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Thread: Non conducting disc?

  1. #1
    Super Member fardeen_gen's Avatar
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    Non conducting disc?

    A non-conducting disc of radius $\displaystyle a$ and uniform positive charge density $\displaystyle \sigma$ is placed on the ground, with its axis vertical. A particle of mass $\displaystyle m$ and positive charge $\displaystyle q$ is dropped, along the axis of the disc, from a height $\displaystyle H$ with zero initial velocity. The particle has $\displaystyle \frac{q}{m} = \frac{4\epsilon_0 g}{\sigma}$

    A) Find the value of H if the particle just reaches the disc.
    Answer:
    Spoiler:
    $\displaystyle H = \frac{4a}{3}$


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

    How to do it?
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  2. #2
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    Voltage at the particle:
    $\displaystyle V_P=\int^{a}_{0}\frac{\sigma 2\pi rdr}{4\pi\epsilon_0\sqrt{r^2+H^2}}=\frac{\sigma}{2 \epsilon_0}(\sqrt{a^2+H^2}-H)$ (1)

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

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

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

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

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

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

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

    Luobo
    Last edited by mr fantastic; Sep 19th 2009 at 12:56 AM. Reason: Restored original reply
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