+ Post New Thread
Page 2 of 2 FirstFirst 1 2
Results 21 to 23 of 23
  1. #21
    Full Member level 4
    Points: 1,390, Level: 8

    Join Date
    Aug 2016
    Posts
    207
    Helped
    31 / 31
    Points
    1,390
    Level
    8

    Re: Better understanding of capacitor charging

    Quote Originally Posted by Akanimo View Post
    The voltage will not drop. Instead, it will go up.

    When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance decreases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
    This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

    Q=CV

    Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a lesser amount of charge to increase the voltage by 1V. Remember?
    Sorry about that. I wrote it the opposite way. Thanks FvM for pointing that out.

    Correction:
    The voltage will drop as depicted by formula:

    Epsilon*A/d

    When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance increases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
    This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

    Q=CV

    Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a higher amount of charge to increase the voltage by 1V. Remember?
    Last edited by Akanimo; 18th December 2018 at 01:40.
    -------------
    --Akanimo.


    1 members found this post helpful.

    •   AltAdvertisment

        
       

  2. #22
    Full Member level 4
    Points: 1,390, Level: 8

    Join Date
    Aug 2016
    Posts
    207
    Helped
    31 / 31
    Points
    1,390
    Level
    8

    Re: Better understanding of capacitor charging

    Quote Originally Posted by Akanimo View Post
    Sorry about that. I wrote it the opposite way. Thanks FvM for pointing that out.

    Correction:
    The voltage will drop as depicted by formula:

    Epsilon*A/d

    When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance increases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
    This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

    Q=CV

    Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a higher amount of charge to increase the voltage by 1V. Remember?
    Correction:
    The voltage will drop.

    When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance increases, as depicted by formula C=Epsilon*A/d, because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
    This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

    Q=CV

    Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance would have a higher voltage for the same amount of charge difference between the two plates . Remember?
    Last edited by Akanimo; 18th December 2018 at 03:54.
    -------------
    --Akanimo.



    •   AltAdvertisment

        
       

  3. #23
    Full Member level 6
    Points: 2,831, Level: 12

    Join Date
    Mar 2012
    Posts
    359
    Helped
    70 / 70
    Points
    2,831
    Level
    12

    Re: Better understanding of capacitor charging

    Quote Originally Posted by Akanimo View Post
    The voltage will not drop. Instead, it will go up.

    When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance decreases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
    This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

    Q=CV

    Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a lesser amount of charge to increase the voltage by 1V. Remember?
    As you can see from this link http://hyperphysics.phy-astr.gsu.edu...ic/pplate.html , the closer the plates, the higher the capacitance and the lower the voltage. The electrostatic field is stronger when the plates are closer, but it it also shorter.

    Ratch

    - - - Updated - - -

    Quote Originally Posted by c_mitra View Post
    SI base units are clearly explained in this page: https://physics.nist.gov/cuu/Units/units.html

    Some of the derived units are also there.
    Then you agree that voltage and temperature are energy densities?

    Ratch
    Hopelessly Pedantic


    1 members found this post helpful.

--[[ ]]--