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 P: 38 Hello!! In an electro-magnetic context, the power that an electric source of field delivers to the field itself may be written as $p_S = - \mathbf{E} \cdot \mathbf{J}$ where $\mathbf{E}$ is the electric field produced by the source and $\mathbf{J}$ is the corrent flowing on the source, forced by the generator. I have some doubts about the procedure to obtain the above expression. The current $J$ may be seen as a charge density $\rho$ which moves in a region with velocity $\mathbf{v}$. If in the region is present an electromagnetic field $\mathbf{E}, \mathbf{H}$, the charge will experience a force per unit volume $\mathbf{f} = \rho \mathbf{E} + \rho \mathbf{v} \times \mathbf{H}$ The charge will also experience an infinitesimal displacement $d\mathbf{r}$, in a time $dt$. If we had $d\mathbf{r} / dt = \mathbf{v}$, then we could say that the power per unit volume from the field is $\mathbf{f} \cdot \mathbf{v} = \rho \mathbf{E}\cdot \mathbf{v} + \rho \mathbf{v} \times \mathbf{H} \cdot \mathbf{v} = \rho \mathbf{E}\cdot \mathbf{v} = \mathbf{E}\cdot \mathbf{J}$ This is the power delivered from the field to the charge density. If we are interested in the power delivered from the charge density to the field, we have just to change the sign and we will obtain $- \mathbf{E}\cdot \mathbf{J}$. But if the moving charge $\rho$ is the source of the field, how can she "suffer" the effect of the field? If the source was an electric dipole, I imagine the field as something which goes away from the source, without interacting with the source. How can a field exercise a force on its source, minute by minute? Moreover, the displacement $d\mathbf{r}$ is due to the generator and not the field. Is it correct to calculate the work, and consequently the power, considering a force and a displacement which is not caused from the force? Thank you for having read it. Emily