A solid insulating sphere has radius R and carries positive...

A solid insulating sphere has radius $R$ and carries positive charge distributed throughout its volume. The charge distribution has spherical symmetry but varies with radial distance $r$ from the center of the sphere. The volume charge density is $\rho(r)=\rho_{0}(1-r / R)$, where $\rho_{0}$ is a constant with units of $\mathrm{C} / \mathrm{m}^{3}$. (a) Derive an expression for the electric field as a function of $r$ for $r<R .$ (b) Repeat part (a) for $r>R .$ (c) At what value of $r$, in terms of $R$, does the electric field have its maximum value?

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is particularly useful for calculating electric fields when the charge distribution has symmetry, as it allows for a direct relationship between the enclosed charge and the field without dealing with the complexities of vector calculus over asymmetrical distributions.

Spherical Symmetry

Spherical symmetry implies that the physical properties of a system, such as the electric field or charge distribution, depend only on the radial distance from the center and not on the direction. This symmetry simplifies the problem significantly because it allows the application of Gauss's Law using a spherical Gaussian surface, resulting in a uniform field magnitude over the surface and an easier integration process.

Volume Charge Distribution

A volume charge distribution describes how electric charge is distributed throughout a three?dimensional region. When the charge density varies with position—often expressed as a function of the radial coordinate in spherically symmetric systems—it requires integrating the charge density over the volume of interest to determine the total charge enclosed within any given radius.

Integration for Enclosed Charge

The calculation of the enclosed charge within a given radius involves integrating the charge density over the volume bounded by that radius. This step is critical as it provides the total charge that contributes to the electric field at that point when using Gauss's Law, and it requires setting up the appropriate integral in spherical coordinates where the volume element is expressed in terms of the radial coordinate.

Optimization of Electric Field

Determining the maximum electric field in a system involves analyzing the expression of the field as a function of radius and finding the point where its derivative with respect to the radius is zero. This process uses techniques from calculus, such as differentiation and solving for critical points, to identify where the field attains a maximum value. It is an important method in understanding the behavior of fields generated by non-uniform charge distributions and in designing systems to optimize field strength.

Transcript

00:01 In this question, we have a solid insulating sphere of radius r.

00:08 Okay, we are given that charge density changes with r according to this function.

00:15 Okay, row not times 1 minus small r divided by big r.

00:20 Okay, and then we want to find the electric fuel inside the sphere, outside the sphere, and find a region finder r that gives the maximum electric field.

00:35 So this question is about gounser's law.

00:40 Okay, so first we write down a few things.

00:46 Okay, gouss law is equal to, it states that the close surface integral of the electric field is equal to the, charge and close divide by epsilon not, okay, or basically the electric flux.

01:10 The total flux through the gaussian surface is equal to the close charge divided by epsilon not.

01:17 And the charge and close is equal to row db.

01:24 So for region that is inside the sphere, okay, you are going to draw a gaussian surface.

01:34 Which is a spherical gaussian surface of radius r small r okay so the left -hand side okay is equal to the total electric flux okay this is going to be e times four pi r square okay by spherical symmetry electric field on the gaussian surface will be the same okay they are equidistant, each point is equidistant from the center, and then so it's constant, and the e and the d .a, they are parallel, so the dot product tends out to be a direct multiplication, and the summation, the integral of the a, which is 4 piar square.

02:23 Okay, so this is the left -hand side, and then the charge and close, is the next thing that we need to calculate, okay, row dv.

02:35 So integrate from 0 to r, row not, 1 minus r prime divide by big r 4 pi, r square, r prime square, the r prime, okay? so i'm using r prime because i don't want to mix with the r on the upper limit.

02:52 So take out the constants, 4 pi row not, integrate from 0 to r, r, r prime square minus r prime x .cube divided by r the r prime okay so after the integration okay this is what you get um one third r cubed minus r to the four divided by four times big r okay so i'm going to take out the r cube okay one third minus r over four r okay so i'm going to equate so by gouss law ke times 4 pi r square is equal to q and close charge and close divided f sot knot so this is 4 pi row not rq divided by epsilon 0 1 3 minus r over 4 r so you get you can cancel the 4 pi can cancel the r square so you end up with e equals to row not r over f .0.

04:20 1 3rd minus r over for r.

04:24 So this is the answer for part a...

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