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 charge $q$ is dropped, along the axis of the disc from a height $H$ with zero initial velocity. It was observed that its velocity becomes zero just when it reaches the disc. The value of $\frac{3H}{a}$ is (​​​​​​​Given $\frac{q}{m}=\frac{4{ϵ}_{o}g}{\sigma }$)

Answer 1

Since the initial and final velocities are both zero, we can say

$\Delta$ K.E = 0

$\Rightarrow$ Work done = 0

i.e Work done by gravity ($W-g$) + Work done by electric field ($W_e$) = 0

i.e $mgH+W_e=0$

Work done by an electric field in moving charge from point A to point B $=-q(V_B-V_A)$

i.e $mgH+(-q(V_o-V_P\left)\right)=0$

i.e $mgH=-q(V_o-V_P)$

Potential due to a disc of radius $a$ at a point at a distance $x$ on its axis

$V=\frac{\sigma }{2{ϵ}_0}(\sqrt{a^2+x^2}-x)$

(see video for derivation)

$V_o=\frac{\sigma }{2{ϵ}_0}a$ and
$V_P=\frac{\sigma }{2{ϵ}_0}(\sqrt{a^2+H^2}-H)$

Substituting in the work energy equation,

$mgH+q\frac{\sigma }{2{ϵ}_0}(a-\sqrt{a^2+H^2}+H)$

$\Rightarrow gH=q\frac{\sigma }{2{ϵ}_{0}m}(a-\sqrt{a^2+H^2}+H)$

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

(Given $\frac{q}{m}=\frac{4{ϵ}_{o}g}{a}$)

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

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

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

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

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