Last edited: 2021-07-11 16:26:31

**The continuity equation is an important tool that one can use whether it is in a fluid dynamics problem or an electromagnetic problem. The continuity equation for charges relates the charge density to the current density in a volume.**

Lets derive the continuity equation for electric charges. If you are just searching for it, this is what it looks like:

$\frac{\partial \rho}{\partial t} = - \nabla \cdot \vec{j},$

where $\rho$ is the charge density and $\vec{j}$ is the current density (current divided by an area). Lets get a few things straight first. When we say the charge is conserved we are talking about the charge in an arbitrary volume. Additional charge can not be created from some sort of source inside the volume. The only way to change the charge is a flow of new charges from outside the volume, that would be our $\vec{j}$. So lets observe an infinitesimal surface element $d \vec{S}$ on the surface of our volume. The surface element is vector that is normal to the surface. So the infinitesimal charge $d q$ that passes through the surface element can be written as:

$dq = \vec{j} \cdot d \vec{S} dt.$

We'll use this expression later. The total charge $q$ for the whole volume $V$ can be written as:

$q = \int_V \rho dV,$

where earlier mentioned $\rho$ is the charge density for the volume. We can now use the conservation of charge and look at the change of the charge through time and flux of charge through the volume's surface $\partial V$. So we have two expressions, firstly:

$\frac{dq}{dt} = \int_V \frac{\partial \rho}{\partial t} dV,$

which is just the derivative of the total charge for the volume. Secondly, we have:

$\frac{dq}{dt} = - \int_{\partial V} \vec{j} \cdot d \vec{S},$

which uses the expression for the charge through a surface element. The reason for the minus sign is because we are calculating the inflow of charges but the surface element vector is pointing outwards from the volume and we have to counteract that. We are almost at our sought after expression but we'll have to use one more trick, the divergence theorem aka Guass's theorem which states:

$\int_V \nabla \cdot \vec{F} dV = \int_{\partial V} \vec{F} \cdot d \vec{S},$

which fits perfectly with our charge flux expression:

$- \int_{\partial V} \vec{j} \cdot d \vec{S} = - \int_V \nabla \cdot \vec{j} dV.$

So now we have:

$\frac{dq}{dt} = \int_V \frac{\partial \rho}{\partial t} dV = - \int_V \nabla \cdot \vec{j} dV.$

As you can see we have two expression with a volume integral for the volume $V$. So we can remove the integral and we finally get our continuity equation for electric charges:

$\frac{\partial \rho}{\partial t} = - \nabla \cdot \vec{j}.$

Now lets see how the continuity equation looks in different coordinate systems.

The continuity equation for charges in Cartesian coordinates is given by:

$\frac{\partial \rho}{\partial t} + \frac{\partial j_x}{\partial x} + \frac{\partial j_y}{\partial y} + \frac{\partial j_z}{\partial z} = 0.$

The continuity equation for charges in cylindrical coordinates is given by:

$\frac{\partial \rho}{\partial t} + \frac{1}{r} \frac{\partial}{\partial r}(r j_r) + \frac{1}{r} \frac{\partial}{\partial \theta}(j_\theta) + \frac{\partial}{\partial z}(j_z) = 0.$

The continuity equation for charges in spherical coordinates is given by:

$\frac{\partial \rho}{\partial t} + \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 j_r) + \frac{1}{r \sin{\theta}} \frac{\partial}{\partial \theta}(j_\theta \sin{\theta}) + \frac{1}{r \sin{\theta}} \frac{\partial}{\partial \varphi}(j_\varphi) = 0.$

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