In free space, let \(\rho_{v}=200 \epsilon_{0} / r^{2.4}\). (a) Use Poisson's equation to find \(V(r)\) if it is assumed that \(r^{2} E_{r} \rightarrow 0\) when \(r \rightarrow 0\), and also that \(V \rightarrow 0\) as \(r \rightarrow \infty\). (b) Now find \(V(r)\) by using Gauss's law and a line integral.

Poisson's equation is a fundamental equation in electromagnetism that relates the charge density distribution within a region to the electric potential in that region. Simply put, it describes how the presence of charges influences the electric potential landscape.

The general form of Poisson's equation is \[\begin{equation}abla^2 V = -\frac{\rho}{\epsilon_0}\end{equation}\]
where \( V \) is the electric potential, \( \rho \) is the charge density, and \( \epsilon_0 \) is the permittivity of free space. The operator \( abla^2 \) is known as the Laplacian, which in spherical coordinates (appropriate for radial symmetry problems like in our original exercise) is written specifically to reflect the geometry of the situation.

In the step-by-step solution of our example, the charge density is given a specific form \( \rho_v = \frac{200\epsilon_0}{r^{2.4}} \), which, when inserted into Poisson's equation, requires integration to solve for the potential \( V(r) \). Students should pay close attention to the boundary conditions given, as they determine the constants of integration when solving differential equations like this one. Understanding how to apply these conditions is key to mastering the use of Poisson's equation in electromagnetic theory.

Gauss's law is a central principle in electromagnetism encapsulated by the equation: \[\begin{equation}\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\mathrm{enc}}}{\epsilon_0}\end{equation}\]
where \( \vec{E} \) is the electric field, \( d\vec{A} \) is a vector representing an infinitesimal area on a closed surface, \( Q_{\mathrm{enc}} \) is the total charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space.

Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed within it. This law is incredibly powerful, especially for symmetric charge distributions, because it allows us to calculate the electric field without knowing the potential. In our step-by-step exercise, Gauss's law is used to deduce the electric field from the known charge density, which is then integrated to find the electric potential \( V(r) \).

It's essential to understand that Gauss's law applies to all closed surfaces, and choosing the right shape of the Gaussian surface is important to simplify the calculations. For spherical charge distributions, a spherical Gaussian surface is appropriate, as seen in the example, simplifying the use of Gauss's law greatly. Recognizing symmetries and choosing the best Gaussian surface are skills that students will develop through practice.

Electric potential, denoted by \( V \), is a scalar quantity that represents the electric potential energy per unit charge at a given point in space. The difference in electric potential between two points is related to the work done in moving a small test charge between these points, without any acceleration.

The electric potential is related to the electric field \( \vec{E} \) through the negative gradient: \[\begin{equation}\vec{E} = -abla V\end{equation}\]
This equation signifies that the electric field is the steepest downhill slope of the potential, and the negative sign indicates that the electric field points in the direction of decreasing potential. In our solved problem, once the electric field was found using Gauss's law, the potential was determined by taking the line integral of the electric field, which essentially means summing the contributions of the electric field along a path from infinity to the point \( r \).

This concept is vital as it allows charges to be characterized by their energy landscape rather than just the forces they experience. It's more engaging to use the potential to understand the 'shape' of an electric field. Moreover, electric potential is often simpler to work with mathematically because it is a scalar field as opposed to the vector field nature of the electric field. For students studying electromagnetism, becoming familiar with electric potential is as crucial as understanding the electric field.