# Gaussian sphere problem, non uniform charge

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

Q=$\int\int\int$$_{v}$(ρ)dV

3. 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