Here is a fourth way of computing the energy of a uniformly charged solid sphere: Assemble it like a snowball, layer by layer, each time bringing in an infinitesimal charge dq from far away and smearing it uniformly over the surface, thereby increasing the radius. How much work dW does it take to

Question

Here is a fourth way of computing the energy of a uniformly charged solid sphere: Assemble it like a snowball, layer by layer, each time bringing in an infinitesimal charge dq from far away and smearing it uniformly over the surface, thereby increasing the radius. How much work dW does it take to build up the radius by an amount dr? Integrate this to find the work necessary to create the entire sphere of radius R and total charge q.

Accepted Answer

[latex]d W=d \bar{q} V=d \bar{q}\left(\frac{1}{4 \pi \epsilon_{0}} ight) \frac{\bar{q}}{r}, \quad(\bar{q}= ext { charge on sphere of radius } r)[/latex] .

[latex]\bar{q}=\frac{4}{3} \pi r^{3} ho=q \frac{r^{3}}{R^{3}} \quad(q= ext { total charge on sphere })[/latex] .

[latex]d \bar{q}=4 \pi r^{2} d r ho=\frac{4 \pi r^{2}}{\frac{4}{3} \pi R^{3}} q d r=\frac{3 q}{R^{3}} r^{2} d r[/latex] .

[latex]d W=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{q r^{3}}{R^{3}} ight) \frac{1}{r}\left(\frac{3 q}{R^{3}} r^{2} d r ight)=\frac{1}{4 \pi \epsilon_{0}} \frac{3 q^{2}}{R^{6}} r^{4} d r[/latex]

[latex]W=\frac{1}{4 \pi \epsilon_{0}} \frac{3 q^{2}}{R^{6}} \int_{0}^{R} r^{4} d r=\frac{1}{4 \pi \epsilon_{0}} \frac{3 q^{2}}{R^{6}} \frac{R^{5}}{5}=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{3}{5} \frac{q^{2}}{R} ight)[/latex] .

Question 2.35: Here is a fourth way of computing the energy of a uniformly ...

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