Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways: (a) Use Eq. 2.43. You found the potential in Prob. 2.21. (b) Use Eq. 2.45. Don’t forget to integrate over all space. (c) Use Eq. 2.44. Take a spherical volume of radius a. What happens

Question

Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways:

(a) Use Eq. 2.43. You found the potential in Prob. 2.21.

[latex]W=\frac{1}{2} \int ho V d au[/latex] (2.43)

(b) Use Eq. 2.45. Don’t forget to integrate over all space.

[latex]W=\frac{\epsilon_{0}}{2} \int E^{2} d au[/latex] (all space). (2.45)

(c) Use Eq. 2.44. Take a spherical volume of radius a. What happens as a→∞?

[latex]W=\frac{\epsilon_{0}}{2}\left(\int\limits_{ u }^{}{} E^{2} d au+\oint\limits_{S}^{}{} V E \cdot d a ight)[/latex] (2.44)

Accepted Answer

(a) [latex]W=\frac{1}{2} \int ho V d au[/latex]. From Prob. 2.21 (or Prob. 2.28): [latex]V=\frac{ ho}{2 \epsilon_{0}}\left(R^{2}-\frac{r^{2}}{3} ight)=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{2 R}\left(3-\frac{r^{2}}{R^{2}} ight)[/latex]

[latex]W=\frac{1}{2} ho \frac{1}{4 \pi \epsilon_{0}} \frac{q}{2 R} \int_{0}^{R}\left(3-\frac{r^{2}}{R^{2}} ight) 4 \pi r^{2} d r=\left.\frac{q ho}{4 \epsilon_{0} R}\left[3 \frac{r^{3}}{3}-\frac{1}{R^{2}} \frac{r^{5}}{5} ight] ight|_{0} ^{R}=\frac{q ho}{4 \epsilon_{0} R}\left(R^{3}-\frac{R^{3}}{5} ight)[/latex]

=[latex]\frac{q ho}{5 \epsilon_{0}} R^{2}=\frac{q R^{2}}{5 \epsilon_{0}} \frac{q}{\frac{4}{3} \pi R^{3}}=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{3}{5} \frac{q^{2}}{R} ight)[/latex] .

(b) [latex]W=\frac{\epsilon_{0}}{2} \int E^{2} d au[/latex]. Outside (r > R) [latex]E =\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}} \hat{ r } ; ext { Inside }(r

[latex] herefore W=\frac{\epsilon_{0}}{2} \frac{1}{\left(4 \pi \epsilon_{0} ight)^{2}} q^{2}\left\{\int_{R}^{\infty} \frac{1}{r^{4}}\left(r^{2} 4 \pi d r ight)+\int_{0}^{R}\left(\frac{r}{R^{3}} ight)^{2}\left(4 \pi r^{2} d r ight) ight\}[/latex]

=[latex]\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2}}{2}\left\{\left.\left(-\frac{1}{r} ight) ight|_{R} ^{\infty}+\left.\frac{1}{R^{6}}\left(\frac{r^{5}}{5} ight) ight|_{0} ^{R} ight\}=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2}}{2}\left(\frac{1}{R}+\frac{1}{5 R} ight)=\frac{1}{4 \pi \epsilon_{0}} \frac{3}{5} \frac{q^{2}}{R}[/latex] .

(c) [latex]W=\frac{\epsilon_{0}}{2}\left\{\oint_{ S } V E \cdot d a +\int_{ u } E^{2} d au ight\}[/latex] , where ν is large enough to enclose all the charge, but otherwise arbitrary. Let’s use a sphere of radius a > R. Here [latex]V=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r}[/latex] .

[latex]W=\frac{\epsilon_{0}}{2}\left\{\int_{r=a}\left(\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r} ight)\left(\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}} ight) r^{2} \sin heta d heta d \phi+\int_{0}^{R} E^{2} d au+\int_{R}^{a}\left(\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}} ight)^{2}\left(4 \pi r^{2} d r ight) ight\}[/latex]

=[latex]\frac{\epsilon_{0}}{2}\left\{\frac{q^{2}}{\left(4 \pi \epsilon_{0} ight)^{2}} \frac{1}{a} 4 \pi+\frac{q^{2}}{\left(4 \pi \epsilon_{0} ight)^{2}} \frac{4 \pi}{5 R}+\left.\frac{1}{\left(4 \pi \epsilon_{0} ight)^{2}} 4 \pi q^{2}\left(-\frac{1}{r} ight) ight|_{R} ^{a} ight\} [/latex]

=[latex]\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2}}{2}\left\{\frac{1}{a}+\frac{1}{5 R}-\frac{1}{a}+\frac{1}{R} ight\}=\frac{1}{4 \pi \epsilon_{0}} \frac{3}{5} \frac{q^{2}}{R}[/latex] .

As a → 1, the contribution from the surface integral [latex]\left(\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2}}{2 a} ight)[/latex] goes to zero, while the volume integral [latex]\left(\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2}}{2 a}\left(\frac{6 a}{5 R}-1 ight) ight)[/latex] picks up the slack.

Question 2.34: Find the energy stored in a uniformly charged solid sphere o...

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