Non conducting disc?

**fardeen_gen** · June 29th 2009, 01:22 PM

A non-conducting disc of radius $a$ and uniform positive 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 $\frac{q}{m} = \frac{4\epsilon_0 g}{\sigma}$

A) Find the value of H if the particle just reaches the disc.
Answer:

Spoiler:

B) Sketch the potential energy of the particle as a function of its height and find its equilibrium position.

How to do it?

**luobo** · August 8th 2009, 12:11 PM

Voltage at the particle:
$V_P=\int^{a}_{0}\frac{\sigma 2\pi rdr}{4\pi\epsilon_0\sqrt{r^2+H^2}}=\frac{\sigma}{2 \epsilon_0}(\sqrt{a^2+H^2}-H)$ (1)

Voltage at the disc center [simply set $H=0$ in (1)]:
$V_O=\frac{\sigma}{2\epsilon_0}{a}$ (2)

Increase in electric potential:
$\Delta U=q(V_O-V_P)=\frac{q\sigma}{2\epsilon_0}(a+H-\sqrt{a^2+H^2})$ (3)

Decrease in gravitational potential:
$\Delta V=mgH$ (4)

Energy conservation:
$\Delta U=\Delta V$ (5)

(3)(4)(5) together give
$\frac{q\sigma}{2\epsilon_0}(a+H-\sqrt{a^2+H^2})=mgH$ (6)

Use $\frac{q}{m}=\frac{4\epsilon_0 g}{\sigma}$ leads to
$H+a-\sqrt{a^2+H^2}=\frac{H}{2}$

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

Luobo

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