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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

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.

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.

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!)

\(\mathrm{~ U n i f o r m ~ f i e l d ~ i n ~ a ~ c a v i t y ~ \cjkstart ? ? ? ? ? ?}\) A sphere has radius \(R_{1}\) and uniform volume charge density \(\rho . \mathrm{A}\) spherical cavity with radius \(R_{2}\) is carved out at an arbitrary location inside the larger sphere. Show that the electric field inside the cavity is uniform (in both magnitude and direction). Hint: Find a vector expression for the field in the interior of a charged sphere, and then use superposition. What are the analogous statements for the lower-dimensional analogs with cylinders and slabs? Are the statements still true?

Intersecting sheets ** (a) Figure \(1.49\) shows the cross section of three infinite sheets intersecting at equal angles. The sheets all have surface charge density \(\sigma .\) By adding up the fields from the sheets, find the electric field at all points in space. (b) Find the field instead by using Gauss's law. You should explain clearly why Gauss's law is in fact useful in this setup. (c) What is the field in the analogous setup where there are \(N\) sheets instead of three? What is your answer in the \(N \rightarrow \infty\) limit? This limit is related to the cylinder in Exercise 1.68.

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