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Chapter 21: Problem 36

What surface charge density on an infinite sheet will produce a 1.4-kN/C electric field?

Short Answer

Expert verified
The surface charge density which would produce a 1.4-kN/C electric field would be approximately \(2.48 脳 10^{-8} C/m^2\).

Step by step solution

01

Understand Gauss's Law

Gauss鈥檚 law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric field due to an infinite sheet of charge is given by \(E = \frac{\sigma}{2\epsilon_0}\), where \(E\) is the electric field, \(\sigma\) is the charge density, and \(\epsilon_0\) is the permittivity of free space.
02

Substitute Given Values

We are asked to find the charge density (\(\sigma\)). So, rearranging the formula, we get \(\sigma = 2 \times E \times \epsilon_0\). The value of \(\epsilon_0\) is a constant, \(8.85 脳 10^{-12} C^2/Nm^2\), and \(E\) is given as 1.4-kN/C, which can be rewritten as \(1.4 脳 10^3 N/C\).
03

Calculate the Surface Charge Density

Substitute these values into your equation: \(\sigma = 2 \times (1.4 脳 10^3 N/C) \times (8.85 脳 10^{-12} C^2/Nm^2)\). After carrying out the multiplication, we will get the value of the surface charge density.

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in electromagnetism. It describes the force that a charged object exerts on other charges around it. Imagine the electric field as an invisible force field that surrounds an electric charge. This field tells you how a positive test charge would be pushed or pulled if placed near the source charge. Some important points about electric fields:
  • Electric fields are represented by vectors, indicating both the direction and magnitude of the force.
  • The unit for measuring electric field strength is Newton per Coulomb (N/C).
  • An electric field can vary in strength; closer to the charge, the stronger it is. In the case of an infinite sheet of charge, the electric field is uniform, meaning it remains constant regardless of distance from the sheet.
This concept is crucial in solving problems involving Gauss's Law, as Gauss's Law helps understand the relationship between the electric field and the charge distribution.
Surface Charge Density
Surface charge density (\(\sigma\)) is a measure of how much electric charge is accumulated on a surface per unit area. You can think of it as determining how closely packed the charges are on a surface, such as a metal plate or an infinite sheet. When there is a high surface charge density:
  • The charges are packed tightly together.
  • This typically results in a stronger electric field nearby.
  • It is measured in Coulombs per square meter (C/mg2).
Calculating surface charge density is essential when predicting how much field is produced by a charged surface under Gauss's Law. For an infinite sheet, the formula used is \(E = \frac{\sigma}{2\epsilon_0}\), clearly showing the relationship between the electric field and surface charge density.
Permittivity of Free Space
Permittivity of free space, represented by the symbol:\(\epsilon_0\), is a physical constant key in the study of electromagnetism. Also known as the "electric constant", it quantifies how much resistance is encountered when forming an electric field in a vacuum. Essentially, it sets the scale for the magnitude of electric forces in the vacuum.Key facts about \(\epsilon_0\):
  • The value is approximately \(8.85 \times 10^{-12} C^2/Nm^2\).
  • This constant is crucial when calculating electric fields in vacuum or air, using Gauss's Law.
  • It directly influences the electric field strength as seen in the formula: \(\sigma = 2 \times E \times \epsilon_0\)
Understanding \(\epsilon_0\) is essential in evaluating how electric fields behave in non-conductive, insulating mediums and in calculations involving infinite systems of charge like in this exercise.

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Most popular questions from this chapter

If the flux of the gravitational field through a closed surface is zero, what can you conclude about the region interior to the surface?

A point charge of \(-2 Q\) is at the center of a spherical shell of radius \(R\) carrying charge \(Q\) spread uniformly over its surface. Find the electric field at (a) \(r=\frac{1}{2} R\) and (b) \(r=2 R .\) (c) How would your answers change if the charge on the shell were doubled?

A solid sphere \(25 \mathrm{cm}\) in radius carries \(14 \mu \mathrm{C},\) distributed uniformly throughout its volume. Find the electric field strength (a) \(15 \mathrm{cm},\) (b) \(25 \mathrm{cm},\) and (c) \(50 \mathrm{cm}\) from its center.

A net charge of \(5.0 \mu \mathrm{C}\) is applied on one side of a solid metal sphere \(2.0 \mathrm{cm}\) in diameter. Once electrostatic equilibrium is reached, and assuming no other conductors or charges nearby, what are (a) the volume charge density inside the sphere and (b) the surface charge density on the sphere?

For a coaxial cable in electrostatic equilibrium carrying equal but opposite charges on its two conductors, there's a nonzero electric field a. only in the space between the wire and shield. b. in the space between wire and shield, and outside the shield. c. inside the metal conducting wire and shield, as well as between the wires and outside the shield. d. only outside the shield.
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