Christina DanielIn Derivation #3, the expression, $ \frac{\partial \beta(t)}{\partial t}$, was written. This is an expression for a derivative of a function $ \beta(t)$. Now that a derivative has been introduced, Maxwell’s equations can be investigated. I start with Gauss’ Law. But first, slightly more information about derivatives is needed.
I can consider the pieces of $ \frac{\partial \beta(t)}{\partial t}$ separately. These pieces are $ \beta(t)$ and $ \frac{\partial}{\partial t}$. The $ \beta(t)$ is, of course, a function while $ \frac{\partial}{\partial t}$ is an operator called a differential operator that forms $ \frac{\partial \beta(t)}{\partial t}$ if $ \frac{\partial}{\partial t}$ is applied to the left of the function $ \beta(t)$.
A differential operator can be made for each of the coordinates in a particular Cartesian coordinate system. For example, I can write $ \frac{\partial}{\partial x}$ for the $ x$-axis, $ \frac{\partial}{\partial y}$ for the $ y$-axis, and $ \frac{\partial}{\partial z}$ for the $ z$-axis.
Next, define three unit vectors. Let $ \vec{e}_x$ point along the positive $ x$-axis, $ \vec{e}_y$ point along the positive $ y$-axis, and $ \vec{e}_z$ point along the positive $ z$-axis. Now there is a unit vector for each of three spatial dimensions.
The differential operators and the unit vectors can be combined to form another operator called the del-operator, $ \vec{\nabla}$.
Definition 1 Define $ \vec{\nabla} \equiv \frac{\partial}{\partial x} \vec{e}_x + \frac{\partial}{\partial y} \vec{e}_y + \frac{\partial}{\partial z} \vec{e}_z $. Equivalently, one can use the notation $ \vec{\nabla} \equiv (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$.
Now it is possible to write Gauss’ Law. Knowing what the end result should be is helpful for the upcoming derivation. It is $ \vec{\nabla} \cdot \vec{E}(\vec{r}) = 4 \pi \rho(\vec{r}).$ In this equation, $ \vec{r} \equiv (x,y,z)$ is the position vector, $ \vec{E}(\vec{r}) \equiv (E_x(\vec{r}), E_y(\vec{r}), E_z(\vec{r}))$ is the electric field, and $ \rho(\vec{r})$ is the charge density, or “the charge per unit volume at any point $ (x,y,z)$” [1].
Hence a logical starting point for this derivation is to consider an electron centered at the origin of a Cartesian coordinate system. Suppose a finite, spherical volume $ V$ centered at the origin $ (0,0,0)$ contains the electron. The electron has a charge of $ -e$, so the charge density $ \rho(\vec{r})$ is $ -\frac{e}{V}$. The next question is, what is the electric field produced by this electron?
The answer to this question is given by analyzing the electric field vectors that exist due to the presence of the electron. Note that these electric field vectors are mathematical entities that are not assumed to have an actual physical existence. In physics, electric field vectors are defined so that a positive charge follows the direction of the arrows of the field vectors. Since unlike charges attract, the arrows of the electric field vectors for this scenario point towards the electron rather than away.
In the following drawing, the electron is schematically represented such that its center of mass is located at the origin. The center of mass is a concept used in engineering. If the electron is a point with zero volume, the center of mass of the electron is still at the origin. Note that if all of the electron’s mass if located at the origin, then the density $ \frac{m_e}{V_e}$ of the electron would be undefined due to division by zero. Here, $ m_e$ is the mass of the electron and $ V_e$ is the volume of the electron. The electron must be spherically symmetric in order for the electric field vectors to not vary in magnitude within the electron’s structure.
The next key point is that the electron exists in three-dimensional space. The electric field vectors are ideally defined at each point in space regardless of the structure of the electron. There is nothing else in the system besides the electron, so there is no reason for the electric field vectors to differ from one position on the sphere to another position on the sphere. A position on the sphere is considered because the magnitude of the electric field vector can change depending on how far away a test charge is from the electron. This is known from the equation $ \vec{F}(\vec{r}) = q_2 \vec{E}(\vec{r})$ in combination with Coulomb’s law, $ \vec{F}(\vec{r}) = \frac{k q_1 q_2}{|\vec{r}|^2} \vec{e}_r$. Here, $ \vec{F}(\vec{r})$ is the force on a test charge $ q_2$ located at $ \vec{r}$ due to the presence of a source charge $ q_1$. Hence the electric field at any position due to a source charge $ q_1$ is $ \vec{E}(\vec{r}) = \frac{\vec{F}(\vec{r})}{q_2} = \frac{k q_1}{|\vec{r}|^2} \vec{e}_r$. Here, $ k$ is a proportionality constant, and $ \vec{e}_r$ is a unit vector pointing from the position of the source charge to the position of the test charge. In this way, $ \vec{e}_r$ points away from the origin if a positively-charged source charge $ q_1$ exists at the origin and a positively-charged test charge $ q_2$ is somewhere away from the origin.
The equation $ \vec{E}(\vec{r}) = \frac{k q_1}{|\vec{r}|^2} \vec{e}_r$ is important for understanding the configuration of the electric field vectors. This equation implies that the magnitude of the electric field is inversely proportional to the square of the distance from the origin, if there is a source charge at the origin.
Recall that the surface area of a sphere is $ 4 \pi R^2$. A sphere is considered because the electric field must have spherical symmetry. There is spherical symmetry because $ \vec{E}(\vec{r}) = \frac{k q_1}{|\vec{r}|^2} \vec{e}_r$ does not depend on any Cartesian coordinate more than another, as $ |\vec{r}|^2 = x^2 + y^2 + z^2$.
The unit vector $ \vec{e}_r$ in this example also always points away from the origin in the direction of the test charge. Since a negative charge is located at the origin and a positive charge is attracted to a negative charge, the electric field vectors should point towards the electronic source charge at the origin.
From this discussion it is clear that the electric field vectors should have equal magnitudes at all points on the sphere, and the the direction of each electric field vector should point radially inward from the point on the sphere towards the center of the sphere. This sphere, by the way, is called a Gaussian surface. Also, I am not considering any electric field vector inside of the electron.
The next question to consider is how does the electric field vary in three-dimensional space? The answer to this question will lead closer to Gauss’ law.
References
[1] Melvin Schwartz. Principles of Electrodynamics. Dover, 1972.