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Two horizontal, infinite, plane sheets of charge are separated by a distance \(d\) . The lower sheet has negative charge with uniform surface charge density \(-\sigma<0 .\) The upper sheet has positive charge with uniform surface charge density \(\sigma>0 .\) What is the electric field (magnitude, and direction if the field is nonzero) (a) above the upper sheet, (b) below the lower sheet, (c) between the sheets?

Short Answer

Expert verified
(a) Zero, (b) Zero, (c) \(\frac{\sigma}{\varepsilon_0}\) downward.

Step by step solution

01

Understanding the Electric Field due to a Single Charged Sheet

The electric field due to an infinite plane sheet of charge with surface charge density \( \sigma \) is given by the formula \( E = \frac{\sigma}{2\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space. If the sheet has a positive charge \( +\sigma \), the direction of the electric field is away from the sheet. If it has a negative charge \( -\sigma \), the direction of the electric field is towards the sheet.
02

Calculating the Electric Field Above the Upper Sheet

Above the upper sheet, which is positively charged, the electric field directly due to the sheet is downward (away from the charge), with magnitude \( \frac{\sigma}{2\varepsilon_0} \). The lower negatively charged sheet is far enough apart and exerts an upward field of the same magnitude. These two fields cancel each other out.
03

Calculating the Electric Field Below the Lower Sheet

Below the negatively charged lower sheet, the field due to the lower sheet is upward with magnitude \( \frac{\sigma}{2\varepsilon_0} \). The field due to the positively charged upper sheet at this position is directed downward, and again, it is of the same magnitude. These two fields cancel each other out.
04

Calculating the Electric Field Between the Sheets

Between the sheets, the field from the positively charged upper sheet is directed downward with magnitude \( \frac{\sigma}{2\varepsilon_0} \), and the field from the negatively charged lower sheet is directed upward with the same magnitude. Hence, the fields add up (both pointing downward), giving a total electric field of magnitude \( \frac{\sigma}{\varepsilon_0} \) directed downward.

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Key Concepts

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

Electric Field
Understanding the concept of an electric field is crucial in electrostatics. The electric field is a vector quantity represented as \ \( E \ \), which describes the force per unit charge exerted on a positive test charge placed in the vicinity of other charges. In the context of an infinite plane sheet with uniform surface charge density \ \( \sigma \ \), the electric field's magnitude is given by the formula \\[ E = \frac{\sigma}{2\varepsilon_0} \\]\where \ \( \varepsilon_0 \ \) is the permittivity of free space.
The direction of the electric field is determined by the type of charge on the sheet:
  • For a positive charge \ \( \sigma>0 \ \), the field is directed away from the sheet.
  • For a negative charge \ \( -\sigma<0 \ \), the field is directed towards the sheet.
Understanding how these fields interact when sheets are arranged, like in the exercise, is crucial to calculate the net electric field in regions surrounding and between the sheets.
Surface Charge Density
Surface charge density, symbolized as \ \( \sigma \ \), refers to the amount of electric charge per unit area on a surface. It is measured in units of coulombs per square meter \ \( \text{C/m}^2 \ \). This concept is essential when analyzing electric fields generated by charged surfaces.
An infinite plane sheet of charge creates a uniform electric field perpendicular to the surface. The value and sign of \ \( \sigma \ \) significantly influence the direction and strength of this field:
  • A positive surface charge density \ \( \sigma>0 \ \) means charges are spread across the surface creating an outward pull.
  • A negative surface charge density \ \( \sigma<0 \ \) makes the field point inward towards the sheet.
In electrostatic problems, understanding these interactions helps predict how combined fields impact surrounding areas, revealing cancellation or enhancement effects depending on respective charge configurations.
Permittivity of Free Space
The permittivity of free space, represented by the symbol \ \( \varepsilon_0 \ \), is a fundamental constant of nature that characterizes the ability of a vacuum to permit electric field lines. Its approximate value is \ \( 8.85 \times 10^{-12} \, \text{F/m} \ \) (farads per meter). This constant plays a crucial role in electrostatics and influences the strength of forces between charges.
In formulas describing electric fields, \ \( \varepsilon_0 \ \) appears as a denominator, signifying its function in moderating electric interactions. Specifically, for an infinite plane sheet:
  • The electric field magnitude is inversely proportional to \ \( \varepsilon_0 \ \), highlighting that a greater permittivity weakens the influence of a given charge density.
This constant is vital for accurately computing electric phenomena in theoretical and practical setups, ensuring that the predictions align with physical reality.

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Most popular questions from this chapter

Two positive point charges \(q\) are placed on the \(y\) -axis at \(y=a\) and \(y=-a .\) A negative point charge \(-Q\) is located at some point on the \(+x\) -axis. (a) In a free-body diagram, show the forces that act on the charge \(-Q .\) (b) Find the \(x\) -and \(y\) -components of the net force that the two positive charges exert on \(-Q\) . (Your answer \(r\) . should involve only \(k, q, Q, a\) and the coordinate \(x\) of the third charge.) (c) What is the net force on the charge \(-Q\) when it is at the origin \((x=0) ?(\mathrm{d})\) Graph the \(x\) -component of the net force on the charge \(-Q\) as a function of \(x\) for values of \(x\) between \(-4 a\) and \(+4 a\) .

A particle has charge \(-3.00 \mathrm{nC}\) (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 \(\mathrm{m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C} ?\)

Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)

Two small aluminum spheres, each having mass 0.0250 \(\mathrm{kg}\) are separated by \(80.0 \mathrm{cm} .\) (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 \(\mathrm{g} / \mathrm{mol}\) , and its atomic number is \(13 . )\) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{N}\) (roughly 1 ton \() ?\) Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

(a) What is the electric field of an iron nucleus at a distance of \(6.00 \times 10^{-10} \mathrm{m}\) from the nucleus? The atomic number of iron is \(26 .\) Assume that the nucleus may be treated as a point charge. (b) What is the electric field of a proton at a distance of \(5.29 \times 10^{-11} \mathrm{m}\) from the proton? (This is the radius of the electron orbit in the Bohr model for the ground state of the hydrogen atom.)

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