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Capacitors and Calculating for Capacitance involving Plates

  1. Oct 30, 2015 #1
    A sheet of gold foil (negligible thickness) is placed between the plates of a capacitor and has the same area as each of the plates. The foil is parallel to the plates, at a position 1/2 of the way from one to the other. Before the foil is inserted, the capacitance is 7.9μF. What is the capacitance after the foil is in place?

    What I tried:
    distance 1 (d1) = 1/2 + 1/2 (d) = 1 d

    distance 2 (d2) = 1/2 d

    C1 = EA/d C2 = EA/ 0.5d

    1/Cs = 1/C1 + 1/C2
    = 1/ (EA/d) + 1/ (EA/0.5d)
    = 3d/ EA
    Since 1/ Co = d/EA
    1/Cs = 3/ 7.9
    Cs = 2.633

    Is this right please?
     
    Last edited: Oct 30, 2015
  2. jcsd
  3. Oct 30, 2015 #2
    Why have you taken distance 1 to be d?
    Although the gold foil is of negligible thickness for the calculation, what is the state of either surface of the foil? How is charge distributed on the foil? Therefore, what effective capacitor system do you have?
     
  4. Oct 30, 2015 #3
    For distance 1, I added the position 1/2 + another 1/2 to make 1 d.
     
  5. Oct 30, 2015 #4
    Think about what I asked you in my previous post. You'll still have 2 sides to the gold foil, how is charge distributed? Once you see this, the rest should be easy! ;) If you're having trouble, let me know.
     
  6. Oct 30, 2015 #5
    I'm not sure of what you are talking about but would it be that the capacitors are connected in series and since there is a negligible thickness, the capacitance before the foil was inserted will be the same as after the foil is in place? Please I am quite confused. If you can just kindly give a brief explanation I'd appreciate. Thanks muscaria..
     
  7. Oct 30, 2015 #6
    No problem Aromire ;).
    Yes precisely! But what you say after
    is not quite correct, I think at least. I think the negligible thickness of the foil is with respect to the distance d between the 2 original plates. Although that ratio is negligible, you still have a "tiny thickness" where one side will be positively charged and the other side will be negatively charged. This is because the gold conducting foil has free charge carriers (electrons) which will build up on one side of the foil under the influence of the electric field, leaving net positive charge on the other side of the foil. Therefore if you originally had two plates: + -, i.e left plate positively charged, right plate negatively. You now have: + gold - and the free electrons of the gold are now moving towards the + plate so that you have a: + -+ -, in other words 2 capacitors in series. Does this make sense? Do you see what you need to do now?
     
  8. Oct 30, 2015 #7
    I'm more lost now. I'm sorry but do we gotta deal with the charges in respect to this question though? Please I need your help... Thanks once again
     
  9. Oct 30, 2015 #8
    I'm sorry about that, that obviously was not my intention! You have to deal with the charges in the gold foil if you want to understand why in the end you have 2 capacitors in series, each with distance d/2. Maybe you could try and let me know what you don't understand? Do you understand how charges in a conductor behave under the influence of an electric field? Does an image like the following make sense?
    https://www.google.co.uk/search?q=e...g_WhSGZFM:&usg=__lKhc5Vd14TCXUMHuERyTLEqW_SQ=
     
  10. Oct 31, 2015 #9
    Yes the images does make sense! Will my final answer be 7.9μF after the foil is in place? Thanks once again.
     
  11. Oct 31, 2015 #10
    If the image makes sense, then replace the circle in the image by a thin rectangle which will be the gold foil, and you have - charge on one side and + charge on the other side. The two original plates with the gold foil become 2 capacitors in series with a distance of d/2 for each one. The final answer will not be 7.9 as you suggested. But don't worry, you're not far from getting it ;)! I'll help you a bit more:
    Can you find the capacitance of each of these capacitors separately? You know the value of the capacitance without the foil (it's 7.9##\mu F##) and its expression in terms of d, A and ##\epsilon##. Now, what is the capacitance of each of the 2 new capacitors (hint: they have a new distance)? Once you have that, you know the value of each of the capacitors. Then you have to calculate the equivalent capacitance of 2 capacitors in series. Do you know how to calculate this?
     
  12. Oct 31, 2015 #11
    The following image is precisely what your problem is about:
    https://www.google.co.uk/search?q=e...r6YJMixRM:&usg=__lKhc5Vd14TCXUMHuERyTLEqW_SQ=
    Where the conductor is the gold foil in your problem, although the gold foil is much much thinner of course! You see now how you are getting 2 capacitors in series, each with a distance of d/2 between their plates? Hope you're not getting too discouraged, take your time :).
     
  13. Oct 31, 2015 #12
    Can you just tell me how to set it up and I'll be okay, please? I know how to solve for capacitance in series; 1/C = 1/c1 + 1/c2 ...
     
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