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22.25. The electric field at a distance of 0.145 \(\mathrm{m}\) from the surface of a solid insulating sphere with radius 0.355 \(\mathrm{m}\) is 1750 \(\mathrm{N} / \mathrm{C}\) . (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? ( 6 ) Calculate the electric field inside the sphere at a distance of 0.200 \(\mathrm{m}\) from the center.

Short Answer

Expert verified
Charge density is approximately \(2.59 \times 10^{-7} \ \mathrm{C}/\mathrm{m}^3\), the electric field inside at 0.200 m is approximately 1956 N/C.

Step by step solution

01

Understand the concepts

To solve this, we need to understand that the charge is uniformly distributed inside a solid insulating sphere. The problem involves finding charge density and the electric field using Gauss's Law for a charge distribution.
02

Calculate total charge using Gauss's Law and known field

Given that the electric field at a distance of 0.500 m (0.145 m from the surface) is 1750 N/C. By Gauss's Law, the electric field outside the sphere is given by:\[E = \frac{kQ}{r^2}\]where - \(E\) is the electric field 1750 N/C,- \(k\) is the Coulomb's constant, \( 8.99 \times 10^9 \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \),- \(r = 0.355 + 0.145 = 0.500\mathrm{m}\) is the total distance from the center.Thus, solve for total charge \(Q\):\[Q = \frac{E \cdot r^2}{k} = \frac{1750 \cdot (0.500)^2}{8.99 \times 10^9}\approx 4.86 \times 10^{-8} \ \mathrm{C}\]
03

Calculate charge density

Charge density \(\rho\) is the total charge divided by the volume of the sphere \(V\). The volume is given by \[V = \frac{4}{3} \pi R^3\]where \(R=0.355\,\mathrm{m}\) is the radius of the sphere.Calculate \(V\):\[V = \frac{4}{3} \pi (0.355)^3 \approx 0.1878 \ \mathrm{m}^3\]Thus, the charge density \(\rho\) is:\[\rho = \frac{Q}{V} = \frac{4.86 \times 10^{-8}}{0.1878} \approx 2.59 \times 10^{-7} \ \mathrm{C}/\mathrm{m}^3\]
04

Use Gauss's law to calculate electric field inside the sphere

When calculating the electric field inside the sphere at a distance of 0.200 m from the center, Gauss's Law states:\[E(inside) = \frac{kQ_{enc}}{r^2}\]where \(Q_{enc}\) is the charge enclosed, calculated from the charge density \(\rho\) and volume of the part of the sphere included (radius up to 0.200 m). Therefore, \(Q_{enc} = \rho \cdot V_{part}\).Calculate \(V_{part}\):\[V_{part} = \frac{4}{3} \pi (0.200)^3 \approx 0.0335\ \mathrm{m}^3\]Find \(Q_{enc}\):\[Q_{enc} = 2.59 \times 10^{-7} \times 0.0335 \approx 8.68 \times 10^{-9} \ \mathrm{C}\]Finally, calculate the electric field:\[E = \frac{8.99 \times 10^9 \times 8.68 \times 10^{-9}}{(0.200)^2} \approx 1956 \ \mathrm{N}/\mathrm{C}\]

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle that helps us understand how electric fields behave. It relates the electric flux passing through a closed surface to the charge enclosed by that surface. Imagine it as a magical tool that allows us to calculate the strength of electric fields without having to know every small detail about how the charges are distributed. The law is particularly convenient for problems with a high degree of symmetry, such as spheres, cylinders, and planes.The general formula is:\[ \Phi = \oint E \, \cdot \, dA = \frac{Q_{enc}}{\varepsilon_0} \]where:- \(\Phi\) is the electric flux,- \(E\) denotes the electric field,- \(dA\) is a small element of the area,- \(Q_{enc}\) indicates the enclosed charge, and- \(\varepsilon_0\) represents the permittivity of free space.For a sphere, like in our problem, it simplifies to calculating the electric field a certain distance from the center by only considering the charge enclosed within that imaginary spherical surface. This makes Gauss's Law highly effective when dealing with symmetrical shapes like an insulating sphere.
Charge Density
Charge density is a measure of how much charge is contained in a specific space. It tells you how compact the charge is within a given volume.In mathematical terms, the charge density \(\rho\) signifies the amount of charge per unit volume. It's calculated by dividing the total charge \(Q\) of an object by its volume \(V\):\[ \rho = \frac{Q}{V} \]When dealing with spherical objects, such as an insulating sphere, you can calculate the volume using the sphere's radius \(R\):\[ V = \frac{4}{3} \pi R^3 \]For uniform charge distribution inside a sphere, this formula helps to determine the charge density easily. Knowing the charge density helps us further analyze how an electric field is influenced inside and outside the sphere.It's crucial in problems like these since understanding the charge density allows us to find the enclosed charge for any subsection of the sphere, which is essential in applying Gauss's Law.
Insulating Sphere
An insulating sphere is a spherical object that doesn’t allow electric charges to move freely through its surface. It's different from a conducting sphere, where charges spread out evenly. In the context of this exercise, the sphere has a radius and contains a uniformly distributed charge. This uniform distribution simplifies calculations because the charge can be thought of as being "smeared out" evenly throughout the sphere. When analyzing electric fields with such a sphere, two zones are crucial: - **Inside the Sphere:** The electric field depends on the distance from the center of the sphere. Gauss's Law applies to find the field by considering the charge contained within a smaller hypothetical sphere inside the actual sphere. - **Outside the Sphere:** The entire charge of the sphere is considered in the calculation, as if it were concentrated at the center, mimicking a point charge. Understanding these principles enables us to calculate how the electric field behaves at any point inside or outside the spherical object.
Coulomb's Constant
Coulomb's Constant is a key factor in calculating the electric field created by a point charge. In the world of electric forces, it is a proportional constant that appears in Coulomb's Law.Coulomb's Law describes how the electric force between two distant charges decreases with the square of the distance between them. The formula is:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:- \(F\) is the force between the charges,- \(q_1\) and \(q_2\) are the charges,- \(r\) is the distance separating the charges, and- \(k = 8.99 \times 10^9 \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2\) is Coulomb's Constant.Coulomb's Constant ensures that the units of electric force calculations match up in the International System of Units (SI). It's integral in calculating the electric field around charged objects by providing a way to translate charge and distance into the field's strength.

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Most popular questions from this chapter

22.54. A Uniformiy Charged Stab. A slab of insulating material has thickness 2\(d\) and is oriented so that its faces are parallel to the \(y z\) -plane and given by the planes \(x=d\) and \(x=-d\) . The \(y\) - and \(z\) -dimensions of the slab are very large compared to \(d\) and may be treated as essentially infinite. The slab has a uniform positive charge density \(\rho\) . (a) Explain why the electric field due to the slab is zero at the center of the slab \((x=0)\) . (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

22.31. A negative charge \(-Q\) is placed inside the cavity of a hol- low metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not?(c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the ciffects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

22\. 13. A \(9.60-\mu\) C point charge is at the center of a cube with sides of length 0.500 \(\mathrm{m}\) (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 \(\mathrm{m}\) long? Explain.

22.19. How many excess electrons must be added to an isolated spherical conductor 32.0 \(\mathrm{cm}\) in diameter to produce an electric field of 1150 \(\mathrm{N} / \mathrm{C}\) just outside the surface?

\(22.66 .\) A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$ \begin{array}{ll}{\rho(r)=\alpha} & {\text { for } r \leq R / 2} \\\ {\rho(r)=2 \alpha(1-r / R)} & {\text { for } R / 2 \leq r \leq R} \\\ {\rho(r)=0} & {\text { for } r \geq R}\end{array} $$ Here \(\alpha\) is a positive constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) . (a) Determine \(\alpha\) in terms of \(Q\) and \(R .\) (b) Using Gauss's law, derive an expression for the magnitude of \(\vec{E}\) as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of the total charge \(Q\) . Be sure to check that your results agree on the boundaries of the regions. (c) What fraction of the total charge is contained within the region \(r \leq R / 2 ?\left(\text { d) If an electron with charge } q^{\prime}=-e \text { is }\right.\) oscillating back and forth about \(r=0\) (the center of the distribution) with an amplitude less than \(R / 2,\) show that the motion is simple harmonic. (Hint: Review the discussion of simple harmonic motion in Section 13.2. If, and only if, the net force on the electron is proportional to its displacement from equilibrium, then the motion is simple harmonic. \()(\mathrm{e})\) What is the period of the motion in part \((\mathrm{d}) ?(\mathrm{f})\) If the amplitude of the motion described in part (e) is not?

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