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Chapter 23: Problem 70

A thin insulating rod is bent into a semicircular are 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 are if the potential is assumed to be zero at infinity.

Short Answer

Expert verified
The potential at the center is \( V = \frac{kQ}{a} \).

Step by step solution

01

Concept Clarification

To find the electric potential at the center of curvature, we need to integrate the contribution from each infinitesimal charge element along the semicircular arc. The potential due to a point charge is given by the formula \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is measured.
02

Define the Charge Distribution

Since the total charge \( Q \) is uniformly distributed along the semicircular rod, the linear charge density \( \lambda \) is given by \( \lambda = \frac{Q}{L} \), where \( L \) is the length of the semicircular arc. The length \( L \) is \( \pi a \), because the rod forms a semicircle of radius \( a \). Therefore, \( \lambda = \frac{Q}{\pi a} \).
03

Consider an Infinitesimal Charge Element

Let \( dq \) be an infinitesimal charge element on the semicircular arc. This charge element can be expressed as \( dq = \lambda \, ds \), where \( ds \) is an infinitesimal arc length.
04

Set Up the Integral for Potential

The potential at the center of curvature due to an infinitesimal charge \( dq \) located at a distance \( a \) is given by \( dV = \frac{k \, dq}{a} \). Substituting \( dq = \lambda \, ds \), we have \( dV = \frac{k \, \lambda \, ds}{a} \).
05

Integrate Over the Semicircle

To find the total potential \( V \) at the center, integrate \( dV \) over the entire arc. Thus, \( V = \int dV = \int \frac{k \, \lambda \, ds}{a} = \frac{k \, \lambda}{a} \int ds \). Since the arc length \( ds \) runs from 0 to \( \pi a \), the integral simplifies to \( V = \frac{k \, \lambda}{a} \cdot \pi a \).
06

Simplify the Expression

Substituting \( \lambda = \frac{Q}{\pi a} \) into the expression from Step 5, we get \( V = \frac{k \, Q}{\pi a} \cdot \pi \), which simplifies to \( V = \frac{kQ}{a} \). This is the potential at the center of curvature.

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Key Concepts

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

Electric Potential
Electric potential is a fundamental concept in electrostatics. It represents the potential energy per unit charge at a specific point in space due to electric fields. Consider it as the ability of an electric field to do work on a charge. The electric potential is measured in volts.
  • A point in space can have an electric potential even if no charge is present there.
  • Typically, the reference point where potential is zero is taken at infinity, making calculations easier.
  • The formula for electric potential due to a point charge is: \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant (approximately \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)), \( Q \) is the charge, and \( r \) is the distance from the charge.
In the context of this problem, the electric potential at the center of curvature of the semicircular rod is calculated by summing the potentials from each infinitesimally small charge along the rod, efficiently done by the method of integration due to the continuous distribution of charge.
Integral Calculus
Integral calculus is a branch of mathematics that deals with sums of infinitely small quantities. It is vital in physics for calculating quantities like electric potential when dealing with continuous distributions of charge. The integral allows us to add up all the infinitesimal contributions to a certain quantity, like electric potential in this case.
  • An infinitesimal charge segment \( dq \) along the semicircular rod produces a small potential \( dV \), given by \( dV = \frac{k \, dq}{a} \).
  • Using the linear charge density \( \lambda = \frac{Q}{\pi a} \), \( dq \) becomes \( \lambda \, ds \), where \( ds \) is a differential arc length.
  • We set up the integral \( V = \int \frac{k \, \lambda \, ds}{a} \). The integration from 0 to \( \pi a \) sums up the potential across the semicircle.
Ultimately, integral calculus provides the framework to evaluate this sum, yielding the total electric potential at the center.
Coulomb's Law
Coulomb's law describes the electrostatic interaction between electric charges. It states that the force between two point charges is directly proportional to the product of their charge magnitudes and inversely proportional to the square of the distance separating them. The law is formally expressed as:
\[ F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} \]Where:
  • \( F \) is the electrostatic force between the charges.
  • \( Q_1 \) and \( Q_2 \) are the two charges.
  • \( r \) is the distance between them.
  • \( k \) is Coulomb's constant.
In problems involving electric potential, Coulomb's law is fundamental to understanding how charges influence each other and generate fields. For the semicircular rod, each infinitesimal charge element contributes to the potential at the center, helping us apply these principles by integrating across the rod's distribution.

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Most popular questions from this chapter

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1000 \(\mathrm{km} / \mathrm{s}\) , measured relative to the earth. Find the maximum electrical force that these protons will exert on each other.

Self-Energy of a Sphere of Charge. A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly through- out its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge \(q\) in a sphere of radius \(r\) , how much energy would it take to add a spherical shell of thickness \(d r\) having charge \(d q ?\) Then integrate to get the total energy.)

A uniformly charged thin ring has radius 15.0 \(\mathrm{cm}\) and total charge \(+24.0 \mathrm{nC}\) . An electron is placed on the ring's axis a dis0tance 30.0 \(\mathrm{cm}\) from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring.

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\overrightarrow{\boldsymbol{E}}\) directed to the left. It is released from rest and moves to the left; after it has moved \(6.00 \mathrm{cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{J}\) . \(\mathrm{J}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of \(\overrightarrow{\boldsymbol{E}} ?\)
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