What Happens to the Voltage Across a Capacitor When the Plates are Separated?

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Discussion Overview

The discussion revolves around the behavior of voltage across a capacitor when the plates are separated after being charged and disconnected from a battery. Participants explore the implications of changing plate distance on voltage, capacitance, and charge, referencing external materials for validation.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that since the charge (Q) remains constant and the capacitance (C) decreases as the distance between the plates increases, the voltage (V) across the capacitor should increase.
  • Others reference an external source that claims the voltage decreases when the plates are separated, questioning the validity of that source and its formula.
  • One participant suggests that the decrease in capacitance due to increased distance requires work to be done on the system, which contributes to an increase in potential energy and thus voltage.
  • There is a consensus among some participants that the formula presented in the external source is incorrect, specifically regarding the relationship between charge, capacitance, and voltage.

Areas of Agreement / Disagreement

Participants express disagreement regarding the behavior of voltage when the plates are separated, with some supporting the idea that voltage increases and others referencing an external source that suggests otherwise. The discussion remains unresolved as to which perspective is correct.

Contextual Notes

Participants reference an external document for validation, but there is uncertainty about the correctness of the claims made in that document. The discussion includes unresolved mathematical relationships and interpretations of physical principles.

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A capacitor has a capacitance C and is charged using a battery. After disconnecting the battery the two plates are separated until they are twice as far apart as originally. What happens to the voltage across the capacitor?
I think answer should be voltage increases . As battery has been disconnected q will remain same . Distance between the plates is increasing , capacitance should decrease.
##Q##= CΔV
As Q is constant and C is decreasing , V should increase.
But according to http://www.bama.ua.edu/~tmewes/PH-106/2008/clicker%20questions/CH26%20Clicker.pdf
Go to page 10
Voltage goes down . It is wrong . And formula is also wrong. Is that correct?

I have gone through all 14 pages of this pdf and found them right except this one , but I want to confirm are all other 13 pages correct?
 
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gracy said:
As Q is constant and C is decreasing , V should increase.
But according to http://www.bama.ua.edu/~tmewes/PH-106/2008/clicker%20questions/CH26%20Clicker.pdf
Go to page 10
Voltage goes down . It is wrong . And formula is also wrong. Is that correct?

I agree with you ... it may have been a genuine highlighting mistake ?it increases because you do work on the system to spread the plates, this results in an increase in potential energyDave
 
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gracy said:
I have gone through all 14 pages of this pdf and found them right except this one , but I want to confirm are all other 13 pages correct?
 
Yes, the formula is wrong on that page.(10)
 
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gracy said:
A capacitor has a capacitance C and is charged using a battery. After disconnecting the battery the two plates are separated until they are twice as far apart as originally. What happens to the voltage across the capacitor?
I think answer should be voltage increases . As battery has been disconnected q will remain same . Distance between the plates is increasing , capacitance should decrease.
##Q= CΔV##
As Q is constant and C is decreasing , V should increase.
But according to http://www.bama.ua.edu/~tmewes/PH-106/2008/clicker%20questions/CH26%20Clicker.pdf
Go to page 10
Voltage goes down . It is wrong . And formula is also wrong. Is that correct?

I have gone through all 14 pages of this pdf and found them right except this one , but I want to confirm are all other 13 pages correct?
You are right.

Solving ##\ Q= C\,ΔV\ ## for ΔV gives ##\displaystyle\ \Delta V = \frac QC \ ##.

That pdf / PowerPoint has the incorrect ##\displaystyle\ \Delta V = \frac CQ \ ##.
 
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