# Gaussian sphere problem, non uniform charge

• Duderonimous
In summary: You see, the sphere of radius r is a sub-sphere of the original sphere of radius R, and its charge is given by the same expression as the charge of the original sphere, but with a different bound of integration.
Duderonimous

## Homework Statement

A solid non conducting sphere of radius R=5.60 cm has a nonuniform charge ditribution ρ=(14.1 pC/m^3)r/R, where r is the radial distance from the spheres center. (a) What is the sphere's total charge? What is the magnitude E of the electric field at (b) r=0, (c) r=R/2.00, and (d) r=R? (e) Sketch a graph of E versus r.

## Homework Equations

$\Phi$= q/$\epsilon$$_{o}$=$\oint$(E$\cdot$$\hat{n}$)dA

Q=$\int\int\int$$_{v}$(ρ)dV

## The Attempt at a Solution

I have zero experience with triple integrals and my professor gave us a explaining. I kind of get it but I don't understand exactly how one defines the bounds of the triple integral. dV is defined in spherical coordinates.

(a) Q=$\int\int\int$$_{v}$(ρ)dV

=$\int$$^{R}_{0}$$\int$$^{\pi}_{0}$$\int$$^{2\pi}_{0}$ (14.1 pC/m^3)r/R sinθdrdθd$\phi$

=(14.1 pC/m^3)/R$\int$$^{R}_{0}$rdr$\int$$^{\pi}_{0}$ sinθdθ$\int$$^{2\pi}_{0}$d$\phi$

=(14.1 pC/m^3)/R [$\frac{r^{2}}{2}$]$^{R}_{0}$[-cosθ]$^{\pi}_{0}$[$\phi$]$^{2\pi}_{0}$

=(14.1 pC/m^3)/R[$\frac{R^{2}}{2}$][2][2$\pi$]

=(14.1 pC/m^3)R[2$\pi$]

=(14.1e-15)(0.056)2$\pi$

= 4.96 fC

Book says 7.78 fC

(b) e-field is zero

For part c I use 7.78 fC for Q, r=R/2

(c) $\Phi$= q/$\epsilon$$_{o}$

$\oint$(E$\cdot$$\hat{n}$)dA= q/$\epsilon$$_{o}$

$\oint$|E||$\hat{n}$||cos0$^{o}$|dA=q/$\epsilon$$_{o}$

$\oint$|E|(1)(1)dA=q/$\epsilon$$_{o}$

|E|4$\pi$r$^{2}$=q/$\epsilon$$_{o}$

|E|= $\frac{1}{4\pi\epsilon_{o}}$ $\frac{q_{in}}{r^{2}}$

$\frac{Q}{V}$=$\frac{q_{in}}{V_{in}}$

$q_{in}$=Q$\frac{V_{in}}{V}$

$q_{in}$=Q$\frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi R^{3}}$

$q_{in}$=Q$\frac{r^{3}}{R^{3}}$

$q_{in}$=Q$\frac{R^{3}/8}{R^{3}}$

$q_{in}$=$\frac{Q}{8}$

|E|= $\frac{1}{4\pi\epsilon_{o}}$ $\frac{\frac{Q}{8}}{r^{2}}$

|E|= $\frac{1}{4\pi\epsilon_{o}}$ $\frac{\frac{Q}{8}}{\frac{R^{2}}{4}}$

|E|= $\frac{1}{4\pi\epsilon_{o}}$ $\frac{Q}{2 R^{2}}$

|E|= (9e9) $\frac{7.78e-15}{2(0.056)^{2}}$ = 0.0112

book says 5.58 mN/C

yeah, spherical triple integrals are difficult to get used to. But they're pretty useful once you get some practice with them. uh, you've written ##dV = \sin(\theta ) dr d\theta d\phi## but this is not quite right.

In part (c), you were going all right till you started to calculate the charge in the sub-sphere of radius ##r##. You assumed that the charge would be proportional to its volume. That is correct only when the distribution is constant and uniform, which is not the case in this problem.

Could apply your results of (a) in (c)?

To BruceW: Yes thank you. I was missing an r$^{2}$ term on the right side. I worked it through and got.

To Voko: should I replace q$_{in}$ with ρV$_{in}$?

Related by the equation

ρ=$\frac{q_{in}}{V_{in}}$

I worked it through and got it wrong but I want to know if I am at least heading in the right direction. Thanks.

In a) you had q(R) as a triple integral that turned out to be a simple integral. Now you need q(R/2). Same integral, different bound.

To BvU: Thanks that helped a lot. I think my problem was the I could not visualize what each integral was integrating. I at least now can see the last integral sums up concentric spheres from r=0 to r=R/2.

Good. This is what I meant by applying (a) to (c).

## 1. What is the Gaussian sphere problem?

The Gaussian sphere problem is a mathematical problem that involves calculating the electric field at any point outside a sphere with a non-uniform distribution of charge. This problem is commonly used in electrostatics and is named after the mathematician Carl Friedrich Gauss.

## 2. How is the electric field calculated in the Gaussian sphere problem?

The electric field is calculated by using Gauss's law, which states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of the medium. In this problem, the closed surface is a sphere and the electric field can be calculated using the charge enclosed by that sphere.

## 3. What is a non-uniform charge distribution?

A non-uniform charge distribution refers to a situation where the charge is not distributed evenly throughout a given area or volume. This means that the charge density, which is the amount of charge per unit area or volume, varies at different points within the region.

## 4. How does the non-uniform charge distribution affect the electric field?

The non-uniform charge distribution affects the electric field by causing it to vary at different points outside the sphere. This is because the electric field is directly proportional to the charge density, so as the charge density changes, the electric field also changes. This is why the Gaussian sphere problem is used to calculate the electric field at any point outside the sphere.

## 5. What are some real-world applications of the Gaussian sphere problem with non-uniform charge?

The Gaussian sphere problem with non-uniform charge has many real-world applications, such as calculating the electric field around a charged particle or molecule, designing electrical circuits, and understanding the behavior of lightning strikes. It is also used in the study of plasma physics, which is important in fields such as astrophysics and nuclear fusion research.

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