# derive-it.com

## Gauss’s Law in One Dimension and in Three Dimensions

Recall the following form of Gauss’s Law from this post:

$\frac{\partial}{\partial x} E_x + \frac{\partial}{\partial y} E_y + \frac{\partial}{\partial z} E_z = \frac{\rho}{\epsilon_0 \epsilon_r}$

### One Dimension

If a problem is “one-dimensional” along the, say, $x$-axis, one has

$\frac{\partial}{\partial x} E_x = \frac{\rho}{\epsilon_0 \epsilon_r}$

In words: the partial derivative with respect to $x$ of the $x$-component of the electric field is equal to the volume charge density divided by $\epsilon_0 \epsilon_r$. In one dimension, sometimes people use a linear charge density $\lambda$ instead of a volume charge density $\rho$.

As the partial derivative with respect to $x$ of $E_x$ increases, the volume charge density increases. Alternative, as the volume charge density increases, the partial derivative with respect to $x$ of $E_x$ increases.

### Three Dimensions

With all three Cartesian coordinates, one has:

$\frac{\partial}{\partial x} E_x + \frac{\partial}{\partial y} E_y + \frac{\partial}{\partial z} E_z = \frac{\rho}{\epsilon_0 \epsilon_r}$

So, if $\rho$ increases, it is unclear whether that increase will affect all three terms on the left in the same way, only one term, only two terms, or all three terms in an unequal way.

But, if one knows that $\frac{\partial}{\partial x} E_x$ increases, it is clear that $\rho$ increases, assuming $\epsilon_0 \epsilon_r$ is held constant.

### Conceptual Comment

I conclude with the rather broad statement that Gauss’s Law is concerned with how an electric field vector changes with respect to spatial coordinates. In this case, those spatial coordinates are Cartesian coordinates.