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Example 2.8. Find the potential of a uniformly charged spherical shell of radius R (Fig. 2.33). Solution This is the same problem we solved in Ex. 2.7, but this time let's do it using Eq. 2.30: $\frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r}da'.$ We might as well set the point P on the $z$ axis and use the law of cosines to express $r^2$: $r^2 = R^2 + z^2 - 2Rz\cos\theta'.$ FIGURE 2.33 An element of surface area on the sphere is $R^2\sin\theta'd\theta'd\phi'$, so $4\pi\epsilon_0V(z) = \sigma\int\frac{R^2\sin\theta'd\theta'd\phi'}{\sqrt{R^2 + z^2 - 2Rz\cos\theta'}}$ $= 2\pi R^2\sigma\int_0^\pi\frac{\sin\theta'}{\sqrt{R^2 + z^2 - 2Rz\cos\theta'}}d\theta'$ $= 2\pi R^2\sigma\left(\frac{1}{Rz}\sqrt{R^2 + z^2 - 2Rz\cos\theta'}\right)|_0^\pi$ $= \frac{2\pi R\sigma}{z}\left(\sqrt{R^2 + z^2 + 2Rz} - \sqrt{R^2 + z^2 - 2Rz}\right)$ $= \frac{2\pi R\sigma}{z}\left[\sqrt{(R + z)^2} - \sqrt{(R - z)^2}\right].$

          Example 2.8. Find the potential of a uniformly charged spherical shell of radius
R (Fig. 2.33).
Solution
This is the same problem we solved in Ex. 2.7, but this time let's do it using
Eq. 2.30:
$\frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r}da'.$
We might as well set the point P on the $z$ axis and use the law of cosines to
express $r^2$:
$r^2 = R^2 + z^2 - 2Rz\cos\theta'.$

FIGURE 2.33
An element of surface area on the sphere is $R^2\sin\theta'd\theta'd\phi'$, so
$4\pi\epsilon_0V(z) = \sigma\int\frac{R^2\sin\theta'd\theta'd\phi'}{\sqrt{R^2 + z^2 - 2Rz\cos\theta'}}$
$= 2\pi R^2\sigma\int_0^\pi\frac{\sin\theta'}{\sqrt{R^2 + z^2 - 2Rz\cos\theta'}}d\theta'$ 
$= 2\pi R^2\sigma\left(\frac{1}{Rz}\sqrt{R^2 + z^2 - 2Rz\cos\theta'}\right)|_0^\pi$
$= \frac{2\pi R\sigma}{z}\left(\sqrt{R^2 + z^2 + 2Rz} - \sqrt{R^2 + z^2 - 2Rz}\right)$
$= \frac{2\pi R\sigma}{z}\left[\sqrt{(R + z)^2} - \sqrt{(R - z)^2}\right].$

Example 2.8. Find the potential of a uniformly charged spherical shell of radius
R (Fig. 2.33).
Solution
This is the same problem we solved in Ex. 2.7, but this time let's do it using
Eq. 2.30:
(1)/(4πϵ0)∫(σ)/(r)da'.
We might as well set the point P on the z axis and use the law of cosines to
express r^2:
r^2 = R^2 + z^2 - 2Rzcosθ'.

FIGURE 2.33
An element of surface area on the sphere is R^2sinθ'dθ'dϕ', so
4πϵ0V(z) = σ∫(R^2sinθ'dθ'dϕ')/(√(R^2 + z^2 - 2Rzcosθ'))
= 2π R^2σ∫0^π(sinθ')/(√(R^2 + z^2 - 2Rzcosθ'))dθ'
= 2π R^2σ((1)/(Rz)√(R^2 + z^2 - 2Rzcosθ'))|0^π
= (2π Rσ)/(z)(√(R^2 + z^2 + 2Rz) - √(R^2 + z^2 - 2Rz))
= (2π Rσ)/(z)[√((R + z)^2) - √((R - z)^2)].

Added by Darren W.

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