Capacitor and Mechanics question, IPhO 2004

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SUMMARY

The discussion focuses on a theoretical problem from the IPhO 2004 regarding the mechanics of a capacitor and the forces acting on a disk within an electric field. Key issues identified include the incorrect calculation of the threshold voltage and the misunderstanding of the electric field's role, specifically that only the upper plate's field exerts force on the disk. The final equation presented, v_s^2(1-\eta ^2)(1+\eta ^2)=2d\eta ^2(\eta ^2a_1+a_2), requires careful consideration of acceleration factors, which were miscalculated in the initial attempts. Participants emphasize the importance of correctly applying the electric field equations and algebraic manipulation to arrive at the correct solution.

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  • Basic concepts of potential difference and threshold voltage in capacitors
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Mantaray
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Homework Statement



It's theoretical question 1 d) I'm having problems with, I'm at a loss as to where I have gone wrong with my approach? The question has been attached.

Homework Equations



see attachment please

The Attempt at a Solution



see attachment please.

It's quite a lengthy problem I suppose, so a hearty thanks to whoever takes the time to take a look!
 

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1/ The threshold voltage is wrong. Hint: the E-field that exerts force on the disk is not V/d. Here we also need to reason a bit :smile:
2/ Have you tried computing it to the final answer? Let me rewrite your final equation as:
v_s^2(1-\eta ^2)(1+\eta ^2)=2d\eta ^2(\eta ^2a_1+a_2)
Oops, what happened to \eta ^4? Where is it? :biggrin:
 
1/ The threshold voltage is indeed wrong, I was thinking that the lower plate would exert a force upwards on the disc, but it does not since there is no electric field under the small plate. Therefore only the field of the upper plate exerts a force on the disc.

2/ I'll be thinking about that for some longer... I maybe should've taken into account the velocities instead of the plain speed, or does it work out? The solutions should be somewhere on the web too, but that would be cheating.
 
Hmm I was able to work your formula out to the final answer provided in the solution manual, but I lost a factor two in my accelerations. I used the force exerted on the disc by the electric field, which is \chiV^{2}/(2dm). Then if I input a_{1} = \chiV^{2}/(2dm) - g and a_{2} = -\chiV^{2}/(2dm) -g, I'm a factor two short in \chi V^{2}/m in the final answer. Or have I just made another mistake in my algebra?
 
Maybe I would be more helpful if I can see your full work :smile: Sorry, I'm lazy :biggrin:
Anyway, I guess there should be something wrong with the algebra.
 

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