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A problem about capacitor

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Two plates (area=s) charged to +q1 and +q2 (q2<q1)are brought closer to form a capacitor of capacitance C.The potential difference across the plates is

    2. Relevant equations

    ##V##=##\frac{Q}{C}##

    3. The attempt at a solution

    I don't know how are these plates going to form a capacitor as both of them have same type of (i.e positive)charge and in order to form a capacitor two conducting plates with equal and opposite charges are required.
     
    Last edited: Nov 19, 2015
  2. jcsd
  3. Nov 19, 2015 #2

    ehild

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    What have you learnt about a parallel-plate capacitor? What is the capacitance if the area of the plates is s and the distance between them is d? Does the capacitance depend on the charge of the plates?
     
  4. Nov 19, 2015 #3

    gneill

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    I would be inclined to look at the net electric field between the plates.
     
  5. Nov 19, 2015 #4
    ##\vec{E}##=##\frac{σ}{ε0}##=##\frac{Q}{ε0A}##
     
  6. Nov 19, 2015 #5

    andrevdh

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    What if one would to consider two capacitors one charged with a q1 charge and the other with a q2 charge?
     
  7. Nov 19, 2015 #6

    gneill

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    That's the idea, but check your formula for the electric field... a sheet of charge has two sides so you need to take that into account when you apply Gauss' Law to find the field.

    edit: See for example: Hyperphysics: Electric Feild, Flat Sheets of Charge

    Now, you have two sheets of charge with different charge densities...
     
  8. Nov 19, 2015 #7
    Two capacitors or two plates of a capacitor?
     
  9. Nov 19, 2015 #8

    BvU

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    Let q1 = 1, q2 = 2 for simplicity.

    If the plates are far apart (far enough that they don't influence each other significantly), can you imagine the charges are evenly distributed over the surfaces ?

    What happens if they are brought closer together ?
     
  10. Nov 19, 2015 #9
    Capacitance increases.
    I don't wanna sound skeptical that's why I avoided question mark!It does not mean I am confident about it.
     
  11. Nov 19, 2015 #10
    My thought process is
    For a parallel plate capacitor
    ##C##=##\frac{Aε0}{d}##
    As d (distance between the two plates )decreases,C i.e Capacitance should increase.
     
  12. Nov 19, 2015 #11
    Am I going in right/correct direction?
     
  13. Nov 19, 2015 #12

    gneill

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    I still feel that you'd be better to look at the net field between the plates. Gauss' Law will give you the answer.
     
  14. Nov 19, 2015 #13
    ##E##=##\frac{σ}{2ε0}##
    Right?
     
  15. Nov 19, 2015 #14

    gneill

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    That will give you the field between the plates due to one plate with net charge density σ. You have two plates with different σ's. Draw a sketch and pay attention to the field directions. How do they sum?
     
  16. Nov 20, 2015 #15
    ETF.png

    Electric field due to one plate having charge density σ1 in between the two plates(we will avoid electric field at other places)
    ##E_1##=##\frac{σ1}{2ε0}##

    ##E_2##=##\frac{σ2}{2ε0}##
    Vector addition of ##E_1## and ##E_2##
     
  17. Nov 20, 2015 #16

    gneill

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    Okay, so assuming that you choose an appropriate direction for the net field, what's an expression for its magnitude?
     
  18. Nov 20, 2015 #17
    ##\frac{σ1}{2ε0}##-##\frac{σ2}{2ε0}##
    as Q1>Q2
    Hence σ1>σ2
     
  19. Nov 20, 2015 #18

    gneill

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    Good. But you can expand that a bit using the problem's given variables: s, q1, q2. Usually you want to express any results using the given parameters. You can pretty it up by factoring out the common variables.

    Next, given a constant electric field, how do you find the potential difference between two locations?
     
  20. Nov 20, 2015 #19
    ##\frac{Q1}{2ε0s}##-##\frac{Q2}{2ε0s}##
     
  21. Nov 20, 2015 #20

    gneill

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    Keep going...
     
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