Question

A non-conducting disc of radius a and uniform positive
surface charge density $\sigma$ is placed on the ground with its axis
vertical. A particle of mass m and positive charge q is
dropped, along the axis of the disc from a height H with zero
initial velocity. The particle has q/m $=4{\epsilon }_{0}g/\sigma .$
(a) Find the value of H if the particle just reaches the disc.
(b) Sketch the potential energy of the particle as a function
of its height and find its equilibrium position.

Answer 1

Potential at a height H on the axis of the disc V (P )
The charge dq contained in the ring shown in figure
$dq=(2\pi rdr)\sigma$
Potential at P due to this ring
$dV=\frac{1}{4\pi {\epsilon }_0}.\frac{dq}{x}wherex=\sqrt{H^2+r^2}$
$dV=\frac{1}{4\pi {\epsilon }_0}.\frac{(2\pi rdr)\sigma }{\sqrt{H^2+r^2}}=\frac{\sigma }{2{\epsilon }_0}\frac{rdr}{\sqrt{H^2+r^2}}$
$\therefore$ Potential due to the complete disc
$V_p={\int }_{r=0}^{r=a}dV$
$=\frac{\sigma }{2{\epsilon }_0}{\int }_{r=0}^{r=a}\frac{rdr}{\sqrt{H^2+r^2}}$
$V_p=\frac{\sigma }{2{\epsilon }_0}[\sqrt{a^2+H^2}-H]$
Potential at centre, (O) will be
$V_0=\frac{\sigma a}{2{\epsilon }_0}H=0$
(a) Particle is released from P and it just reaches point O.
Therefore, from conservation of mechanical energy
Decrease in gravitational potential energy = Increase in
electr ostatic potential energy
$(\Delta KE=0because{K}_i=K_f=0)$
$\therefore mgH=q[V_0-V_p]$
or $gH=\left(\frac{q}{m}\right)\left(\frac{\sigma }{2{\epsilon }_0}\right)[a-\sqrt{a^2+H^2}+H]...(i)$
$\frac{q}{m}=\frac{4{\epsilon }_{0}g}{\sigma }$
$\therefore \frac{q\sigma }{2{\epsilon }_{0}m}=2g$
Substituting in Eq. (i), we get
$gH=2g[a+H-\sqrt{a^2+H^2}]$
or $\frac{H}{2}=(a+H)-\sqrt{a^2+H^2}$
or $\sqrt{a^2+H^2}=a+\frac{H}{2}$
or $a^2+H^2=a^2+\frac{H^2}{4}+aH$
or $\frac{3}{4}{H}^2=aH$
or $H=\frac{4}{3}aandH=0$
$\therefore H=(4/3)a$
(b) Potential energy of the particle at height
// = Electrostatic potential energy + gravitational
potential energy
$\therefore U=qV+mgH$
Here V = Potential at height H
$U=\frac{\sigma q}{2{\epsilon }_0}[\sqrt{a^2+H^2}-H]+mgH...(ii)$
At equilibrium position
$F=\frac{-dU}{dH}=0$
Differentiating Eq. (ii) w.r.t. H
or $mg+\sigma \frac{q}{2{\epsilon }_0}\left[\right(\frac{1}{2})(2H)\frac{1}{\sqrt{a^2+H^2}}-1]=0$
$(\because \frac{\sigma q}{2{\epsilon }_0}=2mg)$
$\therefore mg+2mg[\frac{H}{\sqrt{a^2+H^2}}-1]=0$
or $1+\frac{2H}{\sqrt{a^2+H^2}}-2=0$
or $\frac{2H}{\sqrt{a^2+H^2}}=1$
or $\frac{H^2}{a^2+H^2}=\frac{1}{4}$
or $3H^2=a^{2}orH=\frac{a}{\sqrt{3}}$
From Eq. (ii), we can write
U - H equation as
$U=mg(2\sqrt{a^2+H^2}-H)$
(Parabolic variation)
U = 2 mga at H = 0
and $U=U_{min}=\sqrt{3}mgaatH=\frac{a}{\sqrt{3}}$
Therefore, U - H graph will be as shown
Note that at $H=\frac{a}{\sqrt{3}},Uisminimum.$
Therefore, $H=\frac{a}{\sqrt{3}}$ is stable equilibrium position.