Results 1 to 3 of 3

Math Help - Parallel Plate Electric Fields

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    27

    Smile Parallel Plate Electric Fields

    Hi there, again I just wanted to see if anyone could help with the answer to this past examination question as I don't have the answers so not sure if I'm doing them correctly!

    Thanks in advance.

    An electron is liberated from the lower of two parallel plates seperated by 5mm. The upper plate has a potential of 1000v relative to the lower plate. How long does it take to reach the top plate?
    (Neglect gravity)

    Ok, well I thought it best to work out the field strength so I did E=V/d giving E=2.0x10^5N/C

    Then I said that F=ma=qE so a=qE/m and since the chrage of electron is e, this gives a=eE/m=3.5x10^16!

    Surely this is much to greater acceleration!

    I then used s=ut + 1/2at^2 to get the time taken, which gives t = 3.78x10^-10.

    I'm thinking these figures look way out!

    Any help much appriciated.

    Mike
    Last edited by mikewhant; December 30th 2009 at 11:42 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Dec 2009
    From
    toronto
    Posts
    17
    Quote Originally Posted by mikewhant View Post
    Hi there, again I just wanted to see if anyone could help with the answer to this past examination question as I don't have the answers so not sure if I'm doing them correctly!

    Thanks in advance.

    An electron is liberated from the lower of two parallel plates seperated by 5mm. The upper plate has a potential of 1000v relative to the lower plate. How long does it take to reach the top plate?
    (Neglect gravity)

    Ok, well I thought it best to work out the field strength so I did E=V/d giving E=2.0x10^5N/C

    Then I said that F=ma=qE so a=qE/m and since the chrage of electron is e, this gives a=eE/m=3.5x10^16!

    Surely this is much to greater acceleration!

    I then used s=ut + 1/2at^2 to get the time taken, which gives t = 3.78x10^-10.

    I'm thinking these figures look way out!

    Any help much appriciated.

    Mike
    You would probably notice that the speed of the electron is approaching 3*10^8, which happens to be the upper bound.
    Special Relativity states the following equation to calculate such high objects' speed:
    E^2=p^2*c^2+m^2*c^4 where E and p are the energy and momentum of the electron respectively
    Last edited by sym0110; December 30th 2009 at 08:44 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    27
    Thank you, glad to see I'm doing it correctly.

    Happy new year!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coulombs law and Electric Fields
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: December 30th 2009, 07:55 AM
  2. Parallel plate capacitor?
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: June 30th 2009, 04:28 AM
  3. Capacitance calculation (parallel plate type)
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: December 11th 2008, 01:02 PM
  4. Physics Electric Forces and Electric Fields
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: February 13th 2008, 04:36 AM
  5. Electric fields
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: March 4th 2007, 01:58 PM

Search Tags


/mathhelpforum @mathhelpforum