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Capacitor, with half a dielectric

Posted on 2007-12-03
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A parallel plate capacitor has a capacitance C when there is vacuum between the plates. How does
this change if the gap between the plates is half filled with a dielectric with dielectric constant eps_r?

(The given diagram looks almost as good as this ;)

                                             |
-----------------------------------------------------------------------
                                             +--------------------------------+
                                             |             Dielectric              |
                                             +--------------------------------+
-----------------------------------------------------------------------
                                            |


Here's my idea:

Treat the capacitor as two seperate capacitors, in parallel, like so:

                             |
             +-----------+----------+
             |                              |
---------------------      -------------------
                                     Dielectric
---------------------      -------------------
             |                             |
            +-----------+----------+
                            |

Then through a bit of math, this gets me to:

C = C0*(1+eps_r)/2

I notice that the (1+eps_r)/2 term is the average eps_r value (as eps_r=1 in vacuum); so it would make sense.

However, [as my last question proved], it's better safe than sorry...
So is this the correct method?

(I have a quick follow up question after this)


Thanks
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Question by:InteractiveMind
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5 Comments
 
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Accepted Solution

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d-glitch earned 2000 total points
ID: 20398269
That is correct.
The conducting plates are equipotential surfaces, so you can break them up that way.
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by:d-glitch
ID: 20398597
C_left  +  C_right  =  eps_0 *( A/2)/d  +  eps_r * (A/2)/d  =  (eps_0 + eps_r)*(A/2)/d
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by:InteractiveMind
ID: 20398706
Oh, I thought it was:

C_left  +  C_right  =  eps_0 *( A/2)/d  +  eps_r * eps_0 * (A/2)/d  =  eps_0*(1 + eps_r)*(A/2)/d

?
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Expert Comment

by:d-glitch
ID: 20398822
You arecorrect if you are dealing with relative permitivities.
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Author Comment

by:InteractiveMind
ID: 20398852
Okay, cool.

(Don't worry about the follow up question; I've figured it)

Thank you, d-glitch.
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