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A point charge produces an electric flux of \(+235 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\) through a gaussian sphere of radius \(15.0 \mathrm{~cm}\) centered on the charge. ( \(a\) ) What is the flux through a gaussian sphere with a radius \(27.5 \mathrm{~cm} ?\) (b) What is the magnitude and sign of the charge?

Short Answer

Expert verified
(a) The flux remains +235 Nm²/C. (b) The charge is +2.08 nC.

Step by step solution

01

Understanding Flux from a Point Charge

The electric flux \( \Phi \) through a closed surface is defined by Gauss's Law: \( \Phi = \frac{Q}{\varepsilon_0} \), where \( Q \) is the charge enclosed and \( \varepsilon_0 \) is the permittivity of free space. For any closed surface enclosing the charge, the flux depends only on the enclosed charge and not on the radius of the sphere.
02

Electric Flux Through a New Sphere

Since the flux is determined solely by the enclosed charge, the flux through a Gaussian sphere of any radius (centering the charge) will remain \( +235 \, \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} \), regardless of the sphere's size.
03

Using Flux to Determine Charge

From Gauss's Law \( \Phi = \frac{Q}{\varepsilon_0} \), we can rearrange to solve for \( Q \). Substitute \( \Phi = 235 \, \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2 / (\mathrm{N} \cdot \mathrm{m}^2) \): \( Q = \Phi \times \varepsilon_0 = 235 \times 8.85 \times 10^{-12} \, \mathrm{C} \).
04

Calculate and Determine Sign of Charge

Calculate: \( Q = 235 \times 8.85 \times 10^{-12} \approx 2.08 \times 10^{-9} \, \mathrm{C} \). The flux is positive, indicating a positive charge. So, the charge is \( +2.08 \times 10^{-9} \, \mathrm{C} \).

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Key Concepts

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

Gauss's Law and Its Implications
Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux passing through a closed surface to the charge enclosed by that surface. In mathematical terms, the law is expressed as \( \Phi = \frac{Q}{\varepsilon_0} \), where \( \Phi \) is the electric flux, \( Q \) is the total charge enclosed, and \( \varepsilon_0 \) is the permittivity of free space. This law is crucial because it simplifies the calculation of electric fields for symmetrical charge distributions, such as spheres and cylinders.

The application of Gauss's Law significantly reduces the complexity of understanding electric fields in certain situations. For instance, regardless of the radius of the Gaussian surface, the flux depends solely on the total enclosed charge. Through the use of Gauss's Law, understanding scenarios like the one presented in the exercise becomes much more manageable. The key takeaway is that the flux is independent of the radius of the sphere, which means it remains constant whether you increase or decrease the sphere's size, as long as the charge remains enclosed.
Point Charge and Its Characteristics
A point charge is an idealized model where charge is assumed to be concentrated at a single point in space. This simplification is useful as it allows us to analyze the behavior of electric fields easily and accurately without the complexity of distributed charges. In essence, point charges produce a radial electric field that decreases in strength with the square of the distance from the charge.

In the problem given, a point charge generates an electric flux through a Gaussian sphere. Since the charge's influence is uniformly distributed over the surface of a sphere, the electric field here can be evaluated using Gauss's Law instead of integrating over an entire surface. This exercise clearly demonstrates that the electric flux through any closed spherical surface, regardless of its size, is determined solely by the point charge located at its center.
Permittivity of Free Space: Understanding \(\varepsilon_0\)
The permittivity of free space, denoted as \( \varepsilon_0 \), is a constant that describes how electric fields interact with the vacuum of space. It has a value of approximately \( 8.85 \times 10^{-12} \mathrm{C}^2 / (\mathrm{N} \cdot \mathrm{m}^2) \) and is a critical component in electromagnetic theory, particularly in Gauss's Law. It helps set the scale for electric force and field constants, essentially quantifying the space's ability to permit electric field lines.

In practical terms, \( \varepsilon_0 \) is key for calculating the electric flux due to a given charge, as it essentially acts as a conversion factor between charge and electric flux. It represents the property of the space through which the electric field propagates, influencing how strong or weak the field is at a given distance from the charge. Understanding \( \varepsilon_0 \) is essential for solving problems involving Gauss’s Law and for connecting the concepts of electric flux and charge quantitatively.

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Most popular questions from this chapter

(II) A solid metal sphere of radius \(3.00 \mathrm{~m}\) carries a total charge of \(-5.50 \mu \mathrm{C}\). What is the magnitude of the electric field at a distance from the sphere's center of (a) \(0.250 \mathrm{~m}\), (b) \(2.90 \mathrm{~m},\) (c) \(3.10 \mathrm{~m},\) and \((d) 8.00 \mathrm{~m} ?\) How would the answers differ if the sphere were \((e)\) a thin shell, or \((f)\) a solid nonconductor uniformly charged throughout?

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 \(\left(e=1.6 \times 10^{-19} \mathrm{C}\right)\)

A point charge of \(9.20 \mathrm{nC}\) is located at the origin and a second charge of \(-5.00 \mathrm{nC}\) is located on the \(x\) axis at \(x=2.75 \mathrm{~cm} .\) Calculate the electric flux through a sphere centered at the origin with radius \(1.00 \mathrm{~m} .\) Repeat the calculation for a sphere of radius \(2.00 \mathrm{~m}\).

A solid nonconducting sphere of radius \(r_{0}\) has a total charge \(Q\) which is distributed according to \(\rho_{\mathrm{E}}=b r,\) where \(\rho_{\mathrm{E}}\) is the charge per unit volume, or charge density \(\left(\mathrm{C} / \mathrm{m}^{3}\right)\) and \(b\) is a constant. Determine \((a) b\) in terms of \(Q,(b)\) the electric field at points inside the sphere, and \((c)\) the electric field at points outside the sphere.

(II) An uncharged solid conducting sphere of radius \(r_{0}\) contains two spherical cavities of radii \(r_{1}\) and \(r_{2},\) respectively. Point charge \(Q_{1}\) is then placed within the cavity of radius \(r_{1}\) and point charge \(Q_{2}\) is placed within the cavity of radius \(r_{2}\) (Fig. \(22-38\) ). Determine the resulting electric field (magnitude and direction) at locations outside the solid sphere \(\left(r>r_{0}\right)\) where \(r\) is the distance from its center.

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