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Capacitors with dielectric question

  1. May 17, 2010 #1
    [PLAIN]http://img340.imageshack.us/img340/4058/15244008.jpg [Broken]

    so
    C = epysilon A / d
    C= (8.85x10^-12 x 5.0x10^-3)/0.002 = 2.2125x10^-11
    so the capacitance of the capacitors is ^
    then with a dielectric where k = 3
    can i use this equation

    C = epysilon A k/ d, just putting the k next to epysilon, i just read thats what you do in my book
    but yeh
    when i do that i get C = 6.6375x10^-11
    then i just added them as parralell capacitors
    but the 6.6375x10^-11 doesnt seem right to me, that formula must be wrong, is it?
    i thought the dielectric made the capacitance lower
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 17, 2010 #2

    ehild

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    No, the dielectric makes the capacitance higher. That is why a dielectric is applied!

    ehild
     
  4. May 17, 2010 #3

    tiny-tim

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    Hi Daniiel! :smile:

    (have an epsilon: ε and try using the X2 tag just above the Reply box :wink:)
    No … the electric displacement field in the capacitor is the same, so the electric field (E = D/ε) is reduced, so the voltage is reduced (V = -∫Edx), so the capacitance (C = Q/V) is increased :smile:

    see eg http://en.wikipedia.org/wiki/Dielectric_constant" [Broken] …

    The dielectric constant is an essential piece of information when designing capacitors, and in other circumstances where a material might be expected to introduce capacitance into a circuit. If a material with a high dielectric constant is placed in an electric field, the magnitude of that field will be measurably reduced within the volume of the dielectric. This fact is commonly used to increase the capacitance of a particular capacitor design. The layers beneath etched conductors in printed circuit boards (PCBs) also act as dielectrics.​
     
    Last edited by a moderator: May 4, 2017
  5. May 17, 2010 #4
    ohh the voltageee is decreased
    thanks guys
    so that looks correct?
     
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