1. Not finding help here? Sign up for a free 30min 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!

Another question - Capacitance

  1. Oct 26, 2008 #1
    Each pixel in a device acts like a paralel plate capaciter with the transistor updating at a voltage of 50 Hz. The change in voltage over 1 frame is 1 % and dominated by the electrical resistance of the material. The material has a constant dielectric of 20. Find the electrical resistivty of the material that can be used.

    would the capacitance equation be used here? And is there more than one?

    Thanks
     
  2. jcsd
  3. Oct 26, 2008 #2

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    I am not sure what you mean by the "capacitance equation" (C=Q/V? ) but no, I think it is bit more complicated than that.

    Note that they are asking for the resistivity which is a quantity that is independent of the geometry (area).

    Note also that they mention that each pixel acts as a "parallel plate capacitor" which means you can use the equation [itex]C=\epsilon_r\epsilon_0 A/d[/itex].

    Finally, although they don't tell you the voltage they DO tell you the update frequency and that the voltage changes by 1% each frame.


    Now, if you were to draw an equivalent circuit; what would it look like?
     
  4. Oct 26, 2008 #3

    Thanks for the response. Would C = 20 as that is the dielectric constant (e.g. area within the capacitor)? And what about A and D? where do the values (50 Hz and 1 % in particular) come into this?
     
  5. Oct 26, 2008 #4

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    The dielectric constant is the same thing as [itex] \epsilon_r[/itex] in the equation above.
    Since they are asking for resisitivity (as opposed to resistance) you can be pretty sure that the answer will be independent of A/d.

    The 50 Hz and 1% tells you that if the voltage at a t=0 is equal to V it will drop to 0.99V at some time (which?) later before going up again.
     
  6. Oct 27, 2008 #5
    thanks for the reply. So if A and d do not come into this why are they in the equation you used? it is asking to find the electrical resistivity.
     
  7. Oct 27, 2008 #6

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    I didn't say A and d wouldn't be used when solving the problem, merely that the answer will not depend on A and d (I.e. somewhere in your calculation there will be two A and two d and they will cancel out).
    There is nothing stopping you from using them when writing down the equations you need to solve the problem.
     
  8. Oct 27, 2008 #7
    thanks for the reply. so what equation would be used that takes into account the voltage rate, the voltage over one frame being 1 % and the dielectric constant to work out the electrical resistivity?
     
  9. Oct 27, 2008 #8

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    Well, that is the question you are trying to answer; so I can't tell you, can I:smile:

    I think the reason why you are stuck is that you are trying to solve the whole problem at once; this is rarely a good approach. Divide the problem into smaller pieces, write down the equations that are valid for those pieces and then trust the math so to speak (yes, I know it sounds corny but it is true).

    In this case:

    1. You know how much the voltage changes over one cycle and you know the frequency (meaning you know the time). This is a circuit with an equivalent circuit consisting on one capacitor and one resistor, what kind of equation can you write down that describes how the voltage changes (in this case decays) over time in such a circuit?

    2. The next "small" problem is: Which equations can you write down that describes the capacitance and resistance of geometries like the one described in the problem (hint: I've already given you the equation for capacitance)?

    Once you have done 1 and 2 you have essentially solved the problem; all that remains is to combine the expressions (3 of them), and then use some (simple) mathematical manipulations to get the final expression.

    Don't think too much about it, just try it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another question - Capacitance
  1. Another light question (Replies: 1)

  2. Another wave question (Replies: 3)

Loading...