Find the potential of a uniformly charged spherical shell of radius R (Fig. 2.33).

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Accepted Answer

This is the same problem we solved in Ex. 2.7, but this time let’s do it using Eq. 2.30:

[latex]V=\frac{1}{4\pi \epsilon _{0}}\int{\frac{\lambda (\acute{r})}{\varkappa }d\acute{l} } and \quad V=\frac{1}{4\pi \epsilon _{0}}\int{\frac{\sigma (\acute{r})}{\varkappa }d\acute{a} }[/latex] (2.30)

[latex]V=\frac{1}{4\pi \epsilon _{0}}\int{\frac{\sigma }{\varkappa }d\acute{a} }[/latex]

We might as well set the point P on the z axis and use the law of cosines to express η :

[latex]η^{2}=R^{2}+z^{2}-2Rz\cos \acute{ heta }.[/latex]

An element of surface area on the sphere is:[latex]R^{2}\sin \acute{ heta }d \acute{ heta }d \acute{\phi },[/latex] so

[latex]4\pi \epsilon _{0}V(z)=\sigma \int{\frac{R^{2}\sin \acute{ heta }d \acute{ heta }d \acute{\phi }}{\sqrt{R^{2}+z^{2}-2Rz\cos \acute{ heta }} } } [/latex]

[latex]=2\pi R^{2}\sigma \int_{0}^{\pi } {\frac{\sin \acute{ heta }}{\sqrt{R^{2}+z^{2}-2Rz\cos \acute{ heta }} } } d\acute{ heta }[/latex]

[latex]=2\pi R^{2}\sigma \left({\frac{1}{Rz} {\sqrt{R^{2}+z^{2}-2Rz\cos \acute{ heta }} } } ight) \mid ^{\pi }_{0}[/latex]

[latex]=\frac{2\pi R\sigma}{z} \left({{\sqrt{R^{2}+z^{2}-2Rz} }-{\sqrt{R^{2}+z^{2}-2Rz} } } ight)[/latex]

[latex]=\frac{2\pi R\sigma}{z}\left[\sqrt{\left(R+z ight)^{2} }-\sqrt{\left(R-z ight)^{2} } ight] [/latex]

At this stage, we must be very careful to take the positive root. For points outside
the sphere, z is greater than R, and hence [latex]\sqrt{\left(R-z ight)^{2} }=z-R [/latex]; for points inside the sphere,[latex]\sqrt{\left(R-z ight)^{2} }=R-z [/latex]. Thus,

[latex]V(z)=\frac{R\sigma }{2\epsilon _{0}z}\left[\left(R+z ight)-\left(z-R ight) ight]=\frac{R^{2}\sigma }{\epsilon _{0}z}, [/latex] outside;

[latex]V(z)=\frac{R\sigma }{2\epsilon _{0}z}\left[\left(R+z ight)-\left(R-z ight) ight]=\frac{R\sigma }{\epsilon _{0}}, [/latex] inside.

In terms of r and the total charge on the shell,[latex]q=4\pi R^{2}\sigma [/latex]

[latex]V(r)=\begin{cases}\frac{1}{4\pi \epsilon _{0}}\frac{q}{r} & (r\geq R) \quad \\frac{1}{4\pi \epsilon _{0}}\frac{q}{R} & (r\leq R)\end{cases} [/latex]

Of course, in this particular case, it was easier to get V by using Eq. 2.21 than Eq. 2.30, because Gauss’s law gave us E with so little effort. But if you compare Ex. 2.8 with Prob. 2.7, you will appreciate the power of the potential formulation.

[latex]V(r)\equiv -\int_{\omicron }^{r}{E.dI} [/latex] (2.21)

Question 2.8: Find the potential of a uniformly charged spherical shell of...

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