Spring & Capacitor Homework: Equations & Solutions

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    Capacitor Spring
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Homework Help Overview

The discussion revolves around the forces acting on a capacitor and a spring system, specifically examining the relationship between the charge on the capacitor, the forces between its plates, and the resulting spring force. The subject area includes concepts from electrostatics and mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculations of forces when the capacitor is charged and the switch is opened or closed. There are attempts to derive relationships between the forces and the spring force based on varying charges and distances. Questions arise regarding the influence of changing capacitance and potential on these calculations.

Discussion Status

The discussion is active, with participants providing different interpretations of the force calculations and questioning each other's results. Some guidance has been offered regarding the changes in capacitance and potential, and there is acknowledgment of differing results among participants.

Contextual Notes

Participants note that the capacitance and potential change during the process, which may affect the calculations. There is also mention of the need for clarity in the steps taken to arrive at conclusions.

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Homework Statement



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Homework Equations



Force of attraction between the plates = (1/2)(QV)/d

The Attempt at a Solution



Initial charge on the capacitor Q1=CV
Final charge on the capacitor Q2=2CV

When switch is open force between the plates F1 = (1/2)(Q1V)/d1 = (1/2)(CV2)/d1

When switch is closed force between the plates F2 = (1/2)(Q2V)/d2 = (1/2)(2CV2)/d2

d2 = (3/2)d1

F1/F2 = 3/4 or F2 = (4/3)F1

The spring force also becomes 4/3 times of the initial force i.e (4/3)F0

Is it correct ?

Many Thanks
 

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Vibhor said:
inal charge on the capacitor Q2=2CV
Be careful here, the capacitance changes.
The potential changes as well, which influences the calculation of F2.
 
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mfb said:
Be careful here, the capacitance changes.
The potential changes as well, which influences the calculation of F2.

:sorry:

F1/F2 = 8/9 or F2 = (9/8)F1

The spring force also becomes 9/8 times of the initial force i.e (9/8)F0

Is it correct now?
 
Can you show your steps? I get a different result.
 
Sorry once again .

It should be (16/9)F0 .

If it is wrong , i will surely show you the steps :smile: .
 
That agrees with the answer I got.
 
Thanks mfb .
 
The spring stretches in both the situations . More in the latter case , as attractive force between the plates gets stronger.

Right ?
 
Right. Which also means the system has to be moved a bit, otherwise the solution doesn't make sense.
 

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