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

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

Q=[itex]\int\int\int[/itex][itex]_{v}[/itex](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=[itex]\int\int\int[/itex][itex]_{v}[/itex](ρ)dV

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

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

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

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

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

=(14.1e-15)(0.056)2[itex]\pi[/itex]

= 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) [itex]\Phi[/itex]= q/[itex]\epsilon[/itex][itex]_{o}[/itex]

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

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

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

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

|E|= [itex]\frac{1}{4\pi\epsilon_{o}}[/itex] [itex]\frac{q_{in}}{r^{2}}[/itex]

[itex]\frac{Q}{V}[/itex]=[itex]\frac{q_{in}}{V_{in}}[/itex]

[itex]q_{in}[/itex]=Q[itex]\frac{V_{in}}{V}[/itex]

[itex]q_{in}[/itex]=Q[itex]\frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi R^{3}}[/itex]

[itex]q_{in}[/itex]=Q[itex]\frac{r^{3}}{R^{3}}[/itex]

[itex]q_{in}[/itex]=Q[itex]\frac{R^{3}/8}{R^{3}}[/itex]

[itex]q_{in}[/itex]=[itex]\frac{Q}{8}[/itex]

|E|= [itex]\frac{1}{4\pi\epsilon_{o}}[/itex] [itex]\frac{\frac{Q}{8}}{r^{2}}[/itex]

|E|= [itex]\frac{1}{4\pi\epsilon_{o}}[/itex] [itex]\frac{\frac{Q}{8}}{\frac{R^{2}}{4}}[/itex]

|E|= [itex]\frac{1}{4\pi\epsilon_{o}}[/itex] [itex]\frac{Q}{2 R^{2}}[/itex]

|E|= (9e9) [itex]\frac{7.78e-15}{2(0.056)^{2}}[/itex] = 0.0112

book says 5.58 mN/C

Please help thanks.

http://ift.tt/1bE7rAu

Leave a comment

Your email address will not be published.


*


Show Buttons
Hide Buttons