Question

The largest magnitude the electric field on axis of a uniformly charged ring of radius $3\, m$ is at a distance $'h'$ from its centre. The value of $10\, h$ is________. $\left(\text{ Take } \frac{1}{\sqrt{2}} = 0 . 7\right)$

Answer 1

$E_{\text{axial }}=\frac{Q h}{4 \pi \left(\epsilon \right)_{0} \left(h^{2} + a^{2}\right)^{\frac{3}{2}}}$
where,
$a=$ radius of the ring,
$h=$ distance of point from the centre.
For the maximum value,
$\frac{d E}{d h}=0$
$\therefore \frac{d}{dh}\left[\frac{Qx}{4 \pi \left(\epsilon \right)_{0} \left(h^{2} + a^{2}\right)^{\frac{3}{2}}}\right]=0$
$\therefore \frac{\left(h^{2} + a^{2}\right)^{\frac{3}{2}} - h \left(\frac{3}{2}\right) \left(h^{2} + a^{2}\right)^{\frac{1}{2}} \cdot \left(\right. 2 h \left.\right)}{\left(h^{2} + a^{2}\right)^{3}}=0$
$\therefore \left(h^{2} + a^{2}\right)-3h^{2}=0$
$\therefore h=\pm\frac{a}{\sqrt{2}}$
$=\frac{3}{\sqrt{2}}m\ldots .\left(\right.\because a=3m\left.\right)$
$=3\times 0.7=2.1m$