# Capacitor, with half a dielectric

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
LVL 25
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Commented:
That is correct.
The conducting plates are equipotential surfaces, so you can break them up that way.
0

Commented:
C_left  +  C_right  =  eps_0 *( A/2)/d  +  eps_r * (A/2)/d  =  (eps_0 + eps_r)*(A/2)/d
0

Author Commented:
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

?
0

Commented:
You arecorrect if you are dealing with relative permitivities.
0

Author Commented:
Okay, cool.

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

Thank you, d-glitch.
0
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