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Chapter 1: Problem 65

Building a sheet from rods ** An infinite uniform sheet of charge can be thought of as consisting of an infinite number of adjacent uniformly charged rods. Using the fact that the electric field from an infinite rod is \(\lambda / 2 \pi \epsilon_{0} r\), integrate over these rods to show that the field from an infinite sheet with charge density \(\sigma\) is \(\sigma / 2 \epsilon_{0}\)

Short Answer

Expert verified
The electric field from an infinite sheet with charge density \(\sigma\) is \(\sigma / 2 \epsilon_{0}\). This shows the uniformity of the field created by the infinite sheet with a given charge density.

Step by step solution

01

Define The Charge Density For A Rod

For a rod with uniform charge density, we have \(\lambda\) as the charge per unit length. We will use the fact that the electric field from an infinite rod is \(\lambda / 2 \pi \epsilon_{0} r\) where \(\epsilon_{0}\) is the permittivity of the vacuum and \(r\) is the perpendicular distance from the rod.
02

Demonstrating An Infinite Sheet As An Accumulation Of Infinite Rods

We visualize the infinite sheet as an accumulation of infinite uniformly charged rods placed side by side. Each rod contributes to the electric field. Therefore, the total electric field can be found by integrating the fields from these rods.
03

Determine The Charge Density Of The Sheet

We define \(\sigma\) as the charge density of the infinite sheet. \(\sigma\) is related to \(\lambda\) as \(\sigma = \lambda / w\), where \(w\) is the width over which we integrate.
04

Integrate Over The Infinite Rods

We then integrate the expression for electric field from a rod \(\lambda / 2 \pi \epsilon_{0} r\) over the width \(w\) taking into consideration the relative position of each rod. This gives us the total electric field from the infinite sheet.
05

Relation Between Electric Field And Charge Density Of The Sheet

After performing the integration, we find that the field from an infinite sheet with charge density \(\sigma\) is \(\sigma / 2 \epsilon_{0}\). This means that the effect of the infinite sheet is to spread out the electric field uniformly, determined by the charge density of the sheet shared equally on either side.

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Key Concepts

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

Charge Density
In the context of electrostatics, charge density is a way to describe how charge is distributed over a certain area or length. There are different types of charge densities, but here are the most relevant ones for our discussion:

  • Linear Charge Density (\( \lambda \)): This is the charge per unit length along a rod or a line. For a uniformly charged rod, every segment of the rod has the same charge density, denoted by \( \lambda \).
  • Surface Charge Density (\( \sigma \)): This is the charge per unit area on a surface, like our infinite sheet. It shows how much charge exists on a given surface area.

Understanding charge density helps us determine how electric fields are influenced by different charged objects. It lets us calculate the field's strength and direction produced by specific arrangements like rods or sheets.
Infinite Sheet
An infinite sheet of charge can be visualized as an unending plane charged evenly over its surface. It's a theoretical concept often used to simplify the problem of finding electric fields.

To break it down, imagine this infinite sheet composed of numerous parallel rods, each contributing to the electric field. These rods are identical in charge density and extend infinitely. Since the sheet is infinite, every point in the plane contributes equally, making the problem symmetrical and enabling straightforward calculations.

Because of this uniform distribution, the resulting electric field from an infinite sheet is constant and independent of the distance from the sheet. This is a key feature that distinguishes it from more localized charge distributions, simplifying the integration process during calculations.
Uniformly Charged Rods
A uniformly charged rod means that every part of the rod holds the same amount of charge per unit length. This uniformity is described by the linear charge density, \( \lambda \). Imagine a rod stretching indefinitely in both directions, maintaining this uniformity throughout.

From a single infinite rod, the electric field at a point is calculated using the formula: \( \lambda / 2 \pi \epsilon_{0} r \), where \( r \) is the distance perpendicular from the rod, and \( \epsilon_{0} \) is the permittivity of vacuum.

When rods are arranged side by side in an infinite number, they form an infinite sheet of charge. Each rod's electric field contributes to the overall field of the sheet, and their combined effect can be summed through integration to find the field from the infinite sheet.
Permittivity of Vacuum
The permittivity of vacuum, denoted as \( \epsilon_{0} \), is a fundamental physical constant that describes the ability of a vacuum to allow electric field lines to pass through it. It's essential in calculating electric fields and forces between charges.

In equations like \( \lambda / 2 \pi \epsilon_{0} r \) or \( \sigma / 2 \epsilon_{0} \), \( \epsilon_{0} \) helps determine the strength of the field created by charged objects. It essentially dictates how easily electric fields can permeate space.

The value of \( \epsilon_{0} \) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \) (farads per meter), and plays a key role in the study of electromagnetism, set as a constant for vacuum-based problems to measure and understand electric interactions.

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Most popular questions from this chapter

Zero field inside a cylindrical shell * Consider a distribution of charge in the form of a hollow circular cylinder, like a long charged pipe. In the spirit of Problem 1.17, show that the electric field inside the pipe is zero.

Stable equilibrium in electron jelly ** The task of Exercise \(1.77\) is to find the equilibrium positions of two protons located inside a sphere of electron jelly with total charge \(-2 e\). Show that the equilibria are stable. That is, show that a displacement in any direction will result in a force directed back toward the equilibrium position. (There is no need to know the exact locations of the equilibria, so you can solve this problem without solving Exercise \(1.77\) first.)

Ball in a sphere We know that if a point charge \(q\) is located at radius \(a\) in the interior of a sphere with radius \(R\) and uniform volume charge density \(\rho\), then the force on the point charge is effectively due only to the charge that is located inside radius \(a\). (a) Consider instead a uniform ball of charge located entirely inside a larger sphere of radius \(R\). Let the ball's radius be \(b\), and let its center be located at radius \(a\) in the larger sphere. Its volume charge density is such that its total charge is \(q\). Assume that the ball is superposed on top of the sphere, so that all of the sphere's charge is still present. Can the force on the ball be obtained by treating it like a point charge and considering only the charge in the larger sphere that is inside radius \(a\) ? (b) Would the force change if we instead remove the charge in the larger sphere where the ball is? So now we are looking at the force on the ball due to the sphere with a cavity carved out, which is a more realistic scenario.

Pulling two sheets apart ** Two parallel sheets each have large area \(A\) and are separated by a small distance \(\ell\). The surface charge densities are \(\sigma\) and \(-\sigma\). You wish to pull one of the sheets away from the other, by a small distance \(x\). How much work does this require? Calculate this by: (a) using the relation \(W=\) (force) \(\times\) (distance); (b) calculating the increase in energy stored in the electric field. Show that these two methods give the same result.

Hole in a shell \(*\) Figure \(1.52\) shows a spherical shell of charge, of radius \(a\) and surface density \(\sigma\), from which a small circular piece of radius \(b \ll a\) has been removed. What is the direction and magnitude of the field, at the midpoint of the aperture? There are two ways to get the answer. You can integrate over the remaining charge distribution, to sum the contributions of all elements to the field at the point in question. Or, remembering the superposition principle, you can think about the effect of replacing the piece removed, which itself is practically a little disk. Note the connection of this result with our discussion of the force on a surface charge - perhaps that is a third way in which you might arrive at the answer.
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