Explaining Electric Current Density and Conservation Law - Jeffrey

[jeff1evesque]

Statement:
Couls someone explain the following equation to me:
$$\int \int_{surface} \vec{J} \cdot \vec{ds} = -\frac{\partial Q_{enclosed}}{\partial t},$$ where $$\vec{J_{v}}$$ is the electric current density. Question:
I am more confused with the term on the right side. The left side simply determines the current density on some surface, and that is why it is a surface integral.

But the right side says the surface has a negative current density. Why can't we just take the negative sign off, and assume if the resulting answer is less than the initial conditions, then the charge density in a particular region has decreased? Can someone either explain this to me, or provide an example.Thanks,Jeffrey

 

[Dick]

It's because ds is the OUTWARD pointing normal of the surface. If the current flow is outward from the sphere then the left side is positive because J.ds is positive. But that corresponding to decreasing charge in the sphere.

 

[jeff1evesque]

It's because ds is the OUTWARD pointing normal of the surface. If the current flow is outward from the sphere then the left side is positive because J.ds is positive. But that corresponding to decreasing charge in the sphere.

So the right term corresponds to how much charge from the enclosed surface has left per unit time t. Not how much charge remains in the closed surface after a unit time?

 

[Redbelly98]

Yes, that's correct.

The charge that remains would simply be Q_enclosed.