1. Dismiss Notice
  2. Download free Tapatalk for iPhone or Tapatalk for Android for your phone and follow the SDN forums with push notifications.
    Dismiss Notice
Dismiss Notice
Visit Interview Feedback to view and submit interview information.

electron between charged plate question

Discussion in 'MCAT Study Question Q&A' started by dorjiako, Aug 16, 2011.

  1. dorjiako

    2+ Year Member

    Joined:
    Dec 9, 2010
    Messages:
    71
    Likes Received:
    0
    An electron starts from rest at tone plate of a parallel-plate capacitor and accelerates to the other plate. The plate separation is 2mm, and it takes 1ms for the electron to travel from one side to the other. What is the capacitance of this capacitor if there are 2C of charge stored on the plates? m = 9.11 x 10^-31kg, e=1.6 x 10^-19C.
    The answer is 4.4 x 10^10 F but I can't seem to arrive at same answer. Please Help.
     
  2. Note: SDN Members do not see this ad.

  3. raffster2

    2+ Year Member

    Joined:
    Aug 8, 2011
    Messages:
    21
    Likes Received:
    0
    Status:
    Pre-Medical
  4. Bumbl3b33

    Removed

    Joined:
    Feb 12, 2011
    Messages:
    530
    Likes Received:
    1
    Status:
    Medical Student
    Unlike the person above me, I'll try to help you. I haven't done the calculation to see if this is right, but here is how I would approach it:

    First and foremost, understand that capacitance is how much ENERGY is stored on you capacitor. If you want more help with that, http://en.wikipedia.org/wiki/Capacitance The minute you see energy and objects moving the first thing going in my head is always KE=PE at the end. Thus, kinetic energy gained by the electron as it accelerates is the energy provided by the capacitor (this is the easiest way for me to think about it, sorry). So, I'm guessing they told us the distance and time the e- took to travel so we can figure out the final velocity. Why do we need that? So we can plug it into KE=1/2mv^2.

    So, you can use kinematics to get the final velocity of the electron as it hits the positive plate, right? Displacement=1/2at^2 so you can get acceleration, and then with that acceleration you can get the final velocity by doing finalvelocity^2=2(acceleration)(displacement)...in both cases, the Vinitial is disregarded since it's at rest. Now plug that into KE equation.

    Now, that's our KE, it has to equal to PE. The PE for a capacitor is given by 1/2QV...when you rearrange it by plugging in V=Q/C (from Q=CV) for the V in 1/2QV, you get 1/2(Q^2)/C. So since all the potential energy of the capacitor was used to get to the positive plate, we can set KE=PE=0.5Q^2/C. Now, we know KE, we know Q, the charge of 2 coulombs, and all you have to do is solve for C and that should be the answer. Remember to watch your units when doing the kinematics and stuff. Let me know if this is not right!
     
  5. justhanging

    2+ Year Member

    Joined:
    May 22, 2010
    Messages:
    335
    Likes Received:
    2
    Status:
    Pre-Pharmacy
    You have to set K.E = Vq which is the work done by the electric field of the capacitor to move the electron across. 1/2QV is the energy stored in the capacitor. Another way you could do is that once you find out acceleration from kinematics set it equal to the net force which is just the force of the electric field from the capacitor.

    m(electron) a(kinamatics) = Eq(electron)

    Find out E then use V = Ed(distance between plates)

    With V you could then use Q = CV
     
  6. Bumbl3b33

    Removed

    Joined:
    Feb 12, 2011
    Messages:
    530
    Likes Received:
    1
    Status:
    Medical Student
    He's right. I don't know what I was thinking....my bad! PE=qV is for one charge, like in your scenario.

    From an old forum response:
    "Let's assume that you understand that PE=qV for one charge. Now, for condenser you can move one small charge(q) from plate to plate and work will be qV also. However, when you move the next charge(q2) the V is less, since you total charge on both plates are now Q-q. and so on. When you move the "last" charge V is almost zero. To find the total PE you need to sum all these works. (Integrate) or take the average which will be 0.5*QV (Note, I use the capital Q in this formula) "
     

Share This Page