Electrodynamics Continuity Equation

[CSOleson]

Homework Statement

I am currently studying for a quiz and then following a test in my Electrodynamics test. Right now I am struggling to define the following:
Continuity equation and its physical meaning

Homework Equations

The Continuity Equation is given as the following:
∇J=-∂ρ/∂t

The Attempt at a Solution

There is no solution, I just know that it is the mathematical statement of a local charge conservation (defined by Griffiths). So what is the physical meaning? Can someone help me wrap my head around this concept?

 

[vela]

You have a typo. It should be
$$\frac{\partial\rho}{\partial t} = \nabla\cdot \vec{J}.$$ It might be easier for you to understand in integral form. If you integrate the equation over a volume V bounded by a closed surface S, you get
$$\frac{\partial}{\partial t}\int_V \rho\,dv = \int_V (\nabla\cdot \vec{J})\,dv = \oint_S \vec{J}\cdot d\vec{S}$$ where the last equality is the divergence theorem. Can you interpret that equation?

 

[joshuagobin]

I am in the same position as you my friend. but I think that the closest I could physically interpret it is if I use an analogy of water being displaced. A drop of water can be considered as a current density J and they are saying if the drop of water begins to flow away from its point of rest, then as you sum up all the portions that are flowing away between the time it flows away then you would get the same flow rate as if the entire drop of water were to move from one point to another in the same time frame. If I am lacking in my understanding, can someone please steer me in the right direction.

 

[athrun200]

You have a typo. It should be
$$\frac{\partial\rho}{\partial t} = \nabla\cdot \vec{J}.$$ It might be easier for you to understand in integral form. If you integrate the equation over a volume V bounded by a closed surface S, you get
$$\frac{\partial}{\partial t}\int_V \rho\,dv = \int_V (\nabla\cdot \vec{J})\,dv = \oint_S \vec{J}\cdot d\vec{S}$$ where the last equality is the divergence theorem. Can you interpret that equation?

Is it describing the flowing charge produces a current?

 

[vela]

No. A flow of charge is by definition a current.