/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Charge of uniform volume density... [FREE SOLUTION] | 91Ó°ÊÓ

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Charge of uniform volume density \(\rho=1.2 \mathrm{nC} / \mathrm{m}^{3}\) fills an infinite slab between \(x=-5.0 \mathrm{~cm}\) and \(x=+5.0 \mathrm{~cm}\). What is the magnitude of the electric field at any point with the coordinate (a) \(x=\) \(4.0 \mathrm{~cm}\) and (b) \(x=6.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
At \( x=4.0 \text{ cm} \), \( E = 5.42 \times 10^2 \text{ N/C} \); at \( x=6.0 \text{ cm} \), \( E = 6.78 \times 10^2 \text{ N/C} \).

Step by step solution

01

Define the Parameters

We are given that the volume charge density \( \rho = 1.2 \text{ nC/m}^3 \), which needs to be converted to \( \text{C/m}^3 \). Thus, \( \rho = 1.2 \times 10^{-9} \text{ C/m}^3 \). The slab extends between \( x = -5.0 \text{ cm} \) to \( x = +5.0 \text{ cm} \). We are asked to find the electric field at \( x = 4.0 \text{ cm} \) and \( x = 6.0 \text{ cm} \).
02

Understanding Gauss's Law

Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]. For an infinite plane with uniform charge, the electric field \( E \) due to a slab of charge density \( \rho \) is given by \( E = \frac{\rho x}{\varepsilon_0} \) inside the charge distribution and \( E = \frac{\rho a}{\varepsilon_0} \) outside, where \( a \) is half the thickness of the slab.
03

Calculate Electric Field at x = 4.0 cm (Inside the Slab)

Here, \( x = 4.0 \text{ cm} = 0.04 \text{ m} \). Since \( 4.0 \text{ cm} \) is within the slab, use \( E = \frac{\rho x}{\varepsilon_0} \).\[ E_{inside} = \frac{1.2 \times 10^{-9} \times 0.04}{8.85 \times 10^{-12}} = 5.42 \times 10^2 \text{ N/C} \].
04

Calculate Electric Field at x = 6.0 cm (Outside the Slab)

Here, \( x = 6.0 \text{ cm} = 0.06 \text{ m} \). Since \( 6.0 \text{ cm} \) is outside the slab, use \( E = \frac{\rho a}{\varepsilon_0} \) where \( a = 0.05 \text{ m} \), the half-thickness of the slab. \[ E_{outside} = \frac{1.2 \times 10^{-9} \times 0.05}{8.85 \times 10^{-12}} = 6.78 \times 10^2 \text{ N/C} \].

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Key Concepts

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

Electric Field Calculation
Understanding how to calculate the electric field is a fundamental part of solving physics problems related to charges. In such scenarios, Gauss's Law is a pivotal principle used to find the electric field produced by a charge distribution. This law tells us that the total electric flux through a closed surface is directly proportional to the charge enclosed. Mathematically, it can be expressed as: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \] This equation states that the electric flux \( \Phi_E \) through a closed surface is the integral of the electric field \( \mathbf{E} \) across the surface area \( d\mathbf{A} \). The term \( Q_{enc} \) refers to the total charge enclosed by the surface and \( \varepsilon_0 \) is the permittivity of free space. When calculating the electric field for an infinite slab, as described in the exercise, it's crucial to decide whether the point of interest is inside or outside the slab.
  • Inside the slab: \( E = \frac{\rho x}{\varepsilon_0} \), where \( \rho \) is the uniform volume charge density and \( x \) is the distance from the center of the slab to the point where the field is calculated.
  • Outside the slab: \( E = \frac{\rho a}{\varepsilon_0} \), where \( a \) is half the thickness of the slab.
This bifurcation is key to understanding how the electric field behaves in different regions of space with respect to the slab.
Uniform Volume Charge Density
A charge distribution's uniformity refers to how consistently charge is spread out over a given volume. In our scenario, the term "uniform volume charge density" denotes that the electricity is evenly distributed throughout the infinite slab. In simpler terms, if you take any section of this slab, you'll find the same amount of charge per unit volume. This even spread simplifies calculations and allows us to use consistent mathematical expressions like \[\rho = \frac{Q}{V},\]where \( \rho \) is the volume charge density (in Coulombs per cubic meter), \( Q \) is the total charge, and \( V \) is the volume of the charged region. For the given exercise, the volume charge density is \( 1.2 \, \text{nC/m}^3 \). It is important to convert this to SI units as \( 1.2 \times 10^{-9} \, \text{C/m}^3 \) for accurate calculations. Understanding this uniform distribution is critical for correctly applying Gauss's Law and computing the electric field.
Infinite Slab Electric Field
An infinite slab model is a theoretical construct in physics used to simplify the complexity of electric field calculations. It is essential when dealing with scenarios where the slab's length and width are significantly larger than its thickness. In practical terms, it implies that the slab's influence on the electric field behaves as though the slab is endless in two dimensions. When applying Gauss's Law in this context, the slab generates a distinctive electric field configuration:
  • Inside the Slab: The electric field increases linearly with distance from the central plane (midpoint of the slab) up to the slab's boundary. This behavior is captured in the formula \( E = \frac{\rho x}{\varepsilon_0} \).
  • Outside the Slab: Beyond the slab's surface, the electric field is constant, as denoted by \( E = \frac{\rho a}{\varepsilon_0} \), reflecting the cessation of charge contributions beyond the slab's extent.
Because of these properties, the electric field profile of the infinite slab is distinct from that of a point charge or a finite object. This model helps simplify the complexity associated with calculating electric fields, offering clarity in theoretical and practical exercises alike.

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Most popular questions from this chapter

Two charged concentric spherical shells have radii \(10.0 \mathrm{~cm}\) and \(15.0 \mathrm{~cm} .\) The charge on the inner shell is \(4.00 \times 10^{-8} \mathrm{C}\), and that on the outer shell is \(2.00 \times 10^{-8} \mathrm{C}\). Find the electric field (a) at \(r=12.0 \mathrm{~cm}\) and \((\mathrm{b})\) at \(r=20.0 \mathrm{~cm}\).

A charged particle causes an electric flux of \(-750 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\) to pass through a spherical Gaussian surface of \(10.0 \mathrm{~cm}\) radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the charge of the particle?

Go A charge of uniform linear density \(2.0 \mathrm{nC} / \mathrm{m}\) is distributed along a long, thin, nonconducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius \(=5.0 \mathrm{~cm}\), outer radius \(=\) \(10 \mathrm{~cm})\). The net charge on the shell is zero. (a) What is the magnitude of the electric field \(15 \mathrm{~cm}\) from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?

A uniformly charged conducting sphere of \(1.2 \mathrm{~m}\) diameter has surface charge density \(8.1 \mu \mathrm{C} / \mathrm{m}^{2} .\) Find (a) the net charge on the sphere and (b) the total electric flux leaving the surface.

The electric field at point \(P\) just outside the outer surface of a hollow spherical conductor of inner radius \(10 \mathrm{~cm}\) and outer radius \(20 \mathrm{~cm}\) has magnitude \(450 \mathrm{~N} / \mathrm{C}\) and is directed outward. When a particle of unknown charge \(Q\) is introduced into the center of the sphere, the electric field at \(P\) is still directed outward but is now \(180 \mathrm{~N} / \mathrm{C}\). (a) What was the net charge enclosed by the outer surface before \(Q\) was introduced? (b) What is charge \(Q\) ? After \(Q\) is introduced, what is the charge on the (c) inner and (d) outer surface of the conductor?

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