Electrostatics 4 Question 6
6. For a uniformly charged ring of radius $R$, the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then, value of $h$ is
(Main 2019, 9 Jan I)
(a) $\frac{R}{\sqrt{2}}$
(b) $R \sqrt{2}$
(c) $R$
(d) $\frac{R}{\sqrt{5}}$
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Solution:
- Electric field at a distance ’ $h$ ’ from the centre of uniformly charged ring of total charge $q$ (say) on its axis is given as,
$$ E=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{q h}{\left(h^{2}+R^{2}\right)^{3 / 2}} $$
For the magnitude to be maximum, then
$$ \begin{gathered} \frac{d E}{d h}=0 \\ \Rightarrow \quad \frac{d E}{d h}=\frac{q}{4 \pi \varepsilon _0} \\ \frac{\left(h^{2}+R^{2}\right)^{3 / 2}-h\left[3 / 2\left(h^{2}+R^{2}\right)^{1 / 2} 2 h\right]}{\left(h^{2}+R^{2}\right)^{3}}=0 \\ \Rightarrow \quad 0=\frac{\left(h^{2}+R^{2}\right)^{3 / 2}-3 h^{2}\left(h^{2}+R^{2}\right)^{1 / 2}}{\left(h^{2}+R^{2}\right)^{3}} \\ \Rightarrow\left(h^{2}+R^{2}\right)^{3 / 2}=3 h^{2}\left(h+R^{2}\right)^{1 / 2} \Rightarrow 3 h^{2}=\left(h^{2}+R^{2}\right) \\ 3 h^{2}-h^{2}=R^{2} \\ 2 h^{2}=R^{2} \Rightarrow h= \pm \frac{R}{\sqrt{2}} \end{gathered} $$
$\therefore$ At $\frac{R}{\sqrt{2}}$, the value of electric field associated with a charged ring on its axis has the maximum value.
