Question 4.4: A circular ring of radius a carries a uniform charge ρL C/m ...

(a) Consider the system as shown in Figure 4.9. Again the trick in finding E by using      eq. (4.14)

E=\int_{L}\frac{\rho _{L}dl}{4\pi\varepsilon _{o}R^{2}}a_{R}  (line charge)

is deriving each term in the equation. In this case

dl=ad\phi,      R=a\left(-a_{\rho}\right)+ha_{z}

R=\left|R\right|=\left[a^{2}+h^{2}\right]^{1/2},      a_{R}=\frac{R}{R}

or

\frac{a_{R}}{R^{2}}=\frac{R}{\left|R\right|^{3}}=\frac{-aa_{\rho}+ha_{z}}{\left[a^{2}+h^{2}\right]^{3/2} }

Hence

E=\frac{\rho _{L}}{4\pi\varepsilon _{o}}\int_{\phi=0}^{2\pi}\frac{\left(-aa_{\rho}+ha_{z}\right) }{\left[a^{2}+h^{2}\right]^{3/2}}ad\phi

By symmetry, the contributions along a_{\rho} add up to zero. This is evident from the fact that for every element dl there is a corresponding element diametrically opposite that gives an equal but opposite dE_{\rho} so that the two contributions cancel each other.

Thus we are left with the z-component. That is

E=\frac{\rho _{L}aha_{z}}{4\pi\varepsilon _{o}\left[h^{2}+a^{2}\right]^{3/2}}\int_{0}^{2\pi}d\phi=\frac{\rho _{L}aha_{z}}{2\varepsilon _{o}\left[h^{2}+a^{2}\right]^{3/2}}

as required.

(b)

\frac{d\left|E\right| }{dh}=\frac{\rho _{L}a}{2\varepsilon _{o}}\left\{\frac{\left[h^{2}+a^{2}\right]^{3/2}\left(1\right)-\frac{3}{2}\left(h\right)2h\left[h^{2}+a^{2}\right]^{1/2}}{\left[h^{2}+a^{2}\right]^{3}} \right\}

For maximum E,\frac{d\left|E\right| }{dh}=0, which implies that

\left[h^{2}+a^{2}\right]^{1/2}\left[h^{2}+a^{2}-3h^{2}\right]=0

a^{2}-2h^{2}=0      or      h=\pm\frac{a}{\sqrt{2}}

(c) Since the charge is uniformly distributed, the line charge density is

\rho _{L}=\frac{Q}{2\pi a}

so that

E=\frac{Qh}{4\pi\varepsilon _{o}\left[h^{2}+a^{2}\right]^{3/2}}a_{z}

As a\rightarrow 0

E=\frac{Q}{4\pi\varepsilon _{o}h^{2}}a_{z}

or in general

E=\frac{Q}{4\pi\varepsilon _{o}r^{2}}a_{R}

which is the same as that of a point charge, as one would expect.