A thin wire ring of radius $$r$$ carries a charge $$q$$. Find the magnitude of the electric field strength on the axis of the ring as a function of distance $$l$$ from its centre. Investigate the obtained function at $$l > > r$$. Find the maximum strength magnitude and the corresponding distance $$l$$. Draw the approximate plot of the function $$E(l)$$.
From the symmetry of the condition, it is clear that, the field along the normal will be zero.
i.e. $$E_{n} = 0$$ and $$E = E_{1}$$
Now $$dE_{1} = \dfrac {dq}{4\pi \epsilon_{0}(R^{2} + l^{2})} \cos \theta$$
But $$dq = \dfrac {q}{2\pi R}dx$$ and $$\cos \theta = \dfrac {l}{(R^{2} + l^{2})^{1/2}}$$
Hence
$$E = \int dE_{l} = \int_{0}^{2\pi R} \dfrac {ql}{2\pi R} \cdot \dfrac {dx}{4\pi \epsilon_{0}(R^{2} + l^{2})^{3/2}}$$
or $$E = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {ql}{(l^{2} + R^{2})^{3/2}}$$
and for $$l > > R$$, the ring behaves like a point charge, reducing the field to the value,
$$E = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q}{l^{2}}$$
For $$E_{max}$$, we should have $$\dfrac {dE}{dl} = 0$$
So, $$(l^{2} + R^{2})^{3/2} - \dfrac {3}{2} l(l^{2} + R^{2})^{1/2} 2l = 0$$ or $$l^{2} + R^{2} - 3l^{2} = 0$$
Thus $$l = \dfrac {R}{\sqrt {2}}$$ and $$E_{max} = \dfrac {q}{6\sqrt {3} \pi \epsilon_{0} R^{2}}$$.
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