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Density, density, density. (a) A charge \(-300 e\) is uniformly distributed along a circular arc of radius \(4.00 \mathrm{~cm}\), which subtends an angle of \(40^{\circ}\). What is the linear charge density along the arc? (b) \(\mathrm{A}\) charge \(-300 e\) is uniformly distributed over one face of a circular disk of radius \(2.00 \mathrm{~cm}\). What is the surface charge density over that face? (c) A charge \(-300 e\) is uniformly distributed over the surface of a sphere of radius \(2.00 \mathrm{~cm}\). What is the surface charge density over that surface? (d) A charge \(-300 e\) is uniformly spread through the volume of a sphere of radius \(2.00 \mathrm{~cm} .\) What is the volume charge density in that sphere?

Short Answer

Expert verified
(a) Linearly distributed: \(\lambda = -3.37 \times 10^{-6} \, \text{C/m}\); (b) Disk: \(\sigma = -1.19 \times 10^{-3} \, \text{C/m}^2\); (c) Sphere's surface: \(\sigma = -1.19 \times 10^{-3} \, \text{C/m}^2\); (d) Sphere's volume: \(\rho = -7.14 \times 10^{-2} \, \text{C/m}^3\).

Step by step solution

01

Convert Charge to Coulombs

The charge given is \(-300 e\), where \(e\) is the elementary charge, approximately \(1.602 \times 10^{-19} \, \text{C}\). Therefore, the charge \(Q\) is calculated as \((-300) \times (1.602 \times 10^{-19}) \, \text{C}\).
02

Calculate Linear Charge Density

Linear charge density \(\lambda\) is defined as the charge per unit length. First, calculate the arc length \(L\) using the formula \(L = \theta \cdot r\), where \(\theta = 40^{\circ}\) converted to radians is \(40 \cdot \frac{\pi}{180}\). Using \(r = 4.00 \, \text{cm} = 0.04 \, \text{m}\), find \(L\) and then \(\lambda = \frac{Q}{L}\).
03

Calculate Surface Charge Density on the Disk

Surface charge density \(\sigma\) is charge per unit area. The area \(A\) of a disk is \(\pi r^2\), with \(r = 2.00 \, \text{cm} = 0.02 \, \text{m}\). Calculate \(A\) and then \(\sigma = \frac{Q}{A}\).
04

Calculate Surface Charge Density on the Sphere

The surface area of a sphere is \(4\pi r^2\), using \(r = 2.00 \, \text{cm} = 0.02 \, \text{m}\). Calculate the sphere's surface area and find the surface charge density \(\sigma = \frac{Q}{4\pi r^2}\).
05

Calculate Volume Charge Density in the Sphere

Volume charge density \(\rho\) is charge per unit volume. The volume \(V\) of a sphere is \(\frac{4}{3}\pi r^3\). Using \(r = 0.02 \, \text{m}\), calculate \(V\) and \(\rho = \frac{Q}{V}\).

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Key Concepts

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

Linear Charge Density
Linear charge density, denoted as \( \lambda \), represents the amount of electric charge per unit length along a line or curve.
To understand this concept, think of evenly spreading tiny bits of charge along a wire or arc. Linear charge density tells us how much charge exists in a particular length of that wire or arc.

When calculating the linear charge density along a circular arc, follow these steps:
  • Calculate the arc length \( L \) using the formula \( L = \theta \cdot r \), where \( \theta \) is the angle in radians, and \( r \) is the radius of the circle.
  • In our problem, the arc angle \( 40^{\circ} \) needs to be converted to radians, yielding \( \theta = 40 \cdot \frac{\pi}{180} \).
  • With \( r = 0.04 \text{ m} \), calculate \( L \).
  • Finally, determine \( \lambda = \frac{Q}{L} \), where \( Q \) is the total charge in Coulombs.
Linear charge density is critical in physics as it helps to describe electric fields produced by linear charge distributions.
Surface Charge Density
Surface charge density, symbolized by \( \sigma \), refers to the electric charge per unit area on a surface. Imagine evenly sprinkling charge over a flat surface like a disk or a sphere. Surface charge density tells us how much charge covers one unit of that area.

To compute surface charge density on a disk’s face:
  • Find the area \( A \) using the formula \( A = \pi r^2 \), where \( r \) is the radius.
  • In this exercise, \( r = 0.02 \text{ m} \), so calculate \( A \).
  • The surface charge density \( \sigma \) is then \( \sigma = \frac{Q}{A} \).
Surface charge density is essential to understanding phenomena like the electric field near charged surfaces, which appear in a variety of contexts, from capacitors to everyday objects.
Volume Charge Density
Volume charge density, denoted as \( \rho \), describes how much electric charge exists per unit volume within a 3-dimensional space. Visualize distributing charge evenly throughout a solid object. Volume charge density quantifies the charge in one unit of that volume.

The calculation of volume charge density inside a sphere is as follows:
  • Start by determining the volume \( V \) with the formula \( V = \frac{4}{3}\pi r^3 \).
  • Given \( r = 0.02 \text{ m} \) for the sphere, compute \( V \).
  • Then find \( \rho = \frac{Q}{V} \), with \( Q \) as the total charge.
Understanding volume charge density helps in areas such as solid-state physics and electromagnetism where charge distribution impacts how materials and fields behave.
Spherical Charge Distributions
Spherical charge distributions involve charges that are symmetrically arranged around a central point, such as in spheres. The symmetry simplifies calculations due to the predictability of how charge is spread and its effects.
In the context of our exercise, imagine a charge spread over or inside a sphere.

In such cases:
  • For the surface distribution, use the sphere's surface area formula \( 4\pi r^2 \) and determine surface charge density as \( \frac{Q}{4\pi r^2} \).
  • For volume distribution, calculate volume charge density using the sphere's volume \( \frac{4}{3}\pi r^3 \).
Spherical charge distributions are crucial to solving problems in electrostatics and predicting electric fields due to their evenly distributed nature around the central point.

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Most popular questions from this chapter

An electric dipole consisting of charges of magnitude \(1.50 \mathrm{n} \mathrm{C}\) separated by \(6.20 \mu \mathrm{m}\) is in an electric field of strength 1100 \(\mathrm{N} / \mathrm{C} .\) What are (a) the magnitude of the electric dipole moment and (b) the difference between the potential energies for dipole orientations parallel and antiparallel to \(\vec{E} ?\)

An electron enters a region of uniform electric field with an initial velocity of \(40 \mathrm{~km} / \mathrm{s}\) in the same direction as the electric field, which has magnitude \(E=50 \mathrm{~N} / \mathrm{C}\). (a) What is the speed of the electron \(1.5 \mathrm{~ns}\) after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

An electron on the axis of an electric dipole is \(25 \mathrm{~nm}\) from the center of the dipole. What is the magnitude of the electrostatic force on the electron if the dipole moment is \(3.6 \times 10^{-29} \mathrm{C} \cdot \mathrm{m} ?\) Assume that \(25 \mathrm{~nm}\) is much larger than the separation of the charged particles that form the dipole.

How much work is required to turn an electric dipole \(180^{\circ}\) in a uniform electric field of magnitude \(E=46.0 \mathrm{~N} / \mathrm{C}\) if the dipole moment has a magnitude of \(p=3.02 \times\) \(10^{-25} \mathrm{C} \cdot \mathrm{m}\) and the initial angle is \(64^{\circ} ?\)

An electric dipole with dipole moment $$ \vec{p}=(3.00 \hat{\mathrm{i}}+4.00 \hat{\mathrm{j}})\left(1.24 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\right) $$ is in an electric field \(\vec{E}=(4000 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{i}}\). (a) What is the potential energy of the electric dipole? (b) What is the torque acting on it? (c) If an external agent turns the dipole until its electric dipole moment is $$ \vec{p}=(-4.00 \hat{\mathrm{i}}+3.00 \hat{\mathrm{j}})\left(1.24 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\right) $$ how much work is done by the agent?

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