Understanding the Concept of Capacitor: Q&A

  • Context: Undergrad 
  • Thread starter Thread starter Karan Punjabi
  • Start date Start date
  • Tags Tags
    Capacitor Concept
Click For Summary

Discussion Overview

The discussion revolves around the concept of capacitance in capacitors, specifically addressing the relationship between charge, potential difference, and capacitance. Participants explore theoretical aspects and clarify misconceptions related to these concepts.

Discussion Character

  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether increasing the potential difference while keeping the charge constant decreases capacitance.
  • Another participant asserts that capacitance is a property of the capacitor and is not changed by varying the voltage; they reference the equation Q=CV to explain that increasing voltage leads to an increase in charge.
  • A further reply emphasizes that if the charge is kept constant, the voltage cannot be changed, and if voltage is changed, the charge must increase, providing an analogy involving water pressure to illustrate this point.
  • The analogy suggests that applying a voltage results in charge accumulation until equilibrium is reached, which affects the voltage across the capacitor.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between charge, voltage, and capacitance. There is no consensus on the initial claim regarding capacitance changes, and the discussion remains unresolved.

Contextual Notes

Some assumptions about the definitions of capacitance and the conditions under which charge and voltage are varied may not be fully articulated, leading to potential misunderstandings.

Karan Punjabi
Messages
127
Reaction score
3
Guys, I am learning about the concept of capacitor,but I am confused in some things that if i kept the amount of charge constant between two plates of a capacitor and then increase the potential diff. this means that i decreased its capacitance right? If I am wrong then please explain the given concept that i have uploaded in the pic.
 

Attachments

  • IMG_20160615_244100116.jpg
    IMG_20160615_244100116.jpg
    30.2 KB · Views: 471
Physics news on Phys.org
The capacitance is a property of the circuit element in question - it's usually marked right on it. We have Q=CV, so if you increase the potential difference, the amount of charge will increase.

Your image shows up too small to read on my monitor.
 
pixel said:
The capacitance is a property of the circuit element in question - it's usually marked right on it. We have Q=CV, so if you increase the potential difference, the amount of charge will increase. ...See this

Your image shows up too small to read on my monitor.
 
No, you don't decrease the capacitance.
Why ?
If you keep the amount of charge on the capacitor constant, then you cannot change the voltage across it.
And if you change the voltage across it the charge on the plates will increase.
There is no way of circumventing that.

A little analogy:
If you had a container of water with a tube at the bottom of it, what happens when you increase the pressure in the tube ?
Water flows into the container until some point, at which the pressure due to the height is equal to the pressure in the pipe.
A capacitor works similarly. (this analogy certainly isn't perfect, especially since there are two kinds of charge and only one kind of pressure)
If you apply a voltage to a capacitor, charges will flow onto it until the voltage across it is the same as from the voltage across the source connected to it.

When you accumulate like charges in some space their potential energy increases(since they repell each other).And since voltage is just energy per charge, the (value of) voltage also increases.
So if you stuff more charge on a capacitor the voltage across it will rise, regardless of what it is connected to.
The reason why we can say ##C=\frac{Q}{V}## is that when you apply a voltage to the capacitor the amount of charge ##Q=V*C## will be on it, once it reaches equillibrium.(which it usually does insanely fast, unless you have very high resistors, impedances or capacitances in the circuit)
EDIT: I hope you understand what i am getting at.I realized that i wrote it somewhat strange and confusing.
 
  • Like
Likes   Reactions: Karan Punjabi

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
6K