Problem 70 - University Physics with Modern Physics

Question

A thin insulating rod is bent into a semicircular arc of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.

Step-by-Step Solution

Answer

The electric potential at the center is \(\frac{k_e Q}{a}\).

Step 1: Understanding the Problem

We need to find the electric potential at the center of a semicircular arc made by a uniformly charged insulating rod with a radius of \(a\) and total charge \(Q\).

Step 2: Charge Density Calculation

First, calculate the linear charge density \(\lambda\), which is the charge per unit length, given by \(\lambda = \frac{Q}{L}\). Since the rod is bent into a semicircle, its length \(L\) is half the circumference of a circle, \(L = \pi a\). Thus, \(\lambda = \frac{Q}{\pi a}\).

Step 3: Expressing Potential at a Point

The potential \(V\) due to a small charge element \(dq\) at a distance \(r\) is given by \(dV = \frac{k_e dq}{r}\), where \(k_e\) is Coulomb's constant \(\frac{1}{4\pi\varepsilon_0}\).

Step 4: Setup for Integration

For a semicircular arc, every charge element is at a distance \(a\) from the center of curvature. Therefore, \(dV\) simplifies to \(dV = \frac{k_e dq}{a}\). Substitute \(dq = \lambda dL\), where \(dL\) is a small arc length and \(dL = a d\theta\).

Step 5: Integrating Over the Semicircle

Substitute \(dq = \lambda a d\theta\) into \(dV = \frac{k_e \lambda a d\theta}{a}\) and integrate over the angle \(\theta\) from \(0\) to \(\pi\): \(V = \int_{0}^{\pi} \frac{k_e \lambda a d\theta}{a} = k_e \lambda \theta\bigg|_0^\pi = k_e \lambda \pi\).

Step 6: Substitute Charge Density

Substitute \(\lambda = \frac{Q}{\pi a}\) into \(V = k_e \lambda \pi\) to get \(V = \frac{k_e Q}{a}\). The \(\pi\) terms cancel, resulting in \(V = \frac{k_e Q}{a}\).

Key Concepts

Semicircular Arc

A semicircular arc is essentially half a circle. It represents a shape created when a full circle is divided into two equal parts. In this exercise, you encounter a thin, insulating rod that has been crafted into this semicircle shape.
Consider the radius, labeled as \(a\), which is the constant distance from any point on the arc directly to the center. Here, we refer to this center as the 'center of curvature.' Even though it only represents a half-circle, it plays an essential role in electrostatics.

• A semicircle simplifies symmetry problems in physics.
• Its geometry helps set up integrals to find quantities like potential efficiently.

The arc is half as long as the full circumference of a circle of the same radius, hence its perimeter is \(\pi a\), setting the stage for other calculations.

Uniform Charge Distribution

Uniform charge distribution refers to charges spread evenly across a particular object or length. In this problem, it means the insulating rod carries an equal amount of electric charge \(Q\) along every segment of its semicircular length.
This concept simplifies calculations because the charge is evenly distributed. It allows us to use charge densities to figure out how much charge is present in each segment of the rod.

• Uniform distribution ensures no segment of the rod has more charge than another.
• This predictability makes it easier to calculate things like electric potential or electric field.

Recognizing uniformity is essential in simplifying the mathematical equations and ensuring they are accurate for our use case.

Linear Charge Density

Linear charge density, depicted as \(\lambda\), represents the amount of charge per unit length along the rod. It's a pivotal calculation in problems involving charged rods or wires when dealing with potential or fields.
For a semicircular arc, this means dividing the total charge \(Q\) by the total length of the arc, which is \(\pi a\). Thus, \(\lambda\) becomes \(\frac{Q}{\pi a}\). Essentially, it quantifies how much charge resides on a unit segment of the rod.
This concept is crucial in Coulomb's law problems and electrostatics in general, as it bridges the total charge to how it is physically applied along a structure. It provides a means to break continuous charge distributions into manageable calculations.

Coulomb's Constant

Coulomb's constant, denoted as \(k_e\), is a foundational constant in electrostatics. It is expressed as \(\frac{1}{4 \pi \varepsilon_0}\), where \(\varepsilon_0\) is the vacuum permittivity—a measure of resistance to forming an electric field in a vacuum.
It essentially dictates the strength of the force or potential due to charges. In the context of this exercise, it helps compute the potential created at a point by a charged object.

• \(k_e\) is a constant used to convert charge and distance into force or potential.
• It is paramount to the computations involving electric potentials and fields.

Utilizing Coulomb's constant allows us to precisely calculate how charges interact at a distance, making it a key part of solving various electrostatic problems.