Gaussian sphere problem, non uniform charge

[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

[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

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.

 

[BruceW]

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.

 

[voko]

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)?

 

[Duderonimous]

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

To Voko: should I replace q[itex]_{in}[/itex] with ρV[itex]_{in}[/itex]?

Related by the equation

ρ=[itex]\frac{q_{in}}{V_{in}}[/itex]

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

 

[BvU]

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.

 

[Duderonimous]

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.

 

[voko]

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