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Chapter 23: Problem 70

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.

Short Answer

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

Step by step solution

01

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 \(Q\).
02

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, \(\lambda = \frac{Q}{\pi a}\).
03

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 \(\frac{1}{4\pi\varepsilon_0}\).
04

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 \(dL = a d\theta\).
05

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 \(\pi\): \(V = \int_{0}^{\pi} \frac{k_e \lambda a d\theta}{a} = k_e \lambda \theta\bigg|_0^\pi = k_e \lambda \pi\).
06

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 \(V = \frac{k_e Q}{a}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攁 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.

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Most popular questions from this chapter

Nuclear Fusion in the Sun. The source of the sun's energy is a sequence of nuclear reactions that occur in its core. The first of these reactions involves the collision of two protons, which fuse together to form a heavier nucleus and release energy. For this process, called nuclear fusion, to occur, the two protons must first approach until their surfaces are essentially in contact. (a) Assume both protons are moving with the same speed and they collide head-on. If the radius of the proton is \(1.2 \times 10^{-15} \mathrm{m},\) what is the minimum speed that will allow fusion to occur? The charge distribution within a proton is spherically symmetric, so the electric field and potential outside a proton are the same as if it were a point charge. The mass of the proton is \(1.67 \times 10^{-27} \mathrm{kg} .\) (b) Another nuclear fusion reaction that occurs in the sun's core involves a collision between two helium nuclei, each of which has 2.99 times the mass of the proton, charge \(+2 e\) , and radius \(1.7 \times 10^{-15} \mathrm{m}\) . Assuming the same collision geometry as in part (a), what minimum speed is required for this fusion reaction to take place if the nuclei must approach a center-to-center distance of about \(3.5 \times 10^{-15} \mathrm{m}\) ? As for the proton, the charge of the helium nucleus is uniformly distributed throughout its volume. (c) In Section 18.3 it was shown that the average translational kinetic energy of a particle with mass \(m\) in a gas at absolute temperature \(T\) is \(\frac{3}{2} k T\) , where \(k\) is the Boltzmann constant (given in Appendix F). For two protons with kinetic energy equal to this avprage value to be able to undergo the process described in part (a), what absolute temperature is required? What absolute temperature is required for two average helium nuclei to be able to undergo the process described in part (b)? (At these temperatures, atoms are completely ionized, so nuclei and electrons move separately.) (d) The temperature in the sun's core is about 1.5 \(\times 10^{7}\) K. How does this compare to the temperatures calculated in part (c)? How can the reactions described in parts (a) and (b) occur at all in the interior of the sun? (Hint: See the discussion of the distribution of molecular speeds in Section \(18.5 . )\)

For each of the following arrangements of two point charges, find all the points along the line passing through both charges for which the electric potential \(V\) is zero (take \(V=0\) infinitely far from the charges) and for which the electric field \(E\) is zero: (a) charges \(+Q\) and \(+2 Q\) separated by a distance \(d,\) and \((\mathrm{b})\) charges \(-Q\) and \(+2 Q\) separated by a distance \(d .\) (c) Are both \(V\) and \(E\) zero at the same places? Explain.

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d .\) Two of the point charges are identical and have charge \(q .\) If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A has a radius three times that of sphere \(B\) . Let \(Q_{A}\) and \(Q_{B}\) be the charges on the two spheres, and let \(E_{A}\) and \(E_{B}\) be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio \(Q_{B} / Q_{A}\) and (b) the ratio \(E_{B} / E_{A} ?\)

A very small sphere with positive charge \(q=+8.00 \mu \mathrm{C}\) is released from rest at a point 1.50 \(\mathrm{cm}\) from a very long line of uniform linear charge density \(\lambda=+3.00 \mu \mathrm{Cm} .\) What is the kinetic energy of the sphere when it is 4.50 \(\mathrm{cm}\) from the line of charge if the only force on it is the force exerted by the line of charge?
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