A disk of radius \(a\) carries nonuniform surface charge density \(\sigma=\sigma_{0}(r / a),\) where \(\sigma_{0}\) is a constant. (a) Find the potential at an arbitrary point \(x\) on the disk axis, where \(x=0\) is the disk center. (b) Use the result of (a) to find the electric field on the disk axis, and (c) show that the field reduces to an expected form for \(x \gg a\)

To understand electric fields and potentials, one must first grasp the concept of surface charge density. Surface charge density, denoted as \( \sigma \), represents the amount of electric charge per unit area on a surface. In our disk example, the charge density is not uniform but varies with the radius \( r \) as \( \sigma = \sigma_0 (\frac{r}{a}) \) where \( \sigma_0 \) is a constant and \( a \) is the radius of the disk. This means that the charge density is directly proportional to the distance from the center of the disk.

Understanding surface charge density is crucial because it impacts the resulting electric field and potential. With a non-uniform density, as in this case, calculating the field and potential becomes more complex, necessitating integration to sum the contribution of every point on the disk.

Electric potential at a point in space is the work done in bringing a unit positive charge from infinity to that point, without any acceleration. It's a scalar quantity denoted as \( V \). In our disk problem, we find the potential \( V \) at a point on the disk axis by integrating the contributions from each element of the disk. The potential from a small element is given by \( dV = \frac{k dq}{\sqrt{x^2 + r^2}} \), where \( dq \) is the infinitesimal charge on the element, \( k \) is the Coulomb constant, \( r \) is the distance from the disk's center to the element, and \( x \) is the distance along the axis to the point of interest.

Integrating \( dV \) over the whole disk, considering the varying charge density \( \sigma \) gives us the total potential at \( x \). The electric potential is significant in electric field studies as it relates directly to the force that a charge would experience in the field.

Coulomb's Law is foundational when discussing electric fields and charges. It states that the force \( F \) between two charges is proportional to the product of the charges \( q_1 \) and \( q_2 \) and inversely proportional to the square of their separation \( r \) with \( F = k \frac{q_1 q_2}{r^2} \), where \( k \) is the Coulomb's constant. This principle is a cornerstone in electrostatics and underlies the calculations we perform to find the potential and field in our disk problem.

Coulomb's Law facilitates the understanding of how charges interact and influence the space around them, culminating in what we perceive as electric fields and potentials.

Integration is a mathematical tool frequently used in physics to sum up small contributions to calculate a total effect. It is essential when dealing with quantities that are distributed continuously, such as charge, mass, or probability. In our disk problem, integration is required to calculate the total potential at a point \( x \) on the axis of the disk due to the continuous distribution of charge.

The process of integrating allows us to add up infinitely many infinitesimally small potential contributions \( dV \) from each element of charge \( dq \) over the entire disk. The integral, \( V = \int_0^a \frac{k \sigma_{0} \frac{r}{a} 2\pi r dr}{\sqrt{x^2 + r^2}} \), encapsulates the total electric potential due to the entire surface charge. Integration transforms the many tiny effects into a single, meaningful whole, which is pivotal for predicting physical phenomena.

While integration sums a continuous distribution, differentiation deals with rates of change. In physics, differentiation helps find the rate at which one quantity changes with respect to another. For our disk problem, once we integrate to find the potential \( V \), we then differentiate \( V \) with respect to \( x \)—the distance along the disk axis—to determine the electric field \( E_x \) at that point.

The electric field is defined as the negative gradient of the electric potential, which translates into \( E_x = -\frac{dV}{dx} \). By differentiating the potential, we acquire the rate at which the potential changes, giving us the intensity and direction of the electric field. Differentiation is thus a critical operation for analyzing and understanding dynamic systems in physics.