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Chapter 1: Problem 47

Field from a semicircle * A thin plastic rod bent into a semicircle of radius \(R\) has a charge \(Q\) distributed uniformly over its length. Find the electric field at the center of the semicircle.

Short Answer

Expert verified
The electric field at the center of the semicircular rod is \( kQ/(R^2) \), where \( k \) is Coulomb's constant and direction is upwards.

Step by step solution

01

Express the Charge Density 位

The charge \(Q\) is uniformly distributed over the plastic rod. Consequently, the linear charge density \(\lambda\), is \( Q/( \pi R)\), since the total length of the semicircular rod is \(\pi R\), the radius being \(R\).
02

Set Up the Integral for Electric Field

Visualize the semicircular rod by marking its center (the observation point), and individual test points (length elements \( dl \)) over the rod. The electric field \( dE \) due to an element \( dl \) on the rod is \( dE = k \lambda dl / r^2 \), where \( r \) is the radius of the semicircle. Because of symmetry, the horizontal components of the electric fields due to each charge element cancels the other. Thus, the vertical component of \( dE \) is \( dEy = dE \sin\theta = k \lambda dl \sin\theta / r^2 \). Here \(\theta\) is the angle made by the radius vector with the horizontal.
03

Integrate over the Semicircle

Electric field at the center, which is the vertical component \( Ey \), can be given by the integral \( Ey = \int_{0}^{\pi} dE \sin\theta \) which upon substituting the expression for \( dEy \) and evaluating, gives \( Ey = k \lambda / R \), or \( Ey = kQ/(R^2) \), as \( \lambda \) is \( Q/( \pi R)\). Here, \( k \) is Coulomb's constant.

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Key Concepts

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

Understanding the Semicircle
A semicircle is simply half of a circle. Imagine a thin plastic rod bent into this shape. The semicircle in our exercise has a radius denoted by \( R \). The concept of radius here is crucial because it helps us determine the length of the semicircle, which is \( \pi R \).
  • The length is important when considering how the charge is spread across the rod.
  • Since the rod forms a semicircle, it is symmetric around the vertical axis.
This symmetry becomes a key factor when calculating the electric field, as it dictates how components of the field behave.
Exploring Charge Density
Charge density, denoted as \( \lambda \), represents how much charge \( Q \) is distributed over the length of the rod. Here it is described as linear charge density because it spreads along a line.
  • Given that the charge is uniformly distributed, \( \lambda = \frac{Q}{\pi R} \).
  • This means each small length segment \( dl \) of the rod has the same proportion of charge.
This uniform distribution simplifies calculations, allowing us to focus on how each small part contributes equally to the overall electric field.
Applying Coulomb's Law
Coulomb's law is fundamental in understanding electric fields. It expresses the relationship between electrically charged particles. In this scenario, it helps us calculate the electric field produced by a small element of charge.
  • The formula \( dE = k \lambda \frac{dl}{r^2} \) represents the small electric field from a charge element \( dl \).
  • Here, \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point where the field is calculated, which is just \( R \) in this semicircle.
Coulomb's law gives us a way to quantify the tiny contributions of each part of the rod towards the total field.
Role of Integration
Integration allows us to accumulate small influences over an entire interval. In our exercise, it helps sum up the contributions from each tiny piece of the semicircle.
  • To find the total electric field, calculate \( Ey = \int_{0}^{\pi} dE \sin\theta \).
  • This involves integrating from \( 0 \) to \( \pi \), accounting for the vertical components, as horizontal cancel out symmetrically.
By integrating, we efficiently combine these small parts, yielding a single value representing the whole effect at the center.
Appreciating Symmetry in the Problem
Symmetry is a powerful tool in physics, especially when dealing with configurations like our semicircle.
  • Due to symmetry, horizontal components of electric fields from opposite sides cancel out.
  • All that's left are the vertical components, simplifying our calculations significantly.
Understanding symmetry helps reduce the complexity of a problem, allowing us to focus only on the parts of the field that don't cancel each other out, like the vertical component in this exercise.

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Most popular questions from this chapter

Force in a soap bubble ** Like the charged rubber balloon described at the end of Section 1.14, a charged soap bubble experiences an outward electrical force on every bit of its surface. Given the total charge \(Q\) on a bubble of radius \(R\), what is the magnitude of the resultant force tending to pull any hemispherical half of the bubble away from the other half? (Should this force divided by \(2 \pi R\) exceed the surface tension of the soap film, interesting behavior might be expected!)

Fields at the surfaces Consider the electric field at a point on the surface of (a) a sphere with radius \(R\), (b) a cylinder with radius \(R\) whose length is infinite, and (c) a slab with thickness \(2 R\) whose other two dimensions are infinite. All of the objects have the same volume charge density \(\rho\). Compare the fields in the three cases, and explain physically why the sizes take the order they do.

Work for a rectangle ** Two protons and two electrons are located at the corners of a rectangle with side lengths \(a\) and \(b\). There are two essentially different arrangements. Consider the work required to assemble the system, starting with the particles very far apart. Is it possible for the work to be positive for either of the arrangements? If so, how must \(a\) and \(b\) be related? You will need to solve something numerically.

An equilateral triangle Three positive charges, \(A, B\), and \(C\), of \(3 \cdot 10^{-6}, 2 \cdot 10^{-6}\), and \(2 \cdot 10^{-6}\) coulombs, respectively, are located at the corners of an equilateral triangle of side \(0.2 \mathrm{~m}\). (a) Find the magnitude in newtons of the force on each charge. (b) Find the magnitude in newtons/coulomb of the electric field at the center of the triangle.

Find a geometrical arrangement of one proton and two electrons such that the potential energy of the system is exactly zero. How many such arrangements are there with the three particles on the same straight line? You should find that the ratio of two of the distances involved is the golden ratio.
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