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Capacitors with dielectric question

  • Thread starter Daniiel
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  • #1
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[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
 
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Answers and Replies

  • #2
ehild
Homework Helper
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No, the dielectric makes the capacitance higher. That is why a dielectric is applied!

ehild
 
  • #3
tiny-tim
Science Advisor
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Hi Daniiel! :smile:

(have an epsilon: ε and try using the X2 tag just above the Reply box :wink:)
… 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
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.​
 
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  • #4
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ohh the voltageee is decreased
thanks guys
so that looks correct?
 

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