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!

Capacitance and induced charges

  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data

    Two large parallel metal plates are oriented horizontally and separated by a distance 3d. A grounded conducting wire joins them, and initially each plate carries no charge. Now a third identical plate carrying charge Q is inserted between the two plates, parallel to them and located a distance d from the upper plate (a) What induced charge appears on each of the two original plates? (b) What potential difference appears between the middle plate and each of the other plates? Each plate has area A.

    2. Relevant equations


    3. The attempt at a solution
    I'm not sure how to even attempt the question,

    a) I don't know how the distance d is related to charge induced on the top and bottom since E fields between large parallel plates are uniform. What I do think, is that the magnitude of charge induced is equal to Q, by the principle of conservation of charge, but that's all i can come up with.

    b) My intuition tells me that using the equation C = ∈A/d and C = Q/V allows me to find the potential difference.

    Any help will be greatly appreciated.
     
  2. jcsd
  3. Feb 1, 2017 #2

    ShayanJ

    User Avatar
    Gold Member

    You know the induced charge on each plate. You also can calculate the capacitance of each of the capacitors. So V=Q/C gives you the potential difference for each capacitor. What is it that you don't understand?
     
  4. Feb 1, 2017 #3
    After obtaining the circuit diagram for this, i'm not quite sure why this is a parallel circuit and why PD across all the 2 capacitors are the same, which is the main assumption i need to solve the problem.

    upload_2017-2-1_18-26-21.png

    It just doesn't look very parallel-y to me, it actually looks like 2 capacitors in series to me. So what is it that makes this parallel?
     
  5. Feb 1, 2017 #4

    ShayanJ

    User Avatar
    Gold Member

    Who said the capacitors are in parallel and who said the potential difference is the same for them?
    You're right, they're in series and you still can solve the problem easily. Just look at those formulas you've given in your first post.
     
  6. Feb 1, 2017 #5
    upload_2017-2-1_18-34-10.png

    This was pulled from a PDF i downloaded. If this was in series, won't the charge on each of the conducting plates be the same?
     
  7. Feb 1, 2017 #6
    upload_2017-2-1_18-37-17.png

    This was the diagram from the pdf, I think it omitted the grounding of the wire. Apologies for not putting this up initially.
     
  8. Feb 1, 2017 #7

    ShayanJ

    User Avatar
    Gold Member

    That seems completely wrong to me.
     
  9. Feb 1, 2017 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    There are two potentials. The capacitors each connect the two potentials, so from that perspective they are parallel.
     
  10. Feb 1, 2017 #9
    Could you explain what you meant by 2 potentials?
     
  11. Feb 1, 2017 #10

    gneill

    User Avatar

    Staff: Mentor

    When there are only two components in a loop you can view them as being either in series or in parallel (or both!) since the connection topology satisfies the criteria of both.

    It's tempting to judge a circuit's topology just by the way the components have been arranged relative to each other on paper, but it's the electrical connections that count.

    Here's the same circuit portrayed in two ways simply by rearranging the component positions. All electrical connections (the topology) is exactly the same for each:

    upload_2017-2-1_7-31-2.png

    There are two nodes (potentials) in the circuit: v0 and v1.
     
  12. Feb 1, 2017 #11
    Thanks a lot, I think I'm getting the 2 nodes bit. But how exactly does this make V1 = V0 and PD across the 2 capacitors (splitting the middle plate into top and bottom components) identical?

    Also, how does the fact that there is a grounding line affect this setup? Or is it simply an means of illustrating that the conductors were initially neutral?

    Apologies for being slow with this, I've come to realise Electricity might not be my strongest subject :sorry:
     
    Last edited: Feb 1, 2017
  13. Feb 1, 2017 #12

    gneill

    User Avatar

    Staff: Mentor

    v1 ≠ v0 . They are two separate potentials, and if you look at the diagrams they remain separate and the labels are on the same nodes in both. Here, let me make it clearer by coloring the v0 node:

    upload_2017-2-1_7-49-51.png
    Starting with the portrayal on the left, try to picture the red wire stretching and the blue wire contracting as the upper capacitor is dragged around and over to left hand side of the diagram, resulting in the portrayal on the right. No connections were changed in the process.
    The ground is important because it allows the entry/exit of charge from an external source (ground). It also pins one node at a known potential, since we tend to declare "ground" to be zero potential.
    No worries! It's just a new set of "rules and consequences" that you are unfamiliar with. It can be confusing at first as you assimilate them and get used to thinking about circuits and how they behave. It will get easier, and at some point the concepts will "klick" and you'll wonder why it seemed so tricky before!
     
  14. Feb 1, 2017 #13
    So would I be right in saying:
    The grounding holds the 2 plates at the same potential, and because these 2 plates are exposed to a conductor of a certain potential (sum of energy associated with all the charge pairs divided by total charge) the PD across the 2 'split' capacitors have to be the same?
     
  15. Feb 1, 2017 #14

    gneill

    User Avatar

    Staff: Mentor

    Sure. You could also say that anything joined by a conducting path (wire) must share the same potential. Hence the labels v0 and v1 on the diagrams.
     
  16. Feb 1, 2017 #15
    Your help cleared things up a lot, many thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Capacitance and induced charges
  1. Capacitance and Charge (Replies: 2)

  2. Capacitance and charge (Replies: 2)

  3. Induced charges (Replies: 9)

Loading...