_v \vec v

d \vec l = \vec v dt \Rightarrow\; dW = \vec F \cdot d \vec l = q \vec E \cdot \vec v dt = (\int_{v'} \rho_v dv') \;\vec E \;\cdot \;\vec v dt \;=\; \vec E \;\cdot\; [ (\int_{v'} \rho_v dv') \;\vec v ]\; dt

In order to move \;(\int_{v'} \rho_v dv')\; to combine with \vec v , \; \rho_v \; has to be a constant independent to spatial position. But the book did not specify this. Why?