- #1
jeff1evesque
- 312
- 0
Question:
Can someone remind me why the divergence of the electric flux is equal to the volume charge density,
[tex]\nabla \bullet \vec{D} = \rho_{v}[/tex] (where [tex]\vec{D}[/tex] is the electric flux density).
Thoughts:
The divergence measures the flow of a field out of a region of space. The del operator takes the gradient of the field, which measures the tendency of the field to diverge away in space (or the opposite). So when we take the divergence of the electric flux density, we are measuring how quickly the tendency of the flux to diverge in a given space. But how is that the volume charge density? Isn't charge density entirely different from the divergence of the electric flux?
Thanks,JL
Can someone remind me why the divergence of the electric flux is equal to the volume charge density,
[tex]\nabla \bullet \vec{D} = \rho_{v}[/tex] (where [tex]\vec{D}[/tex] is the electric flux density).
Thoughts:
The divergence measures the flow of a field out of a region of space. The del operator takes the gradient of the field, which measures the tendency of the field to diverge away in space (or the opposite). So when we take the divergence of the electric flux density, we are measuring how quickly the tendency of the flux to diverge in a given space. But how is that the volume charge density? Isn't charge density entirely different from the divergence of the electric flux?
Thanks,JL