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

Short Answer

Expert verified
(a) Net charge: \(3.66 \times 10^{-5} \text{ C}\); (b) Electric flux: \(4.13 \times 10^{6} \text{ N} \cdot \text{m}^2/\text{C}\).

Step by step solution

01

Understanding the Problem and Given Data

We have a conducting sphere with a diameter of \(1.2\, \text{m}\), which means its radius \(r\) is half that: \(0.6\, \text{m}\). The surface charge density \(\sigma\) is given as \(8.1\, \mu \text{C} / \text{m}^2\), which is equivalent to \(8.1 \times 10^{-6}\, \text{C} / \text{m}^2\). We need to find (a) the net charge \(Q\) on the sphere and (b) the total electric flux \(\Phi\) leaving the surface of the sphere.
02

Calculate the Surface Area of the Sphere

The surface area \(A\) of a sphere is calculated using the formula \(A = 4\pi r^2\). Substituting the radius \(r = 0.6\, \text{m}\) into the formula, we get:\[ A = 4\pi (0.6)^2 \approx 4.52\, \text{m}^2. \]
03

Calculate the Net Charge on the Sphere

The net charge \(Q\) can be found using the surface charge density formula \(\sigma = \frac{Q}{A}\). Rearrange this to find \(Q = \sigma \cdot A\). Substitute \(\sigma = 8.1 \times 10^{-6}\, \text{C} / \text{m}^2\) and \(A \approx 4.52\, \text{m}^2\) to get:\[ Q = (8.1 \times 10^{-6}) \times 4.52 \approx 3.66 \times 10^{-5}\, \text{C}. \]
04

Determine the Electric Flux Leaving the Surface

The total electric flux \(\Phi\) leaving the surface of a conductor is determined by Gauss's Law, \(\Phi = \frac{Q}{\varepsilon_0}\), where \(\varepsilon_0 = 8.85 \times 10^{-12} \text{C}^2/\text{N} \cdot \text{m}^2\) is the permittivity of free space. Substituting \(Q = 3.66 \times 10^{-5}\, \text{C}\), we get:\[ \Phi = \frac{3.66 \times 10^{-5}}{8.85 \times 10^{-12}} \approx 4.13 \times 10^{6}\, \text{N} \cdot \text{m}^2/\text{C}. \]

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Key Concepts

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

Conducting Sphere
A conducting sphere is a simple yet profound object in electrostatics. It is made of a conductive material which allows electric charges to move freely across its surface. When a conductor reaches electrostatic equilibrium, the charges distribute themselves uniformly across the surface. This property makes conducting spheres ideal for exploring electrostatic principles like charge distribution and electric fields.
A key characteristic of a conducting sphere is that all excess charge resides entirely on its surface. This happens because like charges repel, forcing them to spread out as far as possible from each other. As a result, the electric field inside a conducting sphere is zero in electrostatic equilibrium, while outside, it acts as if the entire charge were concentrated at the center.
In practical applications, understanding conducting spheres allows us to examine how they can store charges, such as in electrostatic capacitors, and how they affect surrounding electric fields, making them useful for both theoretical and practical purposes.
Surface Charge Density
Surface charge density, denoted by the symbol \( \sigma \), is a measure of how much electric charge is distributed over a particular surface area. For conducting spheres, it tells us how densely the charge is packed on the surface. Its units are usually coulombs per square meter (C/m²).
In the context of our example, the surface charge density is given as \( 8.1 \mu \text{C}/\text{m}^2 \), which is equivalent to \( 8.1 \times 10^{-6} \text{C}/\text{m}^2 \). This means that for every square meter of the sphere's surface, there is \( 8.1 \times 10^{-6} \) coulombs of charge present.
Calculating the total charge on the sphere involves multiplying the surface charge density by the surface area. For example, if the surface area of the sphere is \( 4.52 \text{ m}^2 \), then the net charge is:
  • \( Q = \sigma \cdot A = (8.1 \times 10^{-6}) \times 4.52 \approx 3.66 \times 10^{-5} \text{C} \)
Understanding surface charge density helps us in analyzing how charges distribute themselves on conductors and enhances our ability to solve various electrostatic problems.
Gauss's Law
Gauss's Law is a fundamental equation in electrostatics that relates the electric flux through a closed surface to the charge enclosed by that surface. It is part of Maxwell's equations, essential for understanding electric fields. Gauss’s Law is expressed mathematically as:
  • \( \Phi = \frac{Q}{\varepsilon_0} \)
where \( \Phi \) is the electric flux, \( Q \) is the total enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \)).
In our context, since the net charge \( Q \) of the sphere is known to be \( 3.66 \times 10^{-5} \text{ C} \), we can find the total electric flux leaving the surface as:
  • \( \Phi = \frac{3.66 \times 10^{-5}}{8.85 \times 10^{-12}} \approx 4.13 \times 10^{6} \text{ N} \cdot \text{m}^2/\text{C} \)
This law is especially useful for calculating electric fields for symmetric charge distributions, such as spheres, cylinders, and planes, and simplifies complex electrostatic problems into manageable solutions.
Electric Flux
Electric flux, symbolized by \( \Phi \), is a measure of the number of electric field lines passing through a given area. It gives us a quantitative understanding of the strength of the electric field over a particular surface. The concept of electric flux is deeply connected with Gauss's Law, which determines how electric charges influence the field lines they emit.
The unit of electric flux is Newton meters squared per coulomb (N·m²/C). To calculate electric flux for a conducting sphere, we consider the net charge and apply Gauss’s Law. As mentioned earlier, with the charge \( 3.66 \times 10^{-5} \text{ C} \) and considering the permittivity of space, the electric flux through the surface is \( 4.13 \times 10^{6} \text{ N} \cdot \text{m}^2/\text{C} \).
Why does electric flux matter? It provides insight into the behavior of electric charges and their fields. It helps visualize and quantitatively analyze phenomena like shielding effects, where a charge generates field lines that influence other charges. Understanding electric flux is crucial for engineers and physicists for creating applications like electrical shielding and capacitor design.

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Most popular questions from this chapter

Figure 23-43 shows cross sections through two large, parallel, nonconducting sheets with identical distributions of positive charge with surface charge density \(\sigma=1.77 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}\). In unit- vector notation, what is \(\vec{E}\) at points (a) above the sheets, (b) between them, and (c) below them?

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius \(R\), at what radial distances (a) inside and (b) outside the ball is the magnitude of the ball's electric field equal to \(\frac{1}{4}\) of the maximum magnitude of that field?

A free electron is placed between two large, parallel, nonconducting plates that are horizontal and \(2.3 \mathrm{~cm}\) apart. One plate has a uniform positive charge; the other has an equal amount of uniform negative charge. The force on the electron due to the electric field \(\vec{E}\) between the plates balances the gravitational force on the electron. What are (a) the magnitude of the surface charge density on the plates and (b) the direction (up or down) of \(\vec{E}\) ?

\({ }^{60} 10\) Figure 23-30 shows a closed Gaussian surface in the shape of a cube of edge length \(2.00 \mathrm{~m}\). It lies in a region where the nonuniform electric field is given by \(\vec{E}=(3.00 x+4.00) \hat{\mathrm{i}}+6.00 \hat{j}+7.00 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}\), with \(x\) in meters. What is the net charge contained by the cube?

Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metal satellite \(1.3 \mathrm{~m}\) in diameter accumulates \(2.4\) \(\mu \mathrm{C}\) of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.

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