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
InteractiveMind
Asked: 20071203
The conducting plates are equipotential surfaces, so you can break them up that way.