Question

The largest magnitude the electric field on axis of a uniformly charged ring of radius $3m$ is at a distance ${}^{\prime }{h}^{\prime }$ from its centre. The value of $10h$ is________. $\left(\text{ Take }\frac{1}{\sqrt{2}}=0.7\right)$

Answer 1

Answer: 21

$E_{\text{axial }}=\frac{Qh}{4\pi {\left(ϵ\right)}_0{\left(h^2+a^2\right)}^{\frac{3}{2}}}$
where,
$a=$ radius of the ring,
$h=$ distance of point from the centre.
For the maximum value,
$\frac{dE}{dh}=0$
$\therefore \frac{d}{dh}\left[\frac{Qx}{4\pi {\left(ϵ\right)}_0{\left(h^2+a^2\right)}^{\frac{3}{2}}}\right]=0$
$\therefore \frac{{\left(h^2+a^2\right)}^{\frac{3}{2}}-h\left(\frac{3}{2}\right){\left(h^2+a^2\right)}^{\frac{1}{2}}\cdot (2h)}{{\left(h^2+a^2\right)}^3}=0$
$\therefore \left(h^2+a^2\right)-3h^2=0$
$\therefore h=\pm \frac{a}{\sqrt{2}}$
$=\frac{3}{\sqrt{2}}m\dots .(\because a=3m)$
$=3\times 0.7=2.1m$