/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 A nonuniform, but spherically sy... [FREE SOLUTION] | 91Ó°ÊÓ

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A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho\)(\(r\)) given as follows: $$\rho(r) = \rho_0 \bigg(1 - \frac{r}{R}\bigg) \space \space \space \mathrm{for} \space r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \mathrm{for} \space r \leq R$$ where \(\rho_0 = 3Q/{\pi}R^3\) is a positive constant. (a) Show that the total charge contained in the charge distribution is \(Q\). (b) Show that the electric field in the region \(r \geq R\) is identical to that produced by a point charge \(Q\) at \(r =\) 0. (c) Obtain an expression for the electric field in the region \(r \leq R\). (d) Graph the electric-field magnitude \(E\) as a function of \(r\). (e) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field.

Short Answer

Expert verified
Total charge is \( Q \). The external electric field is \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \). Inside, the maximum electric field occurs at \( r = \frac{R}{3} \) with value \( E_{max} = \frac{Q}{6\pi \varepsilon_0 R^2} \).

Step by step solution

01

Understanding Charge Density

The given charge density for a spherically symmetric charge distribution is \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) for \( r \leq R \), and \( \rho(r) = 0 \) for \( r > R \). Observe that \( \rho_0 = \frac{3Q}{\pi R^3} \). This distribution follows a radial decay from the center.
02

Calculate Total Charge

To find the total charge \( Q \), integrate the charge density over the volume of the sphere: \( Q = \int_0^R \rho(r) \cdot 4\pi r^2 \cdot dr \). Substitute: \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) and calculate the volume integral.
03

Integration for Total Charge

Integrate \( Q = \int_0^R \rho_0 \left(1 - \frac{r}{R}\right) 4\pi r^2 \, dr \). Simplify the integral to: \( Q = 4\pi \rho_0 \left(\int_0^R r^2 \ dr - \frac{1}{R} \int_0^R r^3 \, dr \right) \). Evaluate to find \( Q = \frac{4\pi \rho_0 R^3}{3} \). Substitute \( \rho_0 = \frac{3Q}{\pi R^3} \) to show that \( Q = Q \).
04

Show Electric Field Outside

According to Gauss's Law, for \( r \geq R \), the electric field behaves as if all charge was concentrated at a point at the center. The field is \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \). This is identical to any other point charge \( Q \) centered at \( r = 0 \) for \( r \geq R \).
05

Calculate Electric Field Inside

Inside the sphere \( (r \leq R) \), use Gauss's Law: \( E \cdot 4\pi r^2 = \frac{1}{\varepsilon_0} \cdot \int_0^r \rho(r') \cdot 4\pi {r'}^2 \, dr' \). The integral becomes: \( E \cdot 4\pi r^2 = \frac{3Q}{R^3 \varepsilon_0} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) \). Solve for \( E \).
06

Expression for Electric Field Inside

After simplification, the electric field inside is given by: \( E = \frac{Qr}{4\pi \varepsilon_0 R^3} \left(2 - \frac{3r}{R}\right) \). This expression is valid for \( r \leq R \).
07

Plot Electric Field

Graph \( E \) as a function of \( r \). For \( r \leq R \), use the expression: \( E = \frac{Qr}{4\pi \varepsilon_0 R^3} \left(2 - \frac{3r}{R}\right) \). For \( r > R \), use \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \). The graph should show a peak inside the sphere and decrease with \( r^{-2} \) outside.
08

Finding Maximum Electric Field

To find the maximum field inside \( r \leq R \), differentiate \( E(r) = \frac{Qr}{4\pi \varepsilon_0 R^3} \left(2 - \frac{3r}{R}\right) \) with respect to \( r \) and set the derivative to zero. This gives the maximum field condition: \( 2 - \frac{6r}{R} = 0 \), solving gives \( r = \frac{R}{3} \).
09

Maximum Electric Field Value

Substitute \( r = \frac{R}{3} \) back into \( E(r) \): \( E_{max} = \frac{Q}{4\pi \varepsilon_0 R^3} \left(\frac{R}{3}\right) \left(2 - 1 \right) = \frac{Q}{6\pi \varepsilon_0 R^2} \).

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Key Concepts

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

Charge Distribution
Charge distribution refers to how electric charges are spread out in space. In this exercise, we are dealing with a non-uniform charge distribution that is also spherically symmetric. This means that the charge density, or the amount of charge per unit volume, varies with the distance from the center of the sphere but is the same in every direction from that center.

The charge density \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) decreases linearly from the center, reaching zero at the surface of the sphere. Beyond the radius \( R \), the charge density becomes zero. This implies that all the charge is concentrated within the sphere of radius \( R \).

Understanding charge distribution is crucial because it determines how the electric field behaves both inside and outside the sphere. A non-uniform charge distribution will affect the field differently compared to a uniform one, leading to variations in electric field strength at different points.
Gauss's Law
Gauss's Law is a fundamental principle that relates the electric field to the distribution of electric charge. The law states that the electric flux through a closed surface is proportional to the charge enclosed by the surface. Mathematically, it is given by:

\[ \Phi = \oint E \cdot dA = \frac{Q_{enc}}{\varepsilon_0} \]

Where \( \Phi \) is the electric flux, \( E \) is the electric field, \( dA \) is a differential area on the closed surface, \( Q_{enc} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.

In the context of the problem, Gauss's Law helps us determine the electric field both inside and outside the sphere. For \( r \geq R \), we apply Gauss’s Law to a spherical Gaussian surface that encompasses all the charge \( Q \). The field behaves as if all charge is concentrated at the sphere's center. Inside the sphere, the law is used to account for the charge enclosed within a smaller sphere of radius \( r \). This analysis enables us to derive the expression for the electric field at any point within the sphere.
Spherical Symmetry
Spherical symmetry is a situation where a system looks the same when viewed from any direction around its center. In this problem, the spherical symmetry of the charge distribution simplifies the analysis of the electric field. Because of this symmetry, the electric field at any point outside a spherically symmetric charge distribution depends only on the radial distance from the center and not on the direction.

When using spherical symmetry and Gauss's Law together, you can assume a symmetrical Gaussian surface like a sphere to simplify calculations. This is possible because of the symmetry of the system, leading to a constant electric field magnitude on the surface, making the integration straightforward. The resulting electric field expression applies uniformly in any direction from the center, significantly simplifying how we approach finding the electric field in regions inside and outside the sphere.
Charge Density
Charge density is a measure of how much electric charge is present in a particular space. It is usually expressed in terms of charge per unit volume, \( \rho(r) = \frac{dq}{dV} \). Understanding charge density is crucial because it directly influences the electric force at any point within or around the charged object.

In our exercise, the charge density \( \rho(r) = \rho_0 \left(1 - \frac{r}{R}\right) \) is a function of the radial distance \( r \). This function decreases linearly from the center of the distribution, reflecting a non-uniform distribution. Such a distribution causes a change in the internal electric field variation compared to a uniform one.

Charge density calculations often involve integrating this function over a volume to find the total charge. For example, within the given sphere, integrating the charge density helps us verify that the total charge sums to \( Q \). Comprehending how charge density modifies across the system reveals insights into the resulting electric field dynamics, aiding in efficient analysis of the problem at hand.

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Most popular questions from this chapter

A very long conducting tube (hollow cylinder) has inner radius a and outer radius \(b\). It carries charge per unit length \(+a\), where \(a\) is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+a\). (a) Calculate the electric field in terms of a and the distance r from the axis of the tube for (i) \(r < a; (ii) a < r < b; (iii) r > b\). Show your results in a graph of \(E\) as a function of \(r\). (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

How many excess electrons must be added to an isolated spherical conductor 26.0 cm in diameter to produce an electric field of magnitude 1150 N/C just outside the surface?

A solid conducting sphere with radius \(R\) carries a positive total charge \(Q\). The sphere is surrounded by an insulating shell with inner radius \(R\) and outer radius 2\(R\). The insulating shell has a uniform charge density \(\rho\). (a) Find the value of \(\rho\) so that the net charge of the entire system is zero. (b) If \(\rho\) has the value found in part (a), find the electric field \(\overrightarrow{E}\) (magnitude and direction) in each of the regions 0 \(< r < R, R < r < 2R\), and \(r > 2R\). Graph the radial component of \(\overrightarrow{E}\) as a function of r. (c) As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in part (b) agree with this rule.

A solid conducting sphere carrying charge \(q\) has radius \(a\). It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c\). The hollow sphere has no net charge. (a) Derive expressions for the electricfield magnitude in terms of the distance \(r\) from the center for the regions \(r < a, a < r < b, b < r < c\), and \(r > c\). (b) Graph the magnitude of the electric field as a function of \(r\) from \(r =\) 0 to \(r =\) 2c. (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius 2\(c\).

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately 7.4 \(\times\) 10\(^{-15}\) m. (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about 1.0 \(\times\) 10\(^{-10}\) m? (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

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