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 axis of disc from a height $H$ with zero initial velocity. The particle has $\frac{q}{m}=\frac{4{\epsilon }_{0}g}{\sigma }$. The value of $H$ if the particle just reaches the disc is

• A.
  $\frac{a}{2}$
• B.
  $\frac{a}{3}$
• C.
  
  $\frac{4a}{3}$
• D.
  
  $\frac{3a}{2}$

Answer 1

The correct option is C


$\frac{4a}{3}$
Let us consider a ring element on the disc as shown in figure.
The charge $dq$ contained in the ring shown in figure,
$dq=\left(2\pi rdr\right)\sigma$

Potential of $\text{P}$ due to this ring,
$dV=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{x}$

where, $x=\sqrt{H^2+r^2}$
Substituting values, we get
$dV=\frac{1}{4\pi {\epsilon }_0}\frac{\left(2\pi rdr\right)\sigma }{\sqrt{H^2+r^2}}$

$\Rightarrow dV=\frac{1}{2{\epsilon }_0}\frac{\left(rdr\right)\sigma }{\sqrt{H^2+r^2}}$

So, potential due to the complete disc,
$V_P=\frac{\sigma }{2{\epsilon }_0}\int \frac{rdr}{\sqrt{H^2+r^2}}$

Integrating above between the limit $r=0$ to $r=a$ we get,

$V_p=\frac{\sigma }{2{\epsilon }_0}(\sqrt{a^2+H^2}-H)$

Potential at $\text{O}$ where $H=0$ will be
$V_p=\frac{\sigma a}{2{\epsilon }_0}$

Now, particle is released from $\text{P}$ and it just reaches point $\text{O}$. Therefore, from conservation of mechanical energy

$\text{Decrease in gravitational potential energy=Increase in electrostatic potential energy}$

(Here, kinetic energy is not considered because the initial and final velocity of the particle are zero.)

$\therefore mgH=q(V_O-V_P)$
Substituting the values in the above equation we get,
$gH=\left(\frac{q}{m}\right)\left(\frac{\sigma }{2{\epsilon }_0}\right)[a-\sqrt{a^2+H^2}+H]...\left(1\right)$

Given,
$\frac{q}{m}=\frac{4{\epsilon }_{0}g}{\sigma }$

$\Rightarrow \frac{q\sigma }{2{\epsilon }_{0}m}=2g$

Substituting in equation $\left(1\right)$, we get

$gH=2g[a+H-\sqrt{a^2+H^2}]$

$\Rightarrow \frac{H}{2}=(a+H)-\sqrt{a^2+H^2}$

$\Rightarrow \sqrt{a^2+H^2}=a+\frac{H}{2}$

Squaring on both sides,
$\Rightarrow a^2+H^2=a^2+\frac{H^2}{4}+aH$

$\Rightarrow \frac{3}{4}{H}^2=aH$

$\Rightarrow H(\frac{3}{4}H-a)=0$

$\Rightarrow H=\frac{4}{3}a$ and $H=0$

$\therefore H=\frac{4a}{3}$

Why this quation? Tip: Since particle is moving in conservative field so we can easily apply total energy conservation between initial and final position.