A thin line of charge is aligned along the positive \(y\) -axis from \(0 \leq y \leq L,\) with \(L=4.0 \mathrm{~cm} .\) The charge is not uniformly distributed but has a charge per unit length of \(\lambda=A y,\) with \(A=\) \(8.0 \cdot 10^{-7} \mathrm{C} / \mathrm{m}^{2}\). Assuming that the electric potential is zero at infinite distance, find the electric potential at a point on the \(x\) -axis as a function of \(x\). Give the value of the electric potential at \(x=3.0 \mathrm{~cm} .\)

Understanding the concept of charge distribution is fundamental to comprehending various phenomena in electrostatics. When we talk about charge distribution, we refer to how electric charge is spread out over a certain area or volume. In our exercise, the charge is said to be distributed along a thin line extending from 0 to a length L along the y-axis.

The expression \(\lambda = Ay\), where A is a constant and y is the position along the line, describes that the charge per unit length varies with y. This non-uniform distribution results in varying electric influence at different points in space, affecting the calculation of electric potential. It's crucial to note that different types of charge distributions (continuous, surface, or volume) require different approaches when calculating electric fields or potentials.

The electrostatic constant, denoted by k, is also known as Coulomb's constant. It is a proportionality factor that appears in Coulomb's law and is used in calculations of electric force and electric potential. Its value is approximately \(8.99 \times 10^{9} \, Nm^{2} / C^{2}\).

This constant is essential as it relates the magnitude of the electric force between two point charges to the inverse square of the distance between them and to the product of the charges. In the context of electric potential, which is the focus of our exercise, k is used alongside the amount of charge and the distance from the charge to calculate the potential at any given point due to that charge.

In calculations of the electric potential, integration is a mathematical tool that allows us to sum up infinitesimal contributions from a charge distribution to find the total potential at a point. In the initial problem, we are required to calculate the electric potential due to a non-uniformly distributed charge along a line.

Setting Up the Integral

Starting with the differential electric potential \(dV = \cfrac{k \cdot dq}{r}\), we express the infinitesimal charge \(dq\) in terms of \(\lambda\) and \(dy\) to account for the non-uniform charge distribution. The distance \(r\) from each infinitesimal charge element to the point of interest is \(r = \sqrt{x^2 + y^2}\).

Evaluating the Integral

The total potential \(V(x)\) is determined by integrating over the entire length of the charge distribution. This process involves a definite integral from y = 0 to y = L, which sums the effects of all charge elements. Here, we incorporate both the charge distribution function and the geometry of the setup into the integration bounds and the integrand.

Performing this integral correctly is crucial for finding the precise electric potential at any given point along the x-axis. In more complicated setups, numerical integration may be required, especially when an analytical solution is not available.