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(II) A 15.0-cm-diameter nonconducting sphere carries a total charge of \(2.25 \mu \mathrm{C}\) distributed uniformly throughout its volume. Graph the electric field \(E\) as a function of the distance \(r\) from the center of the sphere from \(r=0\) to \(r=30.0 \mathrm{~cm}\)

Short Answer

Expert verified
Graph shows \( E \) increasing linearly inside and decreasing as \( 1/r^2 \) outside.

Step by step solution

01

Understand the Problem

We are given a sphere of 15.0 cm diameter, which means a radius of 7.5 cm. Charge is uniformly distributed throughout this nonconducting sphere, and we need to graph the electric field (E) as a function of distance (r) from the center, from r=0 to r=30.0 cm.
02

Apply Gauss's Law for r Inside the Sphere

For points inside the sphere, where \( r \leq 7.5 \) cm, use Gauss's Law. The electric field \( E \) at a distance \( r \) inside a uniformly charged sphere is given by the equation: \[ E = \frac{1}{{4\pi \varepsilon_0}} \cdot \frac{Qr}{{R^3}} \] where \( Q \) is the total charge, \( \varepsilon_0 \) is the permittivity of free space, \( R \) is the sphere's radius, and \( r \) is the distance from the center.
03

Calculate Electric Field for r Outside the Sphere

For points outside the sphere, where \( r > 7.5 \) cm, the sphere behaves as if all its charge is concentrated at the center. The electric field \( E \) outside the sphere is given by: \[ E = \frac{1}{{4\pi \varepsilon_0}} \cdot \frac{Q}{r^2} \] where \( r \) is the distance from the center.
04

Determine the Expression for E

For \( r \leq 7.5 \) cm, \[ E = \frac{2.25 \times 10^{-6} \times r}{(4\pi \times 8.85 \times 10^{-12}) \times (0.075)^3} \]. For \( r > 7.5 \) cm, \[ E = \frac{2.25 \times 10^{-6}}{(4\pi \times 8.85 \times 10^{-12}) \times r^2} \].
05

Sketch the Graph

Using the expressions derived in Step 4, plot the electric field \( E \) versus the distance \( r \) from 0 to 30 cm. The field increases linearly with \( r \) inside the sphere and decreases with \( 1/r^2 \) outside the sphere.

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Key Concepts

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

Gauss's Law
To understand electric fields, we need to learn about Gauss's Law. This powerful tool in electromagnetism helps us calculate the electric field created by a symmetric charge distribution. The law states that the electric flux passing through a closed surface is directly proportional to the charge enclosed by the surface. Gauss's Law is expressed mathematically as:\[ \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where:
  • \( \Phi \) represents the electric flux,
  • \( \mathbf{E} \) is the electric field,
  • \( d\mathbf{A} \) is the differential area vector, and
  • \( Q_{enc} \) is the total charge contained within the surface.
Knowing Gauss's Law is especially useful for symmetric objects like spheres and cylinders, where calculating the electric field using other methods might not be as straightforward.
Nonconducting Sphere
The exercise involves a nonconducting sphere, which is crucial in understanding how charge distribution affects the electric field. A nonconducting sphere means that the charges inside are stationary and uniformly spread throughout the sphere's volume.
This uniform distribution helps us utilize symmetry to apply Gauss's Law effectively.
When considering points inside a nonconducting sphere, the electric field behaves differently compared to outside. The interesting part is that inside the sphere, the electric field increases linearly with the distance from the center. This happens because more volume is enclosed as distance increases, leading to more charge and thus a stronger field.
Charge Distribution
Charge distribution is a fundamental concept that explains how electric fields originate from charged objects. Understanding how charge is spread over or within objects helps in calculating the resulting electric field.
In this exercise, the sphere's charge is spread out evenly throughout its volume, called a "uniform charge distribution." This means that any volume within the sphere will have a consistent charge per unit volume.
These uniform charge distributions are vital when applying Gauss's Law, as they simplify calculations. It's important to note:
  • A uniform charge distribution results in symmetric field lines.
  • Inside the sphere, the electric field grows with distance because more charge is enclosed.
  • Outside the sphere, the electric field behaves as if all the charge were concentrated at its center.
Permittivity of Free Space
The permittivity of free space, denoted as \( \varepsilon_0 \), is a constant important in electrical equations, especially Gauss's Law. It plays a crucial role in defining the relationship between electric field and charge.Its value is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \) (farads per meter).
In the context of the electric field equation for both inside and outside the sphere, \( \varepsilon_0 \) appears in the denominator. This indicates that it regulates how easily the electric field can "flow" through space.
Understanding \( \varepsilon_0 \) is essential because it tells us how strongly electric fields interact with the vacuum of space, affecting how observed fields behave across different distances and mediums.

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Most popular questions from this chapter

(I) A long thin wire, hundreds of meters long, carries a uniformly distributed charge of \(-7.2 \mu \mathrm{C}\) per meter of length. Estimate the magnitude and direction of the electric field at points \((a) 5.0 \mathrm{~m}\) and \((b) 1.5 \mathrm{~m}\) perpendicular from the center of the wire.

Neutral hydrogen can be modeled as a positive point charge \(+1.6 \times 10^{-19} \mathrm{C}\) surrounded by a distribution of negative charge with volume density given by \(\rho_{\mathrm{E}}(r)=-A e^{-2 r / a_{0}}\) where \(a_{0}=0.53 \times 10^{-10} \mathrm{~m}\) is called the Bohr radius, \(A\) is a constant such that the total amount of negative charge is \(-1.6 \times 10^{-19} \mathrm{C},\) and \(e=2.718 \cdots\) is the base of the natural log. (a) What is the net charge inside a sphere of radius \(a_{0}\) ? (b) What is the strength of the electric field at a distance \(a_{0}\) from the nucleus? [Hint: Do not confuse the exponential number \(e\) with the elementary charge \(e\) which uses the same symbol but has a completely different meaning and value \(\left(e=1.6 \times 10^{-19} \mathrm{C}\right)\)

(II) A flat ring (inner radius \(R_{0},\) outer radius 4\(R_{0} )\) is uniformly charged. In terms of the total charge \(Q,\) determine the electric field on the axis at points \((a) 0.25 R_{0}\) and (b) 75\(R_{0}\) from the center of the ring. [Hint: The ring can be replaced with two oppositely charged superposed disks.]

(I) The total electric flux from a cubical box \(28.0 \mathrm{~cm}\) on a side is \(1.84 \times 10^{3} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\). What charge is enclosed by the box?

(II) Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors are given equal but opposite uniform surface charge densities \(\pm \sigma .\) Ignore edge effects and use Gauss's law to show \((a)\) that for points far from the edges, the electric field between the plates is \(E=\sigma / \epsilon_{0}\) and (b) that outside the plates on either side the field is zero. (c) How would your results be altered if the two plates were nonconductors?

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