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Chapter 16: Problem 65

In this problem, you can show from Coulomb's law that the constant of proportionality in Gauss's law must be \(1 / \epsilon_{0} .\) Imagine a sphere with its center at a point charge q. (a) Write an expression for the electric flux in terms of the field strength \(E\) and the radius \(r\) of the sphere. [Hint: The field strength \(E\) is the same everywhere on the sphere and the field lines cross the sphere perpendicular to its surface.] (b) Use Gauss's law in the form \(\Phi_{\mathrm{E}}=c q(\text { where } c\) is the constant of proportionality) and the electric field strength given by Coulomb's law to show that $c=1 / \epsilon_{0}$

Short Answer

Expert verified
Answer: The constant of proportionality in Gauss's law derived using Coulomb's law is \(\frac{1}{\epsilon_{0}}\), where \(\epsilon_{0}\) is the vacuum permittivity.

Step by step solution

01

Find the expression for the electric flux

To find the electric flux, we must remember that it is given by the integral of the electric field over a closed surface, in this case, a sphere. Since the electric field strength, \(E\), is the same everywhere on the sphere and is perpendicular to the surface, we can write the electric flux as: \(\Phi_{E} = \int_{S} \vec{E} \cdot d\vec{A}\) As \(\vec{E}\) and \(d\vec{A}\) are both in the same direction, the dot product simplifies to \(|\vec{E}| |d\vec{A}|\). Now, we can write \(\Phi_{E}\) as: \(\Phi_{E} = E\int_{S} dA\)
02

Calculate the surface area of the sphere

The surface area of a sphere with radius r is given by the formula: \(A = 4\pi r^2\). Since E is constant, we can remove it from the integral and we get: \(\Phi_{E} = E\int_{S} dA = E \cdot (4\pi r^2)\)
03

Apply Gauss's law

According to Gauss's law, the electric flux through a closed surface is equal to the constant of proportionality (c) times the total charge (q) enclosed by the surface: \(\Phi_{E} = cq\)
04

Relate the electric field strength with Coulomb's law

Coulomb's law states that the electric field strength is given by: \(E = \frac{k q}{r^2}\), where \(k = \frac{1}{4\pi \epsilon_{0}}\)
05

Find the constant of proportionality

Now, using the expression for electric flux found in step 2 and equating it with Gauss's law, we have: \(E \cdot (4\pi r^2) = cq\) Substituting the expression for E from Coulomb's law, we get: \(\frac{k q}{r^2} \cdot (4\pi r^2) = cq\) Simplifying the equation and substituting \(k = \frac{1}{4\pi \epsilon_{0}}\): \(q\cdot \frac{1}{\epsilon_{0}} = cq\) We can now see that the constant of proportionality, \(c\), equals to \(\frac{1}{\epsilon_{0}}\). This confirms that \(c = \frac{1}{\epsilon_{0}}\) and thus, Gauss's law can be written as: \(\Phi_{E} = \frac{1}{\epsilon_{0}}q\)

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Most popular questions from this chapter

\(\mathrm{A} \mathrm{K}^{+}\) ion and a \(\mathrm{Cl}^{-}\) ion are directly across from each other on opposite sides of a membrane 9.0 nm thick. What is the electric force on the \(\mathrm{K}^{+}\) ion due to the \(\mathrm{Cl}^{-}\) ion? Ignore the presence of other charges.
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
In lab tests it was found that rats can detect electric fields of about $5.0 \mathrm{kN} / \mathrm{C}\( or more. If a point charge of \)1.0 \mu \mathrm{C}$ is sitting in a maze, how close must the rat come to the charge in order to detect it?
A positively charged rod is brought near two uncharged conducting spheres of the same size that are initially touching each other (diagram a). The spheres are moved apart and then the charged rod is removed (diagram b). (a) What is the sign of the net charge on sphere 1 in diagram b? (b) In comparison with the charge on sphere \(1,\) how much and what sign of charge is on sphere \(2 ?\)
A parallel-plate capacitor consists of two flat metal plates of area \(A\) separated by a small distance \(d\). The plates are given equal and opposite net charges \(\pm q\) (a) Sketch the field lines and use your sketch to explain why almost all of the charge is on the inner surfaces of the plates. (b) Use Gauss's law to show that the electric field between the plates and away from the edges is $E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} \cdot(\mathrm{c})$ Does this agree with or contra- dict the result of Problem \(70 ?\) Explain. (d) Use the principle of superposition and the result of Problem 69 to arrive at this same answer. [Hint: The inner surfaces of the two plates are thin, flat sheets of charge.]
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