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!

Capacitors in a circuit as Plates.

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Four parallel metal plates P1, P2, P3, and P4, each of area 7.50 cm2, are separated successively by a distance d = 1.06 mm, as shown in the figure below. P1 is connected to the negative terminal of a battery, and P2 to the positive terminal. The battery maintains a potential difference of 12.0 V.

    Important: http://www.webassign.net/pse/p26-55.gif

    (a) If P3 is connected to the negative terminal, what is the capacitance of the three-plate system P1P2P3?

    (b) What is the charge on P2?

    (c) If P4 is now connected to the positive terminal of the battery, what is the capacitance of the four plate system P1P2P3P4?

    (d) What is the charge on P4?

    2. Relevant equations

    C=A/D

    3. The attempt at a solution

    I tried drawing the circuit in a simple way, i had no idea, i tried it once using parallel (P1P2, P2P3) and failed, i tried it in series, also p1p2 and p2p3... it is wrong... i have no idea how the circuit looks like, i have no clue how to solve it, and it is due in less than 3 hours :P ... any tip would be appreciated
     
  2. jcsd
  3. Nov 1, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Hint: Any two plate surfaces that face each other constitutes a capacitor. This means that a metal plate between two others represents two capacitor plates, one for each of its surfaces (think of the surfaces themselves representing the plates, and the bulk of the metal in between them as just a conductor joining them). Thus one metal plate can in fact be part of two separate capacitors.

    Can you share your attempt at a circuit drawing?
     
  4. Nov 1, 2011 #3
    I drew it where p1p2 are parallel to p3p2, and here are my full results (all wrong):
    125.295 pF
    1503.54 pC
     
  5. Nov 1, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    Well, P1P2 are in parallel to P2P3. So that's okay. But your result is out by an order of magnitude. Perhaps your unit conversions went astray?

    What value you did you get for the basic capacitor consisting of two plates with the given separation?
     
  6. Nov 1, 2011 #5
    epsilon0 * A/D
    around 6.2647555 E-10
     
  7. Nov 1, 2011 #6

    gneill

    User Avatar

    Staff: Mentor

    Seems to be off by a factor of 100 (it's too large by that amount). You'll have to spell out the whole calculation, constant and all.
     
  8. Nov 1, 2011 #7
    Epsilon Node: 8.85...E-12
    A = 0.075 M
    D = 0.00106 M

    it would be ironic if my mistakes are from cm -> m and mm-> :P ... cuz i used google to do that
     
  9. Nov 1, 2011 #8
    i know 10 cm = 1 m
    1000 mm = 1 m
     
  10. Nov 1, 2011 #9
    sorry 100 Cm = 1 m...
     
  11. Nov 1, 2011 #10

    gneill

    User Avatar

    Staff: Mentor

    Your area is off. Probably because it's not a conversion of cm to m, but cm2 to m2.
     
  12. Nov 1, 2011 #11
    oh.. ok, fixed :P ... damn, stupid mistakes... THank you soo much
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Capacitors in a circuit as Plates.
  1. Plate Capacitors (Replies: 3)

  2. Plate Capacitor (Replies: 1)

  3. Plate capacitor (Replies: 2)

Loading...