1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Concept of capacitor

  1. Jun 14, 2016 #1
    Guys, Im learning about the concept of capacitor,but im 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 im wrong then please explain the given concept that i have uploaded in the pic.

    Attached Files:

  2. jcsd
  3. Jun 14, 2016 #2
    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.
  4. Jun 14, 2016 #3
  5. Jun 14, 2016 #4
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted